This page lists diagnostic messages produced by the Dart analyzer, with details about what those messages mean and how you can fix your code. For more information about the analyzer, see Customizing static analysis.

Diagnostics

The analyzer produces the following diagnostics for code that doesn’t conform to the language specification or that might work in unexpected ways.

abi_specific_integer_invalid

Classes extending ‘AbiSpecificInteger’ must have exactly one const constructor, no other members, and no type parameters.

Description

The analyzer produces this diagnostic when a class that extends AbiSpecificInteger doesn’t meet all of the following requirements:

• there must be exactly one constructor
• the constructor must be marked const
• there must not be any members of other than the one constructor
• there must not be any type parameters

Examples

The following code produces this diagnostic because the class C doesn’t define a const constructor:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
}

The following code produces this diagnostic because the constructor isn’t a const constructor:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
  C();
}

The following code produces this diagnostic because the class C defines multiple constructors:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
  const C.zero();
  const C.one();
}

The following code produces this diagnostic because the class C defines a field:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
  final int i;

  const C(this.i);
}

The following code produces this diagnostic because the class C has a type parameter:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C<T> extends AbiSpecificInteger { // type parameters
  const C();
}

Common fixes

Change the class so that it meets the requirements of having no type parameters and a single member that is a const constructor:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
  const C();
}

abi_specific_integer_mapping_extra

Classes extending ‘AbiSpecificInteger’ must have exactly one ‘AbiSpecificIntegerMapping’ annotation specifying the mapping from ABI to a ‘NativeType’ integer with a fixed size.

Description

The analyzer produces this diagnostic when a class that extends AbiSpecificInteger has more than one AbiSpecificIntegerMapping annotation.

Example

The following code produces this diagnostic because there are two AbiSpecificIntegerMapping annotations on the class C:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
@AbiSpecificIntegerMapping({Abi.linuxX64 : Uint16()})
final class C extends AbiSpecificInteger {
  const C();
}

Common fixes

Remove all but one of the annotations, merging the arguments as appropriate:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8(), Abi.linuxX64 : Uint16()})
final class C extends AbiSpecificInteger {
  const C();
}

abi_specific_integer_mapping_missing

Classes extending ‘AbiSpecificInteger’ must have exactly one ‘AbiSpecificIntegerMapping’ annotation specifying the mapping from ABI to a ‘NativeType’ integer with a fixed size.

Description

The analyzer produces this diagnostic when a class that extends AbiSpecificInteger doesn’t have an AbiSpecificIntegerMapping annotation.

Example

The following code produces this diagnostic because there’s no AbiSpecificIntegerMapping annotation on the class C:

import 'dart:ffi';

final class C extends AbiSpecificInteger {
  const C();
}

Common fixes

Add an AbiSpecificIntegerMapping annotation to the class:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
  const C();
}

abi_specific_integer_mapping_unsupported

Invalid mapping to ‘{0}’; only mappings to ‘Int8’, ‘Int16’, ‘Int32’, ‘Int64’, ‘Uint8’, ‘Uint16’, ‘UInt32’, and ‘Uint64’ are supported.

Description

The analyzer produces this diagnostic when a value in the map argument of an AbiSpecificIntegerMapping annotation is anything other than one of the following integer types:

• Int8
• Int16
• Int32
• Int64
• Uint8
• Uint16
• UInt32
• Uint64

Example

The following code produces this diagnostic because the value of the map entry is Array<Uint8>, which isn’t a valid integer type:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Array<Uint8>(4)})
final class C extends AbiSpecificInteger {
  const C();
}

Common fixes

Use one of the valid types as a value in the map:

import 'dart:ffi';

@AbiSpecificIntegerMapping({Abi.macosX64 : Int8()})
final class C extends AbiSpecificInteger {
  const C();
}

abstract_field_initializer

Abstract fields can’t have initializers.

Description

The analyzer produces this diagnostic when a field that has the abstract modifier also has an initializer.

Examples

The following code produces this diagnostic because f is marked as abstract and has an initializer:

abstract class C {
  abstract int f = 0;
}

The following code produces this diagnostic because f is marked as abstract and there’s an initializer in the constructor:

abstract class C {
  abstract int f;

  C() : f = 0;
}

Common fixes

If the field must be abstract, then remove the initializer:

abstract class C {
  abstract int f;
}

If the field isn’t required to be abstract, then remove the keyword:

abstract class C {
  int f = 0;
}

abstract_sealed_class

A ‘sealed’ class can’t be marked ‘abstract’ because it’s already implicitly abstract.

Description

The analyzer produces this diagnostic when a class is declared using both the modifier abstract and the modifier sealed. Sealed classes are implicitly abstract, so explicitly using both modifiers is not allowed.

Example

The following code produces this diagnostic because the class C is declared using both abstract and sealed:

abstract sealed class C {}

Common fixes

If the class should be abstract but not sealed, then remove the sealed modifier:

abstract class C {}

If the class should be both abstract and sealed, then remove the abstract modifier:

sealed class C {}

abstract_super_member_reference

The {0} ‘{1}’ is always abstract in the supertype.

Description

The analyzer produces this diagnostic when an inherited member is referenced using super, but there is no concrete implementation of the member in the superclass chain. Abstract members can’t be invoked.

Example

The following code produces this diagnostic because B doesn’t inherit a concrete implementation of a:

abstract class A {
  int get a;
}
class B extends A {
  int get a => super.a;
}

Common fixes

Remove the invocation of the abstract member, possibly replacing it with an invocation of a concrete member.

ambiguous_export

The name ‘{0}’ is defined in the libraries ‘{1}’ and ‘{2}’.

Description

The analyzer produces this diagnostic when two or more export directives cause the same name to be exported from multiple libraries.

Example

Given a file a.dart containing

class C {}

And a file b.dart containing

class C {}

The following code produces this diagnostic because the name C is being exported from both a.dart and b.dart:

export 'a.dart';
export 'b.dart';

Common fixes

If none of the names in one of the libraries needs to be exported, then remove the unnecessary export directives:

export 'a.dart';

If all of the export directives are needed, then hide the name in all except one of the directives:

export 'a.dart';
export 'b.dart' hide C;

ambiguous_extension_member_access

A member named ‘{0}’ is defined in {1}, and none are more specific.

Description

When code refers to a member of an object (for example, o.m() or o.m or o[i]) where the static type of o doesn’t declare the member (m or [], for example), then the analyzer tries to find the member in an extension. For example, if the member is m, then the analyzer looks for extensions that declare a member named m and have an extended type that the static type of o can be assigned to. When there’s more than one such extension in scope, the extension whose extended type is most specific is selected.

The analyzer produces this diagnostic when none of the extensions has an extended type that’s more specific than the extended types of all of the other extensions, making the reference to the member ambiguous.

Example

The following code produces this diagnostic because there’s no way to choose between the member in E1 and the member in E2:

extension E1 on String {
  int get charCount => 1;
}

extension E2 on String {
  int get charCount => 2;
}

void f(String s) {
  print(s.charCount);
}

Common fixes

If you don’t need both extensions, then you can delete or hide one of them.

If you need both, then explicitly select the one you want to use by using an extension override:

extension E1 on String {
  int get charCount => length;
}

extension E2 on String {
  int get charCount => length;
}

void f(String s) {
  print(E2(s).charCount);
}

ambiguous_import

The name ‘{0}’ is defined in the libraries {1}.

Description

The analyzer produces this diagnostic when a name is referenced that is declared in two or more imported libraries.

Example

Given a library (a.dart) that defines a class (C in this example):

class A {}
class C {}

And a library (b.dart) that defines a different class with the same name:

class B {}
class C {}

The following code produces this diagnostic:

import 'a.dart';
import 'b.dart';

void f(C c1, C c2) {}

Common fixes

If any of the libraries aren’t needed, then remove the import directives for them:

import 'a.dart';

void f(C c1, C c2) {}

If the name is still defined by more than one library, then add a hide clause to the import directives for all except one library:

import 'a.dart' hide C;
import 'b.dart';

void f(C c1, C c2) {}

If you must be able to reference more than one of these types, then add a prefix to each of the import directives, and qualify the references with the appropriate prefix:

import 'a.dart' as a;
import 'b.dart' as b;

void f(a.C c1, b.C c2) {}

ambiguous_set_or_map_literal_both

The literal can’t be either a map or a set because it contains at least one literal map entry or a spread operator spreading a ‘Map’, and at least one element which is neither of these.

Description

Because map and set literals use the same delimiters ({ and }), the analyzer looks at the type arguments and the elements to determine which kind of literal you meant. When there are no type arguments, then the analyzer uses the types of the elements. If all of the elements are literal map entries and all of the spread operators are spreading a Map then it’s a Map. If none of the elements are literal map entries and all of the spread operators are spreading an Iterable, then it’s a Set. If neither of those is true then it’s ambiguous.

The analyzer produces this diagnostic when at least one element is a literal map entry or a spread operator spreading a Map, and at least one element is neither of these, making it impossible for the analyzer to determine whether you are writing a map literal or a set literal.

Example

The following code produces this diagnostic:

union(Map<String, String> a, List<String> b, Map<String, String> c) =>
    {...a, ...b, ...c};

The list b can only be spread into a set, and the maps a and c can only be spread into a map, and the literal can’t be both.

Common fixes

There are two common ways to fix this problem. The first is to remove all of the spread elements of one kind or another, so that the elements are consistent. In this case, that likely means removing the list and deciding what to do about the now unused parameter:

union(Map<String, String> a, List<String> b, Map<String, String> c) =>
    {...a, ...c};

The second fix is to change the elements of one kind into elements that are consistent with the other elements. For example, you can add the elements of the list as keys that map to themselves:

union(Map<String, String> a, List<String> b, Map<String, String> c) =>
    {...a, for (String s in b) s: s, ...c};

ambiguous_set_or_map_literal_either

This literal must be either a map or a set, but the elements don’t have enough information for type inference to work.

Description

Because map and set literals use the same delimiters ({ and }), the analyzer looks at the type arguments and the elements to determine which kind of literal you meant. When there are no type arguments and all of the elements are spread elements (which are allowed in both kinds of literals) then the analyzer uses the types of the expressions that are being spread. If all of the expressions have the type Iterable, then it’s a set literal; if they all have the type Map, then it’s a map literal.

This diagnostic is produced when none of the expressions being spread have a type that allows the analyzer to decide whether you were writing a map literal or a set literal.

Example

The following code produces this diagnostic:

union(a, b) => {...a, ...b};

The problem occurs because there are no type arguments, and there is no information about the type of either a or b.

Common fixes

There are three common ways to fix this problem. The first is to add type arguments to the literal. For example, if the literal is intended to be a map literal, you might write something like this:

union(a, b) => <String, String>{...a, ...b};

The second fix is to add type information so that the expressions have either the type Iterable or the type Map. You can add an explicit cast or, in this case, add types to the declarations of the two parameters:

union(List<int> a, List<int> b) => {...a, ...b};

The third fix is to add context information. In this case, that means adding a return type to the function:

Set<String> union(a, b) => {...a, ...b};

In other cases, you might add a type somewhere else. For example, say the original code looks like this:

union(a, b) {
  var x = {...a, ...b};
  return x;
}

You might add a type annotation on x, like this:

union(a, b) {
  Map<String, String> x = {...a, ...b};
  return x;
}

annotation_on_pointer_field

Fields in a struct class whose type is ‘Pointer’ shouldn’t have any annotations.

Description

The analyzer produces this diagnostic when a field that’s declared in a subclass of Struct and has the type Pointer also has an annotation associated with it.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field p, which has the type Pointer and is declared in a subclass of Struct, has the annotation @Double():

import 'dart:ffi';

final class C extends Struct {
  @Double()
  external Pointer<Int8> p;
}

Common fixes

Remove the annotations from the field:

import 'dart:ffi';

final class C extends Struct {
  external Pointer<Int8> p;
}

argument_must_be_a_constant

Argument ‘{0}’ must be a constant.

Description

The analyzer produces this diagnostic when an invocation of either Pointer.asFunction or DynamicLibrary.lookupFunction has an isLeaf argument whose value isn’t a constant expression.

The analyzer also produces this diagnostic when an invocation of either Pointer.fromFunction or NativeCallable.isolateLocal has an exceptionalReturn argument whose value isn’t a constant expression.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the value of the isLeaf argument is a parameter, and hence isn’t a constant:

import 'dart:ffi';

int Function(int) fromPointer(
    Pointer<NativeFunction<Int8 Function(Int8)>> p, bool isLeaf) {
  return p.asFunction(isLeaf: isLeaf);
}

Common fixes

If there’s a suitable constant that can be used, then replace the argument with a constant:

import 'dart:ffi';

const isLeaf = false;

int Function(int) fromPointer(Pointer<NativeFunction<Int8 Function(Int8)>> p) {
  return p.asFunction(isLeaf: isLeaf);
}

If there isn’t a suitable constant, then replace the argument with a boolean literal:

import 'dart:ffi';

int Function(int) fromPointer(Pointer<NativeFunction<Int8 Function(Int8)>> p) {
  return p.asFunction(isLeaf: true);
}

argument_type_not_assignable

The argument type ‘{0}’ can’t be assigned to the parameter type ‘{1}’. {2}

Description

The analyzer produces this diagnostic when the static type of an argument can’t be assigned to the static type of the corresponding parameter.

Example

The following code produces this diagnostic because a num can’t be assigned to a String:

String f(String x) => x;
String g(num y) => f(y);

Common fixes

If possible, rewrite the code so that the static type is assignable. In the example above you might be able to change the type of the parameter y:

String f(String x) => x;
String g(String y) => f(y);

If that fix isn’t possible, then add code to handle the case where the argument value isn’t the required type. One approach is to coerce other types to the required type:

String f(String x) => x;
String g(num y) => f(y.toString());

Another approach is to add explicit type tests and fallback code:

String f(String x) => x;
String g(num y) => f(y is String ? y : '');

If you believe that the runtime type of the argument will always be the same as the static type of the parameter, and you’re willing to risk having an exception thrown at runtime if you’re wrong, then add an explicit cast:

String f(String x) => x;
String g(num y) => f(y as String);

argument_type_not_assignable_to_error_handler

The argument type ‘{0}’ can’t be assigned to the parameter type ‘{1} Function(Object)’ or ‘{1} Function(Object, StackTrace)’.

Description

The analyzer produces this diagnostic when an invocation of Future.catchError has an argument that is a function whose parameters aren’t compatible with the arguments that will be passed to the function when it’s invoked. The static type of the first argument to catchError is just Function, even though the function that is passed in is expected to have either a single parameter of type Object or two parameters of type Object and StackTrace.

Examples

The following code produces this diagnostic because the closure being passed to catchError doesn’t take any parameters, but the function is required to take at least one parameter:

void f(Future<int> f) {
  f.catchError(() => 0);
}

The following code produces this diagnostic because the closure being passed to catchError takes three parameters, but it can’t have more than two required parameters:

void f(Future<int> f) {
  f.catchError((one, two, three) => 0);
}

The following code produces this diagnostic because even though the closure being passed to catchError takes one parameter, the closure doesn’t have a type that is compatible with Object:

void f(Future<int> f) {
  f.catchError((String error) => 0);
}

Common fixes

Change the function being passed to catchError so that it has either one or two required parameters, and the parameters have the required types:

void f(Future<int> f) {
  f.catchError((Object error) => 0);
}

assert_in_redirecting_constructor

A redirecting constructor can’t have an ‘assert’ initializer.

Description

The analyzer produces this diagnostic when a redirecting constructor (a constructor that redirects to another constructor in the same class) has an assert in the initializer list.

Example

The following code produces this diagnostic because the unnamed constructor is a redirecting constructor and also has an assert in the initializer list:

class C {
  C(int x) : assert(x > 0), this.name();
  C.name() {}
}

Common fixes

If the assert isn’t needed, then remove it:

class C {
  C(int x) : this.name();
  C.name() {}
}

If the assert is needed, then convert the constructor into a factory constructor:

class C {
  factory C(int x) {
    assert(x > 0);
    return C.name();
  }
  C.name() {}
}

asset_directory_does_not_exist

The asset directory ‘{0}’ doesn’t exist.

Description

The analyzer produces this diagnostic when an asset list contains a value referencing a directory that doesn’t exist.

Example

Assuming that the directory assets doesn’t exist, the following code produces this diagnostic because it’s listed as a directory containing assets:

name: example
flutter:
  assets:
    - assets/

Common fixes

If the path is correct, then create a directory at that path.

If the path isn’t correct, then change the path to match the path of the directory containing the assets.

asset_does_not_exist

The asset file ‘{0}’ doesn’t exist.

Description

The analyzer produces this diagnostic when an asset list contains a value referencing a file that doesn’t exist.

Example

Assuming that the file doesNotExist.gif doesn’t exist, the following code produces this diagnostic because it’s listed as an asset:

name: example
flutter:
  assets:
    - doesNotExist.gif

Common fixes

If the path is correct, then create a file at that path.

If the path isn’t correct, then change the path to match the path of the file containing the asset.

asset_field_not_list

The value of the ‘asset’ field is expected to be a list of relative file paths.

Description

The analyzer produces this diagnostic when the value of the asset key isn’t a list.

Example

The following code produces this diagnostic because the value of the assets key is a string when a list is expected:

name: example
flutter:
  assets: assets/

Common fixes

Change the value of the asset list so that it’s a list:

name: example
flutter:
  assets:
    - assets/

asset_not_string

Assets are required to be file paths (strings).

Description

The analyzer produces this diagnostic when an asset list contains a value that isn’t a string.

Example

The following code produces this diagnostic because the asset list contains a map:

name: example
flutter:
  assets:
    - image.gif: true

Common fixes

Change the asset list so that it only contains valid POSIX-style file paths:

name: example
flutter:
  assets:
    - image.gif

assignment_of_do_not_store

‘{0}’ is marked ‘doNotStore’ and shouldn’t be assigned to a field or top-level variable.

Description

The analyzer produces this diagnostic when the value of a function (including methods and getters) that is explicitly or implicitly marked by the doNotStore annotation is stored in either a field or top-level variable.

Example

The following code produces this diagnostic because the value of the function f is being stored in the top-level variable x:

import 'package:meta/meta.dart';

@doNotStore
int f() => 1;

var x = f();

Common fixes

Replace references to the field or variable with invocations of the function producing the value.

assignment_to_const

Constant variables can’t be assigned a value.

Description

The analyzer produces this diagnostic when it finds an assignment to a top-level variable, a static field, or a local variable that has the const modifier. The value of a compile-time constant can’t be changed at runtime.

Example

The following code produces this diagnostic because c is being assigned a value even though it has the const modifier:

const c = 0;

void f() {
  c = 1;
  print(c);
}

Common fixes

If the variable must be assignable, then remove the const modifier:

var c = 0;

void f() {
  c = 1;
  print(c);
}

If the constant shouldn’t be changed, then either remove the assignment or use a local variable in place of references to the constant:

const c = 0;

void f() {
  var v = 1;
  print(v);
}

assignment_to_final

‘{0}’ can’t be used as a setter because it’s final.

Description

The analyzer produces this diagnostic when it finds an invocation of a setter, but there’s no setter because the field with the same name was declared to be final or const.

Example

The following code produces this diagnostic because v is final:

class C {
  final v = 0;
}

f(C c) {
  c.v = 1;
}

Common fixes

If you need to be able to set the value of the field, then remove the modifier final from the field:

class C {
  int v = 0;
}

f(C c) {
  c.v = 1;
}

assignment_to_final_local

The final variable ‘{0}’ can only be set once.

Description

The analyzer produces this diagnostic when a local variable that was declared to be final is assigned after it was initialized.

Example

The following code produces this diagnostic because x is final, so it can’t have a value assigned to it after it was initialized:

void f() {
  final x = 0;
  x = 3;
  print(x);
}

Common fixes

Remove the keyword final, and replace it with var if there’s no type annotation:

void f() {
  var x = 0;
  x = 3;
  print(x);
}

assignment_to_final_no_setter

There isn’t a setter named ‘{0}’ in class ‘{1}’.

Description

The analyzer produces this diagnostic when a reference to a setter is found; there is no setter defined for the type; but there is a getter defined with the same name.

Example

The following code produces this diagnostic because there is no setter named x in C, but there is a getter named x:

class C {
  int get x => 0;
  set y(int p) {}
}

void f(C c) {
  c.x = 1;
}

Common fixes

If you want to invoke an existing setter, then correct the name:

class C {
  int get x => 0;
  set y(int p) {}
}

void f(C c) {
  c.y = 1;
}

If you want to invoke the setter but it just doesn’t exist yet, then declare it:

class C {
  int get x => 0;
  set x(int p) {}
  set y(int p) {}
}

void f(C c) {
  c.x = 1;
}

assignment_to_function

Functions can’t be assigned a value.

Description

The analyzer produces this diagnostic when the name of a function appears on the left-hand side of an assignment expression.

Example

The following code produces this diagnostic because the assignment to the function f is invalid:

void f() {}

void g() {
  f = () {};
}

Common fixes

If the right-hand side should be assigned to something else, such as a local variable, then change the left-hand side:

void f() {}

void g() {
  var x = () {};
  print(x);
}

If the intent is to change the implementation of the function, then define a function-valued variable instead of a function:

void Function() f = () {};

void g() {
  f = () {};
}

assignment_to_method

Methods can’t be assigned a value.

Description

The analyzer produces this diagnostic when the target of an assignment is a method.

Example

The following code produces this diagnostic because f can’t be assigned a value because it’s a method:

class C {
  void f() {}

  void g() {
    f = null;
  }
}

Common fixes

Rewrite the code so that there isn’t an assignment to a method.

assignment_to_type

Types can’t be assigned a value.

Description

The analyzer produces this diagnostic when the name of a type name appears on the left-hand side of an assignment expression.

Example

The following code produces this diagnostic because the assignment to the class C is invalid:

class C {}

void f() {
  C = null;
}

Common fixes

If the right-hand side should be assigned to something else, such as a local variable, then change the left-hand side:

void f() {}

void g() {
  var c = null;
  print(c);
}

async_for_in_wrong_context

The async for-in loop can only be used in an async function.

Description

The analyzer produces this diagnostic when an async for-in loop is found in a function or method whose body isn’t marked as being either async or async*.

Example

The following code produces this diagnostic because the body of f isn’t marked as being either async or async*, but f contains an async for-in loop:

void f(list) {
  await for (var e in list) {
    print(e);
  }
}

Common fixes

If the function should return a Future, then mark the body with async:

Future<void> f(list) async {
  await for (var e in list) {
    print(e);
  }
}

If the function should return a Stream of values, then mark the body with async*:

Stream<void> f(list) async* {
  await for (var e in list) {
    print(e);
  }
}

If the function should be synchronous, then remove the await before the loop:

void f(list) {
  for (var e in list) {
    print(e);
  }
}

await_in_late_local_variable_initializer

The ‘await’ expression can’t be used in a ‘late’ local variable’s initializer.

Description

The analyzer produces this diagnostic when a local variable that has the late modifier uses an await expression in the initializer.

Example

The following code produces this diagnostic because an await expression is used in the initializer for v, a local variable that is marked late:

Future<int> f() async {
  late var v = await 42;
  return v;
}

Common fixes

If the initializer can be rewritten to not use await, then rewrite it:

Future<int> f() async {
  late var v = 42;
  return v;
}

If the initializer can’t be rewritten, then remove the late modifier:

Future<int> f() async {
  var v = await 42;
  return v;
}

await_of_extension_type_not_future

The ‘await’ expression can’t be used for an expression with an extension type that is not a subtype of ‘Future’.

Description

The analyzer produces this diagnostic when the type of the expression in an await expression is an extension type, and the extension type isn’t a subclass of Future.

Example

The following code produces this diagnostic because the extension type E isn’t a subclass of Future:

extension type E(int i) {}

void f(E e) async {
  await e;
}

Common fixes

If the extension type is correctly defined, then remove the await:

extension type E(int i) {}

void f(E e) {
  e;
}

If the extension type is intended to be awaitable, then add Future (or a subtype of Future) to the implements clause (adding an implements clause if there isn’t one already), and make the representation type match:

extension type E(Future<int> i) implements Future<int> {}

void f(E e) async {
  await e;
}

body_might_complete_normally

The body might complete normally, causing ‘null’ to be returned, but the return type, ‘{0}’, is a potentially non-nullable type.

Description

The analyzer produces this diagnostic when a method or function has a return type that’s potentially non-nullable but would implicitly return null if control reached the end of the function.

Examples

The following code produces this diagnostic because the method m has an implicit return of null inserted at the end of the method, but the method is declared to not return null:

class C {
  int m(int t) {
    print(t);
  }
}

The following code produces this diagnostic because the method m has an implicit return of null inserted at the end of the method, but because the class C can be instantiated with a non-nullable type argument, the method is effectively declared to not return null:

class C<T> {
  T m(T t) {
    print(t);
  }
}

Common fixes

If there’s a reasonable value that can be returned, then add a return statement at the end of the method:

class C<T> {
  T m(T t) {
    print(t);
    return t;
  }
}

If the method won’t reach the implicit return, then add a throw at the end of the method:

class C<T> {
  T m(T t) {
    print(t);
    throw '';
  }
}

If the method intentionally returns null at the end, then add an explicit return of null at the end of the method and change the return type so that it’s valid to return null:

class C<T> {
  T? m(T t) {
    print(t);
    return null;
  }
}

body_might_complete_normally_catch_error

This ‘onError’ handler must return a value assignable to ‘{0}’, but ends without returning a value.

Description

The analyzer produces this diagnostic when the closure passed to the onError parameter of the Future.catchError method is required to return a non-null value (because of the Futures type argument) but can implicitly return null.

Example

The following code produces this diagnostic because the closure passed to the catchError method is required to return an int but doesn’t end with an explicit return, causing it to implicitly return null:

void g(Future<int> f) {
  f.catchError((e, st) {});
}

Common fixes

If the closure should sometimes return a non-null value, then add an explicit return to the closure:

void g(Future<int> f) {
  f.catchError((e, st) {
    return -1;
  });
}

If the closure should always return null, then change the type argument of the Future to be either void or Null:

void g(Future<void> f) {
  f.catchError((e, st) {});
}

body_might_complete_normally_nullable

This function has a nullable return type of ‘{0}’, but ends without returning a value.

Description

The analyzer produces this diagnostic when a method or function can implicitly return null by falling off the end. While this is valid Dart code, it’s better for the return of null to be explicit.

Example

The following code produces this diagnostic because the function f implicitly returns null:

String? f() {}

Common fixes

If the return of null is intentional, then make it explicit:

String? f() {
  return null;
}

If the function should return a non-null value along that path, then add the missing return statement:

String? f() {
  return '';
}

break_label_on_switch_member

A break label resolves to the ‘case’ or ‘default’ statement.

Description

The analyzer produces this diagnostic when a break in a case clause inside a switch statement has a label that is associated with another case clause.

Example

The following code produces this diagnostic because the label l is associated with the case clause for 0:

void f(int i) {
  switch (i) {
    l: case 0:
      break;
    case 1:
      break l;
  }
}

Common fixes

If the intent is to transfer control to the statement after the switch, then remove the label from the break statement:

void f(int i) {
  switch (i) {
    case 0:
      break;
    case 1:
      break;
  }
}

If the intent is to transfer control to a different case block, then use continue rather than break:

void f(int i) {
  switch (i) {
    l: case 0:
      break;
    case 1:
      continue l;
  }
}

built_in_identifier_as_type

The built-in identifier ‘{0}’ can’t be used as a type.

Description

The analyzer produces this diagnostic when a built-in identifier is used where a type name is expected.

Example

The following code produces this diagnostic because import can’t be used as a type because it’s a built-in identifier:

import<int> x;

Common fixes

Replace the built-in identifier with the name of a valid type:

List<int> x;

built_in_identifier_in_declaration

The built-in identifier ‘{0}’ can’t be used as a prefix name.

The built-in identifier ‘{0}’ can’t be used as a type name.

The built-in identifier ‘{0}’ can’t be used as a type parameter name.

The built-in identifier ‘{0}’ can’t be used as a typedef name.

The built-in identifier ‘{0}’ can’t be used as an extension name.

The built-in identifier ‘{0}’ can’t be used as an extension type name.

Description

The analyzer produces this diagnostic when the name used in the declaration of a class, extension, mixin, typedef, type parameter, or import prefix is a built-in identifier. Built-in identifiers can’t be used to name any of these kinds of declarations.

Example

The following code produces this diagnostic because mixin is a built-in identifier:

extension mixin on int {}

Common fixes

Choose a different name for the declaration.

case_block_not_terminated

The last statement of the ‘case’ should be ‘break’, ‘continue’, ‘rethrow’, ‘return’, or ‘throw’.

Description

The analyzer produces this diagnostic when the last statement in a case block isn’t one of the required terminators: break, continue, rethrow, return, or throw.

Example

The following code produces this diagnostic because the case block ends with an assignment:

void f(int x) {
  switch (x) {
    case 0:
      x += 2;
    default:
      x += 1;
  }
}

Common fixes

Add one of the required terminators:

void f(int x) {
  switch (x) {
    case 0:
      x += 2;
      break;
    default:
      x += 1;
  }
}

case_expression_type_implements_equals

The switch case expression type ‘{0}’ can’t override the ‘==’ operator.

Description

The analyzer produces this diagnostic when the type of the expression following the keyword case has an implementation of the == operator other than the one in Object.

Example

The following code produces this diagnostic because the expression following the keyword case (C(0)) has the type C, and the class C overrides the == operator:

class C {
  final int value;

  const C(this.value);

  bool operator ==(Object other) {
    return false;
  }
}

void f(C c) {
  switch (c) {
    case C(0):
      break;
  }
}

Common fixes

If there isn’t a strong reason not to do so, then rewrite the code to use an if-else structure:

class C {
  final int value;

  const C(this.value);

  bool operator ==(Object other) {
    return false;
  }
}

void f(C c) {
  if (c == C(0)) {
    // ...
  }
}

If you can’t rewrite the switch statement and the implementation of == isn’t necessary, then remove it:

class C {
  final int value;

  const C(this.value);
}

void f(C c) {
  switch (c) {
    case C(0):
      break;
  }
}

If you can’t rewrite the switch statement and you can’t remove the definition of ==, then find some other value that can be used to control the switch:

class C {
  final int value;

  const C(this.value);

  bool operator ==(Object other) {
    return false;
  }
}

void f(C c) {
  switch (c.value) {
    case 0:
      break;
  }
}

case_expression_type_is_not_switch_expression_subtype

The switch case expression type ‘{0}’ must be a subtype of the switch expression type ‘{1}’.

Description

The analyzer produces this diagnostic when the expression following case in a switch statement has a static type that isn’t a subtype of the static type of the expression following switch.

Example

The following code produces this diagnostic because 1 is an int, which isn’t a subtype of String (the type of s):

void f(String s) {
  switch (s) {
    case 1:
      break;
  }
}

Common fixes

If the value of the case expression is wrong, then change the case expression so that it has the required type:

void f(String s) {
  switch (s) {
    case '1':
      break;
  }
}

If the value of the case expression is correct, then change the switch expression to have the required type:

void f(int s) {
  switch (s) {
    case 1:
      break;
  }
}

cast_from_nullable_always_fails

This cast will always throw an exception because the nullable local variable ‘{0}’ is not assigned.

Description

The analyzer produces this diagnostic when a local variable that has a nullable type hasn’t been assigned and is cast to a non-nullable type. Because the variable hasn’t been assigned it has the default value of null, causing the cast to throw an exception.

Example

The following code produces this diagnostic because the variable x is cast to a non-nullable type (int) when it’s known to have the value null:

void f() {
  num? x;
  x as int;
  print(x);
}

Common fixes

If the variable is expected to have a value before the cast, then add an initializer or an assignment:

void f() {
  num? x = 3;
  x as int;
  print(x);
}

If the variable isn’t expected to be assigned, then remove the cast:

void f() {
  num? x;
  print(x);
}

cast_from_null_always_fails

This cast always throws an exception because the expression always evaluates to ‘null’.

Description

The analyzer produces this diagnostic when an expression whose type is Null is being cast to a non-nullable type.

Example

The following code produces this diagnostic because n is known to always be null, but it’s being cast to a non-nullable type:

void f(Null n) {
  n as int;
}

Common fixes

Remove the unnecessary cast:

void f(Null n) {
  n;
}

cast_to_non_type

The name ‘{0}’ isn’t a type, so it can’t be used in an ‘as’ expression.

Description

The analyzer produces this diagnostic when the name following the as in a cast expression is defined to be something other than a type.

Example

The following code produces this diagnostic because x is a variable, not a type:

num x = 0;
int y = x as x;

Common fixes

Replace the name with the name of a type:

num x = 0;
int y = x as int;

class_used_as_mixin

The class ‘{0}’ can’t be used as a mixin because it’s neither a mixin class nor a mixin.

Description

The analyzer produces this diagnostic when a class that is neither a mixin class nor a mixin is used in a with clause.

Example

The following code produces this diagnostic because the class M is being used as a mixin, but it isn’t defined as a mixin class:

class M {}
class C with M {}

Common fixes

If the class can be a pure mixin, then change class to mixin:

mixin M {}
class C with M {}

If the class needs to be both a class and a mixin, then add mixin:

mixin class M {}
class C with M {}

collection_element_from_deferred_library

Constant values from a deferred library can’t be used as keys in a ‘const’ map literal.

Constant values from a deferred library can’t be used as values in a ‘const’ constructor.

Constant values from a deferred library can’t be used as values in a ‘const’ list literal.

Constant values from a deferred library can’t be used as values in a ‘const’ map literal.

Constant values from a deferred library can’t be used as values in a ‘const’ set literal.

Description

The analyzer produces this diagnostic when a collection literal that is either explicitly (because it’s prefixed by the const keyword) or implicitly (because it appears in a constant context) a constant contains a value that is declared in a library that is imported using a deferred import. Constants are evaluated at compile time, and values from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines the constant zero:

const zero = 0;

The following code produces this diagnostic because the constant list literal contains a.zero, which is imported using a deferred import:

import 'a.dart' deferred as a;

var l = const [a.zero];

Common fixes

If the collection literal isn’t required to be constant, then remove the const keyword:

import 'a.dart' deferred as a;

var l = [a.zero];

If the collection is required to be constant and the imported constant must be referenced, then remove the keyword deferred from the import:

import 'a.dart' as a;

var l = const [a.zero];

If you don’t need to reference the constant, then replace it with a suitable value:

var l = const [0];

compound_implements_finalizable

The class ‘{0}’ can’t implement Finalizable.

Description

The analyzer produces this diagnostic when a subclass of either Struct or Union implements Finalizable.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class S implements Finalizable:

import 'dart:ffi';

final class S extends Struct implements Finalizable {
  external Pointer notEmpty;
}

Common fixes

Try removing the implements clause from the class:

import 'dart:ffi';

final class S extends Struct {
  external Pointer notEmpty;
}

concrete_class_has_enum_superinterface

Concrete classes can’t have ‘Enum’ as a superinterface.

Description

The analyzer produces this diagnostic when a concrete class indirectly has the class Enum as a superinterface.

Example

The following code produces this diagnostic because the concrete class B has Enum as a superinterface as a result of implementing A:

abstract class A implements Enum {}

class B implements A {}

Common fixes

If the implemented class isn’t the class you intend to implement, then change it:

abstract class A implements Enum {}

class B implements C {}

class C {}

If the implemented class can be changed to not implement Enum, then do so:

abstract class A {}

class B implements A {}

If the implemented class can’t be changed to not implement Enum, then remove it from the implements clause:

abstract class A implements Enum {}

class B {}

concrete_class_with_abstract_member

‘{0}’ must have a method body because ‘{1}’ isn’t abstract.

Description

The analyzer produces this diagnostic when a member of a concrete class is found that doesn’t have a concrete implementation. Concrete classes aren’t allowed to contain abstract members.

Example

The following code produces this diagnostic because m is an abstract method but C isn’t an abstract class:

class C {
  void m();
}

Common fixes

If it’s valid to create instances of the class, provide an implementation for the member:

class C {
  void m() {}
}

If it isn’t valid to create instances of the class, mark the class as being abstract:

abstract class C {
  void m();
}

conflicting_constructor_and_static_member

‘{0}’ can’t be used to name both a constructor and a static field in this class.

‘{0}’ can’t be used to name both a constructor and a static getter in this class.

‘{0}’ can’t be used to name both a constructor and a static method in this class.

‘{0}’ can’t be used to name both a constructor and a static setter in this class.

Description

The analyzer produces this diagnostic when a named constructor and either a static method or static field have the same name. Both are accessed using the name of the class, so having the same name makes the reference ambiguous.

Examples

The following code produces this diagnostic because the static field foo and the named constructor foo have the same name:

class C {
  C.foo();
  static int foo = 0;
}

The following code produces this diagnostic because the static method foo and the named constructor foo have the same name:

class C {
  C.foo();
  static void foo() {}
}

Common fixes

Rename either the member or the constructor.

conflicting_generic_interfaces

The {0} ‘{1}’ can’t implement both ‘{2}’ and ‘{3}’ because the type arguments are different.

Description

The analyzer produces this diagnostic when a class attempts to implement a generic interface multiple times, and the values of the type arguments aren’t the same.

Example

The following code produces this diagnostic because C is defined to implement both I<int> (because it extends A) and I<String> (because it implementsB), but int and String aren’t the same type:

class I<T> {}
class A implements I<int> {}
class B implements I<String> {}
class C extends A implements B {}

Common fixes

Rework the type hierarchy to avoid this situation. For example, you might make one or both of the inherited types generic so that C can specify the same type for both type arguments:

class I<T> {}
class A<S> implements I<S> {}
class B implements I<String> {}
class C extends A<String> implements B {}

conflicting_type_variable_and_container

‘{0}’ can’t be used to name both a type variable and the class in which the type variable is defined.

‘{0}’ can’t be used to name both a type variable and the enum in which the type variable is defined.

‘{0}’ can’t be used to name both a type variable and the extension in which the type variable is defined.

‘{0}’ can’t be used to name both a type variable and the mixin in which the type variable is defined.

Description

The analyzer produces this diagnostic when a class, mixin, or extension declaration declares a type parameter with the same name as the class, mixin, or extension that declares it.

Example

The following code produces this diagnostic because the type parameter C has the same name as the class C of which it’s a part:

class C<C> {}

Common fixes

Rename either the type parameter, or the class, mixin, or extension:

class C<T> {}

conflicting_type_variable_and_member

‘{0}’ can’t be used to name both a type variable and a member in this class.

‘{0}’ can’t be used to name both a type variable and a member in this enum.

‘{0}’ can’t be used to name both a type variable and a member in this extension.

‘{0}’ can’t be used to name both a type variable and a member in this mixin.

Description

The analyzer produces this diagnostic when a class, mixin, or extension declaration declares a type parameter with the same name as one of the members of the class, mixin, or extension that declares it.

Example

The following code produces this diagnostic because the type parameter T has the same name as the field T:

class C<T> {
  int T = 0;
}

Common fixes

Rename either the type parameter or the member with which it conflicts:

class C<T> {
  int total = 0;
}

constant_pattern_never_matches_value_type

The matched value type ‘{0}’ can never be equal to this constant of type ‘{1}’.

Description

The analyzer produces this diagnostic when a constant pattern can never match the value it’s being tested against because the type of the constant is known to never match the type of the value.

Example

The following code produces this diagnostic because the type of the constant pattern (true) is bool, and the type of the value being matched (x) is int, and a Boolean can never match an integer:

void f(int x) {
  if (x case true) {}
}

Common fixes

If the type of the value is correct, then rewrite the pattern to be compatible:

void f(int x) {
  if (x case 3) {}
}

If the type of the constant is correct, then rewrite the value to be compatible:

void f(bool x) {
  if (x case true) {}
}

constant_pattern_with_non_constant_expression

The expression of a constant pattern must be a valid constant.

Description

The analyzer produces this diagnostic when a constant pattern has an expression that isn’t a valid constant.

Example

The following code produces this diagnostic because the constant pattern i isn’t a constant:

void f(int e, int i) {
  switch (e) {
    case i:
      break;
  }
}

Common fixes

If the value that should be matched is known, then replace the expression with a constant:

void f(int e, int i) {
  switch (e) {
    case 0:
      break;
  }
}

If the value that should be matched isn’t known, then rewrite the code to not use a pattern:

void f(int e, int i) {
  if (e == i) {}
}

const_constructor_param_type_mismatch

A value of type ‘{0}’ can’t be assigned to a parameter of type ‘{1}’ in a const constructor.

Description

The analyzer produces this diagnostic when the runtime type of a constant value can’t be assigned to the static type of a constant constructor’s parameter.

Example

The following code produces this diagnostic because the runtime type of i is int, which can’t be assigned to the static type of s:

class C {
  final String s;

  const C(this.s);
}

const dynamic i = 0;

void f() {
  const C(i);
}

Common fixes

Pass a value of the correct type to the constructor:

class C {
  final String s;

  const C(this.s);
}

const dynamic i = 0;

void f() {
  const C('$i');
}

const_constructor_with_field_initialized_by_non_const

Can’t define the ‘const’ constructor because the field ‘{0}’ is initialized with a non-constant value.

Description

The analyzer produces this diagnostic when a constructor has the keyword const, but a field in the class is initialized to a non-constant value.

Example

The following code produces this diagnostic because the field s is initialized to a non-constant value:

String x = '3';
class C {
  final String s = x;
  const C();
}

Common fixes

If the field can be initialized to a constant value, then change the initializer to a constant expression:

class C {
  final String s = '3';
  const C();
}

If the field can’t be initialized to a constant value, then remove the keyword const from the constructor:

String x = '3';
class C {
  final String s = x;
  C();
}

const_constructor_with_non_const_super

A constant constructor can’t call a non-constant super constructor of ‘{0}’.

Description

The analyzer produces this diagnostic when a constructor that is marked as const invokes a constructor from its superclass that isn’t marked as const.

Example

The following code produces this diagnostic because the const constructor in B invokes the constructor nonConst from the class A, and the superclass constructor isn’t a const constructor:

class A {
  const A();
  A.nonConst();
}

class B extends A {
  const B() : super.nonConst();
}

Common fixes

If it isn’t essential to invoke the superclass constructor that is currently being invoked, then invoke a constant constructor from the superclass:

class A {
  const A();
  A.nonConst();
}

class B extends A {
  const B() : super();
}

If it’s essential that the current constructor be invoked and if you can modify it, then add const to the constructor in the superclass:

class A {
  const A();
  const A.nonConst();
}

class B extends A {
  const B() : super.nonConst();
}

If it’s essential that the current constructor be invoked and you can’t modify it, then remove const from the constructor in the subclass:

class A {
  const A();
  A.nonConst();
}

class B extends A {
  B() : super.nonConst();
}

const_constructor_with_non_final_field

Can’t define a const constructor for a class with non-final fields.

Description

The analyzer produces this diagnostic when a constructor is marked as a const constructor, but the constructor is defined in a class that has at least one non-final instance field (either directly or by inheritance).

Example

The following code produces this diagnostic because the field x isn’t final:

class C {
  int x;

  const C(this.x);
}

Common fixes

If it’s possible to mark all of the fields as final, then do so:

class C {
  final int x;

  const C(this.x);
}

If it isn’t possible to mark all of the fields as final, then remove the keyword const from the constructor:

class C {
  int x;

  C(this.x);
}

const_deferred_class

Deferred classes can’t be created with ‘const’.

Description

The analyzer produces this diagnostic when a class from a library that is imported using a deferred import is used to create a const object. Constants are evaluated at compile time, and classes from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

The following code produces this diagnostic because it attempts to create a const instance of a class from a deferred library:

import 'dart:convert' deferred as convert;

const json2 = convert.JsonCodec();

Common fixes

If the object isn’t required to be a constant, then change the code so that a non-constant instance is created:

import 'dart:convert' deferred as convert;

final json2 = convert.JsonCodec();

If the object must be a constant, then remove deferred from the import directive:

import 'dart:convert' as convert;

const json2 = convert.JsonCodec();

const_initialized_with_non_constant_value

Const variables must be initialized with a constant value.

Description

The analyzer produces this diagnostic when a value that isn’t statically known to be a constant is assigned to a variable that’s declared to be a const variable.

Example

The following code produces this diagnostic because x isn’t declared to be const:

var x = 0;
const y = x;

Common fixes

If the value being assigned can be declared to be const, then change the declaration:

const x = 0;
const y = x;

If the value can’t be declared to be const, then remove the const modifier from the variable, possibly using final in its place:

var x = 0;
final y = x;

const_initialized_with_non_constant_value_from_deferred_library

Constant values from a deferred library can’t be used to initialize a ‘const’ variable.

Description

The analyzer produces this diagnostic when a const variable is initialized using a const variable from a library that is imported using a deferred import. Constants are evaluated at compile time, and values from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

The following code produces this diagnostic because the variable pi is being initialized using the constant math.pi from the library dart:math, and dart:math is imported as a deferred library:

import 'dart:math' deferred as math;

const pi = math.pi;

Common fixes

If you need to reference the value of the constant from the imported library, then remove the keyword deferred:

import 'dart:math' as math;

const pi = math.pi;

If you don’t need to reference the imported constant, then remove the reference:

const pi = 3.14;

const_instance_field

Only static fields can be declared as const.

Description

The analyzer produces this diagnostic when an instance field is marked as being const.

Example

The following code produces this diagnostic because f is an instance field:

class C {
  const int f = 3;
}

Common fixes

If the field needs to be an instance field, then remove the keyword const, or replace it with final:

class C {
  final int f = 3;
}

If the field really should be a const field, then make it a static field:

class C {
  static const int f = 3;
}

const_map_key_not_primitive_equality

The type of a key in a constant map can’t override the ‘==’ operator, or ‘hashCode’, but the class ‘{0}’ does.

Description

The analyzer produces this diagnostic when the class of object used as a key in a constant map literal implements either the == operator, the getter hashCode, or both. The implementation of constant maps uses both the == operator and the hashCode getter, so any implementation other than the ones inherited from Object requires executing arbitrary code at compile time, which isn’t supported.

Examples

The following code produces this diagnostic because the constant map contains a key whose type is C, and the class C overrides the implementation of ==:

class C {
  const C();

  bool operator ==(Object other) => true;
}

const map = {C() : 0};

The following code produces this diagnostic because the constant map contains a key whose type is C, and the class C overrides the implementation of hashCode:

class C {
  const C();

  int get hashCode => 3;
}

const map = {C() : 0};

Common fixes

If you can remove the implementation of == and hashCode from the class, then do so:

class C {
  const C();
}

const map = {C() : 0};

If you can’t remove the implementation of == and hashCode from the class, then make the map non-constant:

class C {
  const C();

  bool operator ==(Object other) => true;
}

final map = {C() : 0};

const_not_initialized

The constant ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a variable that is declared to be a constant doesn’t have an initializer.

Example

The following code produces this diagnostic because c isn’t initialized:

const c;

Common fixes

Add an initializer:

const c = 'c';

const_set_element_not_primitive_equality

(Previously known as const_set_element_type_implements_equals)

An element in a constant set can’t override the ‘==’ operator, or ‘hashCode’, but the type ‘{0}’ does.

Description

The analyzer produces this diagnostic when the class of object used as an element in a constant set literal implements either the == operator, the getter hashCode, or both. The implementation of constant sets uses both the == operator and the hashCode getter, so any implementation other than the ones inherited from Object requires executing arbitrary code at compile time, which isn’t supported.

Example

The following code produces this diagnostic because the constant set contains an element whose type is C, and the class C overrides the implementation of ==:

class C {
  const C();

  bool operator ==(Object other) => true;
}

const set = {C()};

The following code produces this diagnostic because the constant set contains an element whose type is C, and the class C overrides the implementation of hashCode:

class C {
  const C();

  int get hashCode => 3;
}

const map = {C()};

Common fixes

If you can remove the implementation of == and hashCode from the class, then do so:

class C {
  const C();
}

const set = {C()};

If you can’t remove the implementation of == and hashCode from the class, then make the set non-constant:

class C {
  const C();

  bool operator ==(Object other) => true;
}

final set = {C()};

const_spread_expected_list_or_set

A list or a set is expected in this spread.

Description

The analyzer produces this diagnostic when the expression of a spread operator in a constant list or set evaluates to something other than a list or a set.

Example

The following code produces this diagnostic because the value of list1 is null, which is neither a list nor a set:

const List<int> list1 = null;
const List<int> list2 = [...list1];

Common fixes

Change the expression to something that evaluates to either a constant list or a constant set:

const List<int> list1 = [];
const List<int> list2 = [...list1];

const_spread_expected_map

A map is expected in this spread.

Description

The analyzer produces this diagnostic when the expression of a spread operator in a constant map evaluates to something other than a map.

Example

The following code produces this diagnostic because the value of map1 is null, which isn’t a map:

const Map<String, int> map1 = null;
const Map<String, int> map2 = {...map1};

Common fixes

Change the expression to something that evaluates to a constant map:

const Map<String, int> map1 = {};
const Map<String, int> map2 = {...map1};

const_with_non_const

The constructor being called isn’t a const constructor.

Description

The analyzer produces this diagnostic when the keyword const is used to invoke a constructor that isn’t marked with const.

Example

The following code produces this diagnostic because the constructor in A isn’t a const constructor:

class A {
  A();
}

A f() => const A();

Common fixes

If it’s desirable and possible to make the class a constant class (by making all of the fields of the class, including inherited fields, final), then add the keyword const to the constructor:

class A {
  const A();
}

A f() => const A();

Otherwise, remove the keyword const:

class A {
  A();
}

A f() => A();

const_with_non_constant_argument

Arguments of a constant creation must be constant expressions.

Description

The analyzer produces this diagnostic when a const constructor is invoked with an argument that isn’t a constant expression.

Example

The following code produces this diagnostic because i isn’t a constant:

class C {
  final int i;
  const C(this.i);
}
C f(int i) => const C(i);

Common fixes

Either make all of the arguments constant expressions, or remove the const keyword to use the non-constant form of the constructor:

class C {
  final int i;
  const C(this.i);
}
C f(int i) => C(i);

const_with_type_parameters

A constant constructor tearoff can’t use a type parameter as a type argument.

A constant creation can’t use a type parameter as a type argument.

A constant function tearoff can’t use a type parameter as a type argument.

Description

The analyzer produces this diagnostic when a type parameter is used as a type argument in a const invocation of a constructor. This isn’t allowed because the value of the type parameter (the actual type that will be used at runtime) can’t be known at compile time.

Example

The following code produces this diagnostic because the type parameter T is being used as a type argument when creating a constant:

class C<T> {
  const C();
}

C<T> newC<T>() => const C<T>();

Common fixes

If the type that will be used for the type parameter can be known at compile time, then remove the use of the type parameter:

class C<T> {
  const C();
}

C<int> newC() => const C<int>();

If the type that will be used for the type parameter can’t be known until runtime, then remove the keyword const:

class C<T> {
  const C();
}

C<T> newC<T>() => C<T>();

continue_label_invalid

(Previously known as continue_label_on_switch)

The label used in a ‘continue’ statement must be defined on either a loop or a switch member.

Description

The analyzer produces this diagnostic when the label in a continue statement resolves to a label on a switch statement.

Example

The following code produces this diagnostic because the label l, used to label a switch statement, is used in the continue statement:

void f(int i) {
  l: switch (i) {
    case 0:
      continue l;
  }
}

Common fixes

Find a different way to achieve the control flow you need; for example, by introducing a loop that re-executes the switch statement.

creation_of_struct_or_union

Subclasses of ‘Struct’ and ‘Union’ are backed by native memory, and can’t be instantiated by a generative constructor.

Description

The analyzer produces this diagnostic when a subclass of either Struct or Union is instantiated using a generative constructor.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class C is being instantiated using a generative constructor:

import 'dart:ffi';

final class C extends Struct {
  @Int32()
  external int a;
}

void f() {
  C();
}

Common fixes

If you need to allocate the structure described by the class, then use the ffi package to do so:

import 'dart:ffi';
import 'package:ffi/ffi.dart';

final class C extends Struct {
  @Int32()
  external int a;
}

void f() {
  final pointer = calloc.allocate<C>(4);
  final c = pointer.ref;
  print(c);
  calloc.free(pointer);
}

creation_with_non_type

The name ‘{0}’ isn’t a class.

Description

The analyzer produces this diagnostic when an instance creation using either new or const specifies a name that isn’t defined as a class.

Example

The following code produces this diagnostic because f is a function rather than a class:

int f() => 0;

void g() {
  new f();
}

Common fixes

If a class should be created, then replace the invalid name with the name of a valid class:

int f() => 0;

void g() {
  new Object();
}

If the name is the name of a function and you want that function to be invoked, then remove the new or const keyword:

int f() => 0;

void g() {
  f();
}

dead_code

Dead code.

Description

The analyzer produces this diagnostic when code is found that won’t be executed because execution will never reach the code.

Example

The following code produces this diagnostic because the invocation of print occurs after the function has returned:

void f() {
  return;
  print('here');
}

Common fixes

If the code isn’t needed, then remove it:

void f() {
  return;
}

If the code needs to be executed, then either move the code to a place where it will be executed:

void f() {
  print('here');
  return;
}

Or, rewrite the code before it, so that it can be reached:

void f({bool skipPrinting = true}) {
  if (skipPrinting) {
    return;
  }
  print('here');
}

dead_code_catch_following_catch

Dead code: Catch clauses after a ‘catch (e)’ or an ‘on Object catch (e)’ are never reached.

Description

The analyzer produces this diagnostic when a catch clause is found that can’t be executed because it’s after a catch clause of the form catch (e) or on Object catch (e). The first catch clause that matches the thrown object is selected, and both of those forms will match any object, so no catch clauses that follow them will be selected.

Example

The following code produces this diagnostic:

void f() {
  try {
  } catch (e) {
  } on String {
  }
}

Common fixes

If the clause should be selectable, then move the clause before the general clause:

void f() {
  try {
  } on String {
  } catch (e) {
  }
}

If the clause doesn’t need to be selectable, then remove it:

void f() {
  try {
  } catch (e) {
  }
}

dead_code_on_catch_subtype

Dead code: This on-catch block won’t be executed because ‘{0}’ is a subtype of ‘{1}’ and hence will have been caught already.

Description

The analyzer produces this diagnostic when a catch clause is found that can’t be executed because it is after a catch clause that catches either the same type or a supertype of the clause’s type. The first catch clause that matches the thrown object is selected, and the earlier clause always matches anything matchable by the highlighted clause, so the highlighted clause will never be selected.

Example

The following code produces this diagnostic:

void f() {
  try {
  } on num {
  } on int {
  }
}

Common fixes

If the clause should be selectable, then move the clause before the general clause:

void f() {
  try {
  } on int {
  } on num {
  }
}

If the clause doesn’t need to be selectable, then remove it:

void f() {
  try {
  } on num {
  }
}

dead_null_aware_expression

The left operand can’t be null, so the right operand is never executed.

Description

The analyzer produces this diagnostic in two cases.

The first is when the left operand of an ?? operator can’t be null. The right operand is only evaluated if the left operand has the value null, and because the left operand can’t be null, the right operand is never evaluated.

The second is when the left-hand side of an assignment using the ??= operator can’t be null. The right-hand side is only evaluated if the left-hand side has the value null, and because the left-hand side can’t be null, the right-hand side is never evaluated.

Examples

The following code produces this diagnostic because x can’t be null:

int f(int x) {
  return x ?? 0;
}

The following code produces this diagnostic because f can’t be null:

class C {
  int f = -1;

  void m(int x) {
    f ??= x;
  }
}

Common fixes

If the diagnostic is reported for an ?? operator, then remove the ?? operator and the right operand:

int f(int x) {
  return x;
}

If the diagnostic is reported for an assignment, and the assignment isn’t needed, then remove the assignment:

class C {
  int f = -1;

  void m(int x) {
  }
}

If the assignment is needed, but should be based on a different condition, then rewrite the code to use = and the different condition:

class C {
  int f = -1;

  void m(int x) {
    if (f < 0) {
      f = x;
    }
  }
}

default_list_constructor

The default ‘List’ constructor isn’t available when null safety is enabled.

Description

The analyzer produces this diagnostic when it finds a use of the default constructor for the class List in code that has opted in to null safety.

Example

Assuming the following code is opted in to null safety, it produces this diagnostic because it uses the default List constructor:

var l = List<int>();

Common fixes

If no initial size is provided, then convert the code to use a list literal:

var l = <int>[];

If an initial size needs to be provided and there is a single reasonable initial value for the elements, then use List.filled:

var l = List.filled(3, 0);

If an initial size needs to be provided but each element needs to be computed, then use List.generate:

var l = List.generate(3, (i) => i);

default_value_in_function_type

Parameters in a function type can’t have default values.

Description

The analyzer produces this diagnostic when a function type associated with a parameter includes optional parameters that have a default value. This isn’t allowed because the default values of parameters aren’t part of the function’s type, and therefore including them doesn’t provide any value.

Example

The following code produces this diagnostic because the parameter p has a default value even though it’s part of the type of the parameter g:

void f(void Function([int p = 0]) g) {
}

Common fixes

Remove the default value from the function-type’s parameter:

void f(void Function([int p]) g) {
}

default_value_in_redirecting_factory_constructor

Default values aren’t allowed in factory constructors that redirect to another constructor.

Description

The analyzer produces this diagnostic when a factory constructor that redirects to another constructor specifies a default value for an optional parameter.

Example

The following code produces this diagnostic because the factory constructor in A has a default value for the optional parameter x:

class A {
  factory A([int x = 0]) = B;
}

class B implements A {
  B([int x = 1]) {}
}

Common fixes

Remove the default value from the factory constructor:

class A {
  factory A([int x]) = B;
}

class B implements A {
  B([int x = 1]) {}
}

Note that this fix might change the value used when the optional parameter is omitted. If that happens, and if that change is a problem, then consider making the optional parameter a required parameter in the factory method:

class A {
 factory A(int x) = B;
}

class B implements A {
  B([int x = 1]) {}
}

default_value_on_required_parameter

Required named parameters can’t have a default value.

Description

The analyzer produces this diagnostic when a named parameter has both the required modifier and a default value. If the parameter is required, then a value for the parameter is always provided at the call sites, so the default value can never be used.

Example

The following code generates this diagnostic:

void log({required String message = 'no message'}) {}

Common fixes

If the parameter is really required, then remove the default value:

void log({required String message}) {}

If the parameter isn’t always required, then remove the required modifier:

void log({String message = 'no message'}) {}

deferred_import_of_extension

Imports of deferred libraries must hide all extensions.

Description

The analyzer produces this diagnostic when a library that is imported using a deferred import declares an extension that is visible in the importing library. Extension methods are resolved at compile time, and extensions from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines a named extension:

class C {}

extension E on String {
  int get size => length;
}

The following code produces this diagnostic because the named extension is visible to the library:

import 'a.dart' deferred as a;

void f() {
  a.C();
}

Common fixes

If the library must be imported as deferred, then either add a show clause listing the names being referenced or add a hide clause listing all of the named extensions. Adding a show clause would look like this:

import 'a.dart' deferred as a show C;

void f() {
  a.C();
}

Adding a hide clause would look like this:

import 'a.dart' deferred as a hide E;

void f() {
  a.C();
}

With the first fix, the benefit is that if new extensions are added to the imported library, then the extensions won’t cause a diagnostic to be generated.

If the library doesn’t need to be imported as deferred, or if you need to make use of the extension method declared in it, then remove the keyword deferred:

import 'a.dart' as a;

void f() {
  a.C();
}

definitely_unassigned_late_local_variable

The late local variable ‘{0}’ is definitely unassigned at this point.

Description

The analyzer produces this diagnostic when definite assignment analysis shows that a local variable that’s marked as late is read before being assigned.

Example

The following code produces this diagnostic because x wasn’t assigned a value before being read:

void f(bool b) {
  late int x;
  print(x);
}

Common fixes

Assign a value to the variable before reading from it:

void f(bool b) {
  late int x;
  x = b ? 1 : 0;
  print(x);
}

dependencies_field_not_map

The value of the ‘{0}’ field is expected to be a map.

Description

The analyzer produces this diagnostic when the value of either the dependencies or dev_dependencies key isn’t a map.

Example

The following code produces this diagnostic because the value of the top-level dependencies key is a list:

name: example
dependencies:
  - meta

Common fixes

Use a map as the value of the dependencies key:

name: example
dependencies:
  meta: ^1.0.2

deprecated_colon_for_default_value

Using a colon as the separator before a default value is deprecated and will not be supported in language version 3.0 and later.

Description

The analyzer produces this diagnostic when a colon (:) is used as the separator before the default value of an optional named parameter. While this syntax is allowed, it is deprecated in favor of using an equal sign (=).

Example

The following code produces this diagnostic because a colon is being used before the default value of the optional parameter i:

void f({int i : 0}) {}

Common fixes

Replace the colon with an equal sign.

void f({int i = 0}) {}

deprecated_export_use

The ability to import ‘{0}’ indirectly is deprecated.

Description

The analyzer produces this diagnostic when one library imports a name from a second library, and the second library exports the name from a third library but has indicated that it won’t export the third library in the future.

Example

Given a library a.dart defining the class A:

class A {}

And a second library b.dart that exports a.dart but has marked the export as being deprecated:

import 'a.dart';

@deprecated
export 'a.dart';

The following code produces this diagnostic because the class A won’t be exported from b.dart in some future version:

import 'b.dart';

A? a;

Common fixes

If the name is available from a different library that you can import, then replace the existing import with an import for that library (or add an import for the defining library if you still need the old import):

import 'a.dart';

A? a;

If the name isn’t available, then look for instructions from the library author or contact them directly to find out how to update your code.

deprecated_field

The ‘{0}’ field is no longer used and can be removed.

Description

The analyzer produces this diagnostic when a key is used in a pubspec.yaml file that was deprecated. Unused keys take up space and might imply semantics that are no longer valid.

Example

The following code produces this diagnostic because the author key is no longer being used:

name: example
author: 'Dash'

Common fixes

Remove the deprecated key:

name: example

deprecated_member_use

‘{0}’ is deprecated and shouldn’t be used.

‘{0}’ is deprecated and shouldn’t be used. {1}

Description

The analyzer produces this diagnostic when a deprecated library or class member is used in a different package.

Example

If the method m in the class C is annotated with @deprecated, then the following code produces this diagnostic:

void f(C c) {
  c.m();
}

Common fixes

The documentation for declarations that are annotated with @deprecated should indicate what code to use in place of the deprecated code.

deprecated_member_use_from_same_package

‘{0}’ is deprecated and shouldn’t be used.

‘{0}’ is deprecated and shouldn’t be used. {1}

Description

The analyzer produces this diagnostic when a deprecated library member or class member is used in the same package in which it’s declared.

Example

The following code produces this diagnostic because x is deprecated:

@deprecated
var x = 0;
var y = x;

Common fixes

The fix depends on what’s been deprecated and what the replacement is. The documentation for deprecated declarations should indicate what code to use in place of the deprecated code.

deprecated_new_in_comment_reference

Using the ‘new’ keyword in a comment reference is deprecated.

Description

The analyzer produces this diagnostic when a comment reference (the name of a declaration enclosed in square brackets in a documentation comment) uses the keyword new to refer to a constructor. This form is deprecated.

Examples

The following code produces this diagnostic because the unnamed constructor is being referenced using new C:

/// See [new C].
class C {
  C();
}

The following code produces this diagnostic because the constructor named c is being referenced using new C.c:

/// See [new C.c].
class C {
  C.c();
}

Common fixes

If you’re referencing a named constructor, then remove the keyword new:

/// See [C.c].
class C {
  C.c();
}

If you’re referencing the unnamed constructor, then remove the keyword new and append .new after the class name:

/// See [C.new].
class C {
  C.c();
}

deprecated_subtype_of_function

Extending ‘Function’ is deprecated.

Implementing ‘Function’ has no effect.

Mixing in ‘Function’ is deprecated.

Description

The analyzer produces this diagnostic when the class Function is used in either the extends, implements, or with clause of a class or mixin. Using the class Function in this way has no semantic value, so it’s effectively dead code.

Example

The following code produces this diagnostic because Function is used as the superclass of F:

class F extends Function {}

Common fixes

Remove the class Function from whichever clause it’s in, and remove the whole clause if Function is the only type in the clause:

class F {}

disallowed_type_instantiation_expression

Only a generic type, generic function, generic instance method, or generic constructor can have type arguments.

Description

The analyzer produces this diagnostic when an expression with a value that is anything other than one of the allowed kinds of values is followed by type arguments. The allowed kinds of values are:

• generic types,
• generic constructors, and
• generic functions, including top-level functions, static and instance members, and local functions.

Example

The following code produces this diagnostic because i is a top-level variable, which isn’t one of the allowed cases:

int i = 1;

void f() {
  print(i<int>);
}

Common fixes

If the referenced value is correct, then remove the type arguments:

int i = 1;

void f() {
  print(i);
}

division_optimization

The operator x ~/ y is more efficient than (x / y).toInt().

Description

The analyzer produces this diagnostic when the result of dividing two numbers is converted to an integer using toInt. Dart has a built-in integer division operator that is both more efficient and more concise.

Example

The following code produces this diagnostic because the result of dividing x and y is converted to an integer using toInt:

int divide(num x, num y) => (x / y).toInt();

Common fixes

Use the integer division operator (~/):

int divide(num x, num y) => x ~/ y;

duplicate_constructor

The constructor with name ‘{0}’ is already defined.

The unnamed constructor is already defined.

Description

The analyzer produces this diagnostic when a class declares more than one unnamed constructor or when it declares more than one constructor with the same name.

Examples

The following code produces this diagnostic because there are two declarations for the unnamed constructor:

class C {
  C();

  C();
}

The following code produces this diagnostic because there are two declarations for the constructor named m:

class C {
  C.m();

  C.m();
}

Common fixes

If there are multiple unnamed constructors and all of the constructors are needed, then give all of them, or all except one of them, a name:

class C {
  C();

  C.n();
}

If there are multiple unnamed constructors and all except one of them are unneeded, then remove the constructors that aren’t needed:

class C {
  C();
}

If there are multiple named constructors and all of the constructors are needed, then rename all except one of them:

class C {
  C.m();

  C.n();
}

If there are multiple named constructors and all except one of them are unneeded, then remove the constructors that aren’t needed:

class C {
  C.m();
}

duplicate_definition

The name ‘{0}’ is already defined.

Description

The analyzer produces this diagnostic when a name is declared, and there is a previous declaration with the same name in the same scope.

Example

The following code produces this diagnostic because the name x is declared twice:

int x = 0;
int x = 1;

Common fixes

Choose a different name for one of the declarations.

int x = 0;
int y = 1;

duplicate_export

Duplicate export.

Description

The analyzer produces this diagnostic when an export directive is found that is the same as an export before it in the file. The second export doesn’t add value and should be removed.

Example

The following code produces this diagnostic because the same library is being exported twice:

export 'package:meta/meta.dart';
export 'package:meta/meta.dart';

Common fixes

Remove the unnecessary export:

export 'package:meta/meta.dart';

duplicate_field_formal_parameter

The field ‘{0}’ can’t be initialized by multiple parameters in the same constructor.

Description

The analyzer produces this diagnostic when there’s more than one initializing formal parameter for the same field in a constructor’s parameter list. It isn’t useful to assign a value that will immediately be overwritten.

Example

The following code produces this diagnostic because this.f appears twice in the parameter list:

class C {
  int f;

  C(this.f, this.f) {}
}

Common fixes

Remove one of the initializing formal parameters:

class C {
  int f;

  C(this.f) {}
}

duplicate_field_name

The field name ‘{0}’ is already used in this record.

Description

The analyzer produces this diagnostic when either a record literal or a record type annotation contains a field whose name is the same as a previously declared field in the same literal or type.

Examples

The following code produces this diagnostic because the record literal has two fields named a:

var r = (a: 1, a: 2);

The following code produces this diagnostic because the record type annotation has two fields named a, one a positional field and the other a named field:

void f((int a, {int a}) r) {}

Common fixes

Rename one or both of the fields:

var r = (a: 1, b: 2);

duplicate_hidden_name

Duplicate hidden name.

Description

The analyzer produces this diagnostic when a name occurs multiple times in a hide clause. Repeating the name is unnecessary.

Example

The following code produces this diagnostic because the name min is hidden more than once:

import 'dart:math' hide min, min;

var x = pi;

Common fixes

If the name was mistyped in one or more places, then correct the mistyped names:

import 'dart:math' hide max, min;

var x = pi;

If the name wasn’t mistyped, then remove the unnecessary name from the list:

import 'dart:math' hide min;

var x = pi;

duplicate_ignore

The diagnostic ‘{0}’ doesn’t need to be ignored here because it’s already being ignored.

Description

The analyzer produces this diagnostic when a diagnostic name appears in an ignore comment, but the diagnostic is already being ignored, either because it’s already included in the same ignore comment or because it appears in an ignore-in-file comment.

Examples

The following code produces this diagnostic because the diagnostic named unused_local_variable is already being ignored for the whole file so it doesn’t need to be ignored on a specific line:

// ignore_for_file: unused_local_variable
void f() {
  // ignore: unused_local_variable
  var x = 0;
}

The following code produces this diagnostic because the diagnostic named unused_local_variable is being ignored twice on the same line:

void f() {
  // ignore: unused_local_variable, unused_local_variable
  var x = 0;
}

Common fixes

Remove the ignore comment, or remove the unnecessary diagnostic name if the ignore comment is ignoring more than one diagnostic:

// ignore_for_file: unused_local_variable
void f() {
  var x = 0;
}

duplicate_import

Duplicate import.

Description

The analyzer produces this diagnostic when an import directive is found that is the same as an import before it in the file. The second import doesn’t add value and should be removed.

Example

The following code produces this diagnostic:

import 'package:meta/meta.dart';
import 'package:meta/meta.dart';

@sealed class C {}

Common fixes

Remove the unnecessary import:

import 'package:meta/meta.dart';

@sealed class C {}

duplicate_named_argument

The argument for the named parameter ‘{0}’ was already specified.

Description

The analyzer produces this diagnostic when an invocation has two or more named arguments that have the same name.

Example

The following code produces this diagnostic because there are two arguments with the name a:

void f(C c) {
  c.m(a: 0, a: 1);
}

class C {
  void m({int a, int b}) {}
}

Common fixes

If one of the arguments should have a different name, then change the name:

void f(C c) {
  c.m(a: 0, b: 1);
}

class C {
  void m({int a, int b}) {}
}

If one of the arguments is wrong, then remove it:

void f(C c) {
  c.m(a: 1);
}

class C {
  void m({int a, int b}) {}
}

duplicate_part

The library already contains a part with the URI ‘{0}’.

Description

The analyzer produces this diagnostic when a single file is referenced in multiple part directives.

Example

Given a file part.dart containing

part of lib;

The following code produces this diagnostic because the file part.dart is included multiple times:

library lib;

part 'part.dart';
part 'part.dart';

Common fixes

Remove all except the first of the duplicated part directives:

library lib;

part 'part.dart';

duplicate_pattern_assignment_variable

The variable ‘{0}’ is already assigned in this pattern.

Description

The analyzer produces this diagnostic when a single pattern variable is assigned a value more than once in the same pattern assignment.

Example

The following code produces this diagnostic because the variable a is assigned twice in the pattern (a, a):

int f((int, int) r) {
  int a;
  (a, a) = r;
  return a;
}

Common fixes

If you need to capture all of the values, then use a unique variable for each of the subpatterns being matched:

int f((int, int) r) {
  int a, b;
  (a, b) = r;
  return a + b;
}

If some of the values don’t need to be captured, then use a wildcard pattern _ to avoid having to bind the value to a variable:

int f((int, int) r) {
  int a;
  (_, a) = r;
  return a;
}

duplicate_pattern_field

The field ‘{0}’ is already matched in this pattern.

Description

The analyzer produces this diagnostic when a record pattern matches the same field more than once, or when an object pattern matches the same getter more than once.

Examples

The following code produces this diagnostic because the record field a is matched twice in the same record pattern:

void f(({int a, int b}) r) {
  switch (r) {
    case (a: 1, a: 2):
      return;
  }
}

The following code produces this diagnostic because the getter f is matched twice in the same object pattern:

void f(Object o) {
  switch (o) {
    case C(f: 1, f: 2):
      return;
  }
}
class C {
  int? f;
}

Common fixes

If the pattern should match for more than one value of the duplicated field, then use a logical-or pattern:

void f(({int a, int b}) r) {
  switch (r) {
    case (a: 1, b: _) || (a: 2, b: _):
      break;
  }
}

If the pattern should match against multiple fields, then change the name of one of the fields:

void f(({int a, int b}) r) {
  switch (r) {
    case (a: 1, b: 2):
      return;
  }
}

duplicate_rest_element_in_pattern

At most one rest element is allowed in a list or map pattern.

Description

The analyzer produces this diagnostic when there’s more than one rest pattern in either a list or map pattern. A rest pattern will capture any values unmatched by other subpatterns, making subsequent rest patterns unnecessary because there’s nothing left to capture.

Example

The following code produces this diagnostic because there are two rest patterns in the list pattern:

void f(List<int> x) {
  if (x case [0, ..., ...]) {}
}

Common fixes

Remove all but one of the rest patterns:

void f(List<int> x) {
  if (x case [0, ...]) {}
}

duplicate_shown_name

Duplicate shown name.

Description

The analyzer produces this diagnostic when a name occurs multiple times in a show clause. Repeating the name is unnecessary.

Example

The following code produces this diagnostic because the name min is shown more than once:

import 'dart:math' show min, min;

var x = min(2, min(0, 1));

Common fixes

If the name was mistyped in one or more places, then correct the mistyped names:

import 'dart:math' show max, min;

var x = max(2, min(0, 1));

If the name wasn’t mistyped, then remove the unnecessary name from the list:

import 'dart:math' show min;

var x = min(2, min(0, 1));

duplicate_variable_pattern

The variable ‘{0}’ is already defined in this pattern.

Description

The analyzer produces this diagnostic when a branch of a logical-and pattern declares a variable that is already declared in an earlier branch of the same pattern.

Example

The following code produces this diagnostic because the variable a is declared in both branches of the logical-and pattern:

void f((int, int) r) {
  if (r case (var a, 0) && (0, var a)) {
    print(a);
  }
}

Common fixes

If you need to capture the matched value in multiple branches, then change the names of the variables so that they are unique:

void f((int, int) r) {
  if (r case (var a, 0) && (0, var b)) {
    print(a + b);
  }
}

If you only need to capture the matched value on one branch, then remove the variable pattern from all but one branch:

void f((int, int) r) {
  if (r case (var a, 0) && (0, _)) {
    print(a);
  }
}

empty_map_pattern

A map pattern must have at least one entry.

Description

The analyzer produces this diagnostic when a map pattern is empty.

Example

The following code produces this diagnostic because the map pattern is empty:

void f(Map<int, String> x) {
  if (x case {}) {}
}

Common fixes

If the pattern should match any map, then replace it with an object pattern:

void f(Map<int, String> x) {
  if (x case Map()) {}
}

If the pattern should only match an empty map, then check the length in the pattern:

void f(Map<int, String> x) {
  if (x case Map(isEmpty: true)) {}
}

empty_record_literal_with_comma

A record literal without fields can’t have a trailing comma.

Description

The analyzer produces this diagnostic when a record literal that has no fields has a trailing comma. Empty record literals can’t contain a comma.

Example

The following code produces this diagnostic because the empty record literal has a trailing comma:

var r = (,);

Common fixes

If the record is intended to be empty, then remove the comma:

var r = ();

If the record is intended to have one or more fields, then add the expressions used to compute the values of those fields:

var r = (3, 4);

empty_record_type_named_fields_list

The list of named fields in a record type can’t be empty.

Description

The analyzer produces this diagnostic when a record type has an empty list of named fields.

Example

The following code produces this diagnostic because the record type has an empty list of named fields:

void f((int, int, {}) r) {}

Common fixes

If the record is intended to have named fields, then add the types and names of the fields:

void f((int, int, {int z}) r) {}

If the record isn’t intended to have named fields, then remove the curly braces:

void f((int, int) r) {}

empty_record_type_with_comma

A record type without fields can’t have a trailing comma.

Description

The analyzer produces this diagnostic when a record type that has no fields has a trailing comma. Empty record types can’t contain a comma.

Example

The following code produces this diagnostic because the empty record type has a trailing comma:

void f((,) r) {}

Common fixes

If the record type is intended to be empty, then remove the comma:

void f(() r) {}

If the record type is intended to have one or more fields, then add the types of those fields:

void f((int, int) r) {}

empty_struct

The class ‘{0}’ can’t be empty because it’s a subclass of ‘{1}’.

Description

The analyzer produces this diagnostic when a subclass of Struct or Union doesn’t have any fields. Having an empty Struct or Union isn’t supported.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class C, which extends Struct, doesn’t declare any fields:

import 'dart:ffi';

final class C extends Struct {}

Common fixes

If the class is intended to be a struct, then declare one or more fields:

import 'dart:ffi';

final class C extends Struct {
  @Int32()
  external int x;
}

If the class is intended to be used as a type argument to Pointer, then make it a subclass of Opaque:

import 'dart:ffi';

final class C extends Opaque {}

If the class isn’t intended to be a struct, then remove or change the extends clause:

class C {}

enum_constant_same_name_as_enclosing

The name of the enum constant can’t be the same as the enum’s name.

Description

The analyzer produces this diagnostic when an enum constant has the same name as the enum in which it’s declared.

Example

The following code produces this diagnostic because the enum constant E has the same name as the enclosing enum E:

enum E {
  E
}

Common fixes

If the name of the enum is correct, then rename the constant:

enum E {
  e
}

If the name of the constant is correct, then rename the enum:

enum F {
  E
}

enum_constant_with_non_const_constructor

The invoked constructor isn’t a ‘const’ constructor.

Description

The analyzer produces this diagnostic when an enum constant is being created using either a factory constructor or a generative constructor that isn’t marked as being const.

Example

The following code produces this diagnostic because the enum constant e is being initialized by a factory constructor:

enum E {
  e();

  factory E() => e;
}

Common fixes

Use a generative constructor marked as const:

enum E {
  e._();

  factory E() => e;

  const E._();
}

enum_mixin_with_instance_variable

Mixins applied to enums can’t have instance variables.

Description

The analyzer produces this diagnostic when a mixin that’s applied to an enum declares one or more instance variables. This isn’t allowed because the enum constants are constant, and there isn’t any way for the constructor in the enum to initialize any of the mixin’s fields.

Example

The following code produces this diagnostic because the mixin M defines the instance field x:

mixin M {
  int x = 0;
}

enum E with M {
  a
}

Common fixes

If you need to apply the mixin, then change all instance fields into getter and setter pairs and implement them in the enum if necessary:

mixin M {
  int get x => 0;
}

enum E with M {
  a
}

If you don’t need to apply the mixin, then remove it:

enum E {
  a
}

enum_with_abstract_member

‘{0}’ must have a method body because ‘{1}’ is an enum.

Description

The analyzer produces this diagnostic when a member of an enum is found that doesn’t have a concrete implementation. Enums aren’t allowed to contain abstract members.

Example

The following code produces this diagnostic because m is an abstract method and E is an enum:

enum E {
  e;

  void m();
}

Common fixes

Provide an implementation for the member:

enum E {
  e;

  void m() {}
}

enum_with_name_values

The name ‘values’ is not a valid name for an enum.

Description

The analyzer produces this diagnostic when an enum is declared to have the name values. This isn’t allowed because the enum has an implicit static field named values, and the two would collide.

Example

The following code produces this diagnostic because there’s an enum declaration that has the name values:

enum values {
  c
}

Common fixes

Rename the enum to something other than values.

equal_elements_in_const_set

Two elements in a constant set literal can’t be equal.

Description

The analyzer produces this diagnostic when two elements in a constant set literal have the same value. The set can only contain each value once, which means that one of the values is unnecessary.

Example

The following code produces this diagnostic because the string 'a' is specified twice:

const Set<String> set = {'a', 'a'};

Common fixes

Remove one of the duplicate values:

const Set<String> set = {'a'};

Note that literal sets preserve the order of their elements, so the choice of which element to remove might affect the order in which elements are returned by an iterator.

equal_elements_in_set

Two elements in a set literal shouldn’t be equal.

Description

The analyzer produces this diagnostic when an element in a non-constant set is the same as a previous element in the same set. If two elements are the same, then the second value is ignored, which makes having both elements pointless and likely signals a bug.

Example

The following code produces this diagnostic because the element 1 appears twice:

const a = 1;
const b = 1;
var s = <int>{a, b};

Common fixes

If both elements should be included in the set, then change one of the elements:

const a = 1;
const b = 2;
var s = <int>{a, b};

If only one of the elements is needed, then remove the one that isn’t needed:

const a = 1;
var s = <int>{a};

Note that literal sets preserve the order of their elements, so the choice of which element to remove might affect the order in which elements are returned by an iterator.

equal_keys_in_const_map

Two keys in a constant map literal can’t be equal.

Description

The analyzer produces this diagnostic when a key in a constant map is the same as a previous key in the same map. If two keys are the same, then the second value would overwrite the first value, which makes having both pairs pointless.

Example

The following code produces this diagnostic because the key 1 is used twice:

const map = <int, String>{1: 'a', 2: 'b', 1: 'c', 4: 'd'};

Common fixes

If both entries should be included in the map, then change one of the keys to be different:

const map = <int, String>{1: 'a', 2: 'b', 3: 'c', 4: 'd'};

If only one of the entries is needed, then remove the one that isn’t needed:

const map = <int, String>{1: 'a', 2: 'b', 4: 'd'};

Note that literal maps preserve the order of their entries, so the choice of which entry to remove might affect the order in which keys and values are returned by an iterator.

equal_keys_in_map

Two keys in a map literal shouldn’t be equal.

Description

The analyzer produces this diagnostic when a key in a non-constant map is the same as a previous key in the same map. If two keys are the same, then the second value overwrites the first value, which makes having both pairs pointless and likely signals a bug.

Example

The following code produces this diagnostic because the keys a and b have the same value:

const a = 1;
const b = 1;
var m = <int, String>{a: 'a', b: 'b'};

Common fixes

If both entries should be included in the map, then change one of the keys:

const a = 1;
const b = 2;
var m = <int, String>{a: 'a', b: 'b'};

If only one of the entries is needed, then remove the one that isn’t needed:

const a = 1;
var m = <int, String>{a: 'a'};

Note that literal maps preserve the order of their entries, so the choice of which entry to remove might affect the order in which the keys and values are returned by an iterator.

equal_keys_in_map_pattern

Two keys in a map pattern can’t be equal.

Description

The analyzer produces this diagnostic when a map pattern contains more than one key with the same name. The same key can’t be matched twice.

Example

The following code produces this diagnostic because the key 'a' appears twice:

void f(Map<String, int> x) {
  if (x case {'a': 1, 'a': 2}) {}
}

Common fixes

If you are trying to match two different keys, then change one of the keys in the pattern:

void f(Map<String, int> x) {
  if (x case {'a': 1, 'b': 2}) {}
}

If you are trying to match the same key, but allow any one of multiple patterns to match, the use a logical-or pattern:

void f(Map<String, int> x) {
  if (x case {'a': 1 || 2}) {}
}

expected_one_list_pattern_type_arguments

List patterns require one type argument or none, but {0} found.

Description

The analyzer produces this diagnostic when a list pattern has more than one type argument. List patterns can have either zero type arguments or one type argument, but can’t have more than one.

Example

The following code produces this diagnostic because the list pattern ([0]) has two type arguments:

void f(Object x) {
  if (x case <int, int>[0]) {}
}

Common fixes

Remove all but one of the type arguments:

void f(Object x) {
  if (x case <int>[0]) {}
}

expected_one_list_type_arguments

List literals require one type argument or none, but {0} found.

Description

The analyzer produces this diagnostic when a list literal has more than one type argument.

Example

The following code produces this diagnostic because the list literal has two type arguments when it can have at most one:

var l = <int, int>[];

Common fixes

Remove all except one of the type arguments:

var l = <int>[];

expected_one_set_type_arguments

Set literals require one type argument or none, but {0} were found.

Description

The analyzer produces this diagnostic when a set literal has more than one type argument.

Example

The following code produces this diagnostic because the set literal has three type arguments when it can have at most one:

var s = <int, String, int>{0, 'a', 1};

Common fixes

Remove all except one of the type arguments:

var s = <int>{0, 1};

expected_two_map_pattern_type_arguments

Map patterns require two type arguments or none, but {0} found.

Description

The analyzer produces this diagnostic when a map pattern has either one type argument or more than two type arguments. Map patterns can have either two type arguments or zero type arguments, but can’t have any other number.

Example

The following code produces this diagnostic because the map pattern (<int>{}) has one type argument:

void f(Object x) {
  if (x case <int>{0: _}) {}
}

Common fixes

Add or remove type arguments until there are two, or none:

void f(Object x) {
  if (x case <int, int>{0: _}) {}
}

expected_two_map_type_arguments

Map literals require two type arguments or none, but {0} found.

Description

The analyzer produces this diagnostic when a map literal has either one or more than two type arguments.

Example

The following code produces this diagnostic because the map literal has three type arguments when it can have either two or zero:

var m = <int, String, int>{};

Common fixes

Remove all except two of the type arguments:

var m = <int, String>{};

export_internal_library

The library ‘{0}’ is internal and can’t be exported.

Description

The analyzer produces this diagnostic when it finds an export whose dart: URI references an internal library.

Example

The following code produces this diagnostic because _interceptors is an internal library:

export 'dart:_interceptors';

Common fixes

Remove the export directive.

export_legacy_symbol

The symbol ‘{0}’ is defined in a legacy library, and can’t be re-exported from a library with null safety enabled.

Description

The analyzer produces this diagnostic when a library that was opted in to null safety exports another library, and the exported library is opted out of null safety.

Example

Given a library that is opted out of null safety:

// @dart = 2.8
String s;

The following code produces this diagnostic because it’s exporting symbols from an opted-out library:

export 'optedOut.dart';

class C {}

Common fixes

If you’re able to do so, migrate the exported library so that it doesn’t need to opt out:

String? s;

If you can’t migrate the library, then remove the export:

class C {}

If the exported library (the one that is opted out) itself exports an opted-in library, then it’s valid for your library to indirectly export the symbols from the opted-in library. You can do so by adding a hide combinator to the export directive in your library that hides all of the names declared in the opted-out library.

export_of_non_library

The exported library ‘{0}’ can’t have a part-of directive.

Description

The analyzer produces this diagnostic when an export directive references a part rather than a library.

Example

Given a file part.dart containing

part of lib;

The following code produces this diagnostic because the file part.dart is a part, and only libraries can be exported:

library lib;

export 'part.dart';

Common fixes

Either remove the export directive, or change the URI to be the URI of the library containing the part.

expression_in_map

Expressions can’t be used in a map literal.

Description

The analyzer produces this diagnostic when the analyzer finds an expression, rather than a map entry, in what appears to be a map literal.

Example

The following code produces this diagnostic:

var map = <String, int>{'a': 0, 'b': 1, 'c'};

Common fixes

If the expression is intended to compute either a key or a value in an entry, fix the issue by replacing the expression with the key or the value. For example:

var map = <String, int>{'a': 0, 'b': 1, 'c': 2};

extends_non_class

Classes can only extend other classes.

Description

The analyzer produces this diagnostic when an extends clause contains a name that is declared to be something other than a class.

Example

The following code produces this diagnostic because f is declared to be a function:

void f() {}

class C extends f {}

Common fixes

If you want the class to extend a class other than Object, then replace the name in the extends clause with the name of that class:

void f() {}

class C extends B {}

class B {}

If you want the class to extend Object, then remove the extends clause:

void f() {}

class C {}

extension_as_expression

Extension ‘{0}’ can’t be used as an expression.

Description

The analyzer produces this diagnostic when the name of an extension is used in an expression other than in an extension override or to qualify an access to a static member of the extension. Because classes define a type, the name of a class can be used to refer to the instance of Type representing the type of the class. Extensions, on the other hand, don’t define a type and can’t be used as a type literal.

Example

The following code produces this diagnostic because E is an extension:

extension E on int {
  static String m() => '';
}

var x = E;

Common fixes

Replace the name of the extension with a name that can be referenced, such as a static member defined on the extension:

extension E on int {
  static String m() => '';
}

var x = E.m();

extension_conflicting_static_and_instance

An extension can’t define static member ‘{0}’ and an instance member with the same name.

Description

The analyzer produces this diagnostic when an extension declaration contains both an instance member and a static member that have the same name. The instance member and the static member can’t have the same name because it’s unclear which member is being referenced by an unqualified use of the name within the body of the extension.

Example

The following code produces this diagnostic because the name a is being used for two different members:

extension E on Object {
  int get a => 0;
  static int a() => 0;
}

Common fixes

Rename or remove one of the members:

extension E on Object {
  int get a => 0;
  static int b() => 0;
}

extension_declares_abstract_member

Extensions can’t declare abstract members.

Description

The analyzer produces this diagnostic when an abstract declaration is declared in an extension. Extensions can declare only concrete members.

Example

The following code produces this diagnostic because the method a doesn’t have a body:

extension E on String {
  int a();
}

Common fixes

Either provide an implementation for the member or remove it.

extension_declares_constructor

Extensions can’t declare constructors.

Description

The analyzer produces this diagnostic when a constructor declaration is found in an extension. It isn’t valid to define a constructor because extensions aren’t classes, and it isn’t possible to create an instance of an extension.

Example

The following code produces this diagnostic because there is a constructor declaration in E:

extension E on String {
  E() : super();
}

Common fixes

Remove the constructor or replace it with a static method.

extension_declares_instance_field

Extensions can’t declare instance fields

Description

The analyzer produces this diagnostic when an instance field declaration is found in an extension. It isn’t valid to define an instance field because extensions can only add behavior, not state.

Example

The following code produces this diagnostic because s is an instance field:

extension E on String {
  String s;
}

Common fixes

Remove the field, make it a static field, or convert it to be a getter, setter, or method.

extension_declares_member_of_object

Extensions can’t declare members with the same name as a member declared by ‘Object’.

Description

The analyzer produces this diagnostic when an extension declaration declares a member with the same name as a member declared in the class Object. Such a member can never be used because the member in Object is always found first.

Example

The following code produces this diagnostic because toString is defined by Object:

extension E on String {
  String toString() => this;
}

Common fixes

Remove the member or rename it so that the name doesn’t conflict with the member in Object:

extension E on String {
  String displayString() => this;
}

extension_override_access_to_static_member

An extension override can’t be used to access a static member from an extension.

Description

The analyzer produces this diagnostic when an extension override is the receiver of the invocation of a static member. Similar to static members in classes, the static members of an extension should be accessed using the name of the extension, not an extension override.

Example

The following code produces this diagnostic because m is static:

extension E on String {
  static void m() {}
}

void f() {
  E('').m();
}

Common fixes

Replace the extension override with the name of the extension:

extension E on String {
  static void m() {}
}

void f() {
  E.m();
}

extension_override_argument_not_assignable

The type of the argument to the extension override ‘{0}’ isn’t assignable to the extended type ‘{1}’.

Description

The analyzer produces this diagnostic when the argument to an extension override isn’t assignable to the type being extended by the extension.

Example

The following code produces this diagnostic because 3 isn’t a String:

extension E on String {
  void method() {}
}

void f() {
  E(3).method();
}

Common fixes

If you’re using the correct extension, then update the argument to have the correct type:

extension E on String {
  void method() {}
}

void f() {
  E(3.toString()).method();
}

If there’s a different extension that’s valid for the type of the argument, then either replace the name of the extension or unwrap the argument so that the correct extension is found.

extension_override_without_access

An extension override can only be used to access instance members.

Description

The analyzer produces this diagnostic when an extension override is found that isn’t being used to access one of the members of the extension. The extension override syntax doesn’t have any runtime semantics; it only controls which member is selected at compile time.

Example

The following code produces this diagnostic because E(i) isn’t an expression:

extension E on int {
  int get a => 0;
}

void f(int i) {
  print(E(i));
}

Common fixes

If you want to invoke one of the members of the extension, then add the invocation:

extension E on int {
  int get a => 0;
}

void f(int i) {
  print(E(i).a);
}

If you don’t want to invoke a member, then unwrap the argument:

extension E on int {
  int get a => 0;
}

void f(int i) {
  print(i);
}

extension_override_with_cascade

Extension overrides have no value so they can’t be used as the receiver of a cascade expression.

Description

The analyzer produces this diagnostic when an extension override is used as the receiver of a cascade expression. The value of a cascade expression e..m is the value of the receiver e, but extension overrides aren’t expressions and don’t have a value.

Example

The following code produces this diagnostic because E(3) isn’t an expression:

extension E on int {
  void m() {}
}
f() {
  E(3)..m();
}

Common fixes

Use . rather than ..:

extension E on int {
  void m() {}
}
f() {
  E(3).m();
}

If there are multiple cascaded accesses, you’ll need to duplicate the extension override for each one.

extension_type_constructor_with_super_formal_parameter

Extension type constructors can’t declare super formal parameters.

Description

The analyzer produces this diagnostic when a constructor in an extension type has a super parameter. Super parameters aren’t valid because extension types don’t have a superclass.

Example

The following code produces this diagnostic because the named constructor n contains a super parameter:

extension type E(int i) {
  E.n(this.i, super.foo);
}

Common fixes

If you need the parameter, replace the super parameter with a normal parameter:

extension type E(int i) {
  E.n(this.i, String foo);
}

If you don’t need the parameter, remove the super parameter:

extension type E(int i) {
  E.n(this.i);
}

extension_type_constructor_with_super_invocation

Extension type constructors can’t include super initializers.

Description

The analyzer produces this diagnostic when a constructor in an extension type includes an invocation of a super constructor in the initializer list. Because extension types don’t have a superclass, there’s no constructor to invoke.

Example

The following code produces this diagnostic because the constructor E.n invokes a super constructor in its initializer list:

extension type E(int i) {
  E.n() : i = 0, super.n();
}

Common fixes

Remove the invocation of the super constructor:

extension type E(int i) {
  E.n() : i = 0;
}

extension_type_declares_instance_field

Extension types can’t declare instance fields.

Description

The analyzer produces this diagnostic when there’s a field declaration in the body of an extension type declaration.

Example

The following code produces this diagnostic because the extension type E declares a field named f:

extension type E(int i) {
  final int f = 0;
}

Common fixes

If you don’t need the field, then remove it or replace it with a getter and/or setter:

extension type E(int i) {
  int get f => 0;
}

If you need the field, then convert the extension type into a class:

class E {
  final int i;

  final int f = 0;

  E(this.i);
}

extension_type_declares_member_of_object

Extension types can’t declare members with the same name as a member declared by ‘Object’.

Description

The analyzer produces this diagnostic when the body of an extension type declaration contains a member with the same name as one of the members declared by Object.

Example

The following code produces this diagnostic because the class Object already defines a member named hashCode:

extension type E(int i) {
  int get hashCode => 0;
}

Common fixes

If you need a member with the implemented semantics, then rename the member:

extension type E(int i) {
  int get myHashCode => 0;
}

If you don’t need a member with the implemented semantics, then remove the member:

extension type E(int i) {}

extension_type_implements_disallowed_type

Extension types can’t implement ‘{0}’.

Description

The analyzer produces this diagnostic when an extension type implements a type that it isn’t allowed to implement.

Example

The following code produces this diagnostic because extension types can’t implement the type dynamic:

extension type A(int i) implements dynamic {}

Common fixes

Remove the disallowed type from the implements clause:

extension type A(int i) {}

extension_type_implements_itself

The extension type can’t implement itself.

Description

The analyzer produces this diagnostic when an extension type implements itself, either directly or indirectly.

Example

The following code produces this diagnostic because the extension type A directly implements itself:

extension type A(int i) implements A {}

The following code produces this diagnostic because the extension type A indirectly implements itself (through B):

extension type A(int i) implements B {}

extension type B(int i) implements A {}

Common fixes

Break the cycle by removing a type from the implements clause of at least one of the types involved in the cycle:

extension type A(int i) implements B {}

extension type B(int i) {}

extension_type_implements_not_supertype

‘{0}’ is not a supertype of ‘{1}’, the representation type.

Description

The analyzer produces this diagnostic when an extension type implements a type that isn’t a supertype of the representation type.

Example

The following code produces this diagnostic because the extension type A implements String, but String isn’t a supertype of the representation type int:

extension type A(int i) implements String {}

Common fixes

If the representation type is correct, then remove or replace the type in the implements clause:

extension type A(int i) {}

If the representation type isn’t correct, then replace it with the correct type:

extension type A(String s) implements String {}

extension_type_implements_representation_not_supertype

‘{0}’, the representation type of ‘{1}’, is not a supertype of ‘{2}’, the representation type of ‘{3}’.

Description

The analyzer produces this diagnostic when an extension type implements another extension type, and the representation type of the implemented extension type isn’t a subtype of the representation type of the implementing extension type.

Example

The following code produces this diagnostic because the extension type B implements A, but the representation type of A (num) isn’t a subtype of the representation type of B (String):

extension type A(num i) {}

extension type B(String s) implements A {}

Common fixes

Either change the representation types of the two extension types so that the representation type of the implemented type is a supertype of the representation type of the implementing type:

extension type A(num i) {}

extension type B(int n) implements A {}

Or remove the implemented type from the implements clause:

extension type A(num i) {}

extension type B(String s) {}

extension_type_inherited_member_conflict

The extension type ‘{0}’ has more than one distinct member named ‘{1}’ from implemented types.

Description

The analyzer produces this diagnostic when an extension type implements two or more other types, and at least two of those types declare a member with the same name.

Example

The following code produces this diagnostic because the extension type C implements both A and B, and both declare a member named m:

class A {
  void m() {}
}

extension type B(A a) {
  void m() {}
}

extension type C(A a) implements A, B {}

Common fixes

If the extension type doesn’t need to implement all of the listed types, then remove all but one of the types introducing the conflicting members:

class A {
  void m() {}
}

extension type B(A a) {
  void m() {}
}

extension type C(A a) implements A {}

If the extension type needs to implement all of the listed types but you can rename the members in those types, then give the conflicting members unique names:

class A {
  void m() {}
}

extension type B(A a) {
  void n() {}
}

extension type C(A a) implements A, B {}

extension_type_representation_depends_on_itself

The extension type representation can’t depend on itself.

Description

The analyzer produces this diagnostic when an extension type has a representation type that depends on the extension type itself, either directly or indirectly.

Example

The following code produces this diagnostic because the representation type of the extension type A depends on A directly:

extension type A(A a) {}

The following two code examples produce this diagnostic because the representation type of the extension type A depends on A indirectly through the extension type B:

extension type A(B b) {}

extension type B(A a) {}

extension type A(List<B> b) {}

extension type B(List<A> a) {}

Common fixes

Remove the dependency by choosing a different representation type for at least one of the types in the cycle:

extension type A(String s) {}

external_with_initializer

External fields can’t have initializers.

External variables can’t have initializers.

Description

The analyzer produces this diagnostic when a field or variable marked with the keyword external has an initializer, or when an external field is initialized in a constructor.

Examples

The following code produces this diagnostic because the external field x is assigned a value in an initializer:

class C {
  external int x;
  C() : x = 0;
}

The following code produces this diagnostic because the external field x has an initializer:

class C {
  external final int x = 0;
}

The following code produces this diagnostic because the external top level variable x has an initializer:

external final int x = 0;

Common fixes

Remove the initializer:

class C {
  external final int x;
}

extra_annotation_on_struct_field

Fields in a struct class must have exactly one annotation indicating the native type.

Description

The analyzer produces this diagnostic when a field in a subclass of Struct has more than one annotation describing the native type of the field.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field x has two annotations describing the native type of the field:

import 'dart:ffi';

final class C extends Struct {
  @Int32()
  @Int16()
  external int x;
}

Common fixes

Remove all but one of the annotations:

import 'dart:ffi';
final class C extends Struct {
  @Int32()
  external int x;
}

extra_positional_arguments

Too many positional arguments: {0} expected, but {1} found.

Description

The analyzer produces this diagnostic when a method or function invocation has more positional arguments than the method or function allows.

Example

The following code produces this diagnostic because f defines 2 parameters but is invoked with 3 arguments:

void f(int a, int b) {}
void g() {
  f(1, 2, 3);
}

Common fixes

Remove the arguments that don’t correspond to parameters:

void f(int a, int b) {}
void g() {
  f(1, 2);
}

extra_positional_arguments_could_be_named

Too many positional arguments: {0} expected, but {1} found.

Description

The analyzer produces this diagnostic when a method or function invocation has more positional arguments than the method or function allows, but the method or function defines named parameters.

Example

The following code produces this diagnostic because f defines 2 positional parameters but has a named parameter that could be used for the third argument:

void f(int a, int b, {int c}) {}
void g() {
  f(1, 2, 3);
}

Common fixes

If some of the arguments should be values for named parameters, then add the names before the arguments:

void f(int a, int b, {int c}) {}
void g() {
  f(1, 2, c: 3);
}

Otherwise, remove the arguments that don’t correspond to positional parameters:

void f(int a, int b, {int c}) {}
void g() {
  f(1, 2);
}

extra_size_annotation_carray

‘Array’s must have exactly one ‘Array’ annotation.

Description

The analyzer produces this diagnostic when a field in a subclass of Struct has more than one annotation describing the size of the native array.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field a0 has two annotations that specify the size of the native array:

import 'dart:ffi';

final class C extends Struct {
  @Array(4)
  @Array(8)
  external Array<Uint8> a0;
}

Common fixes

Remove all but one of the annotations:

import 'dart:ffi';

final class C extends Struct {
  @Array(8)
  external Array<Uint8> a0;
}

field_initialized_by_multiple_initializers

The field ‘{0}’ can’t be initialized twice in the same constructor.

Description

The analyzer produces this diagnostic when the initializer list of a constructor initializes a field more than once. There is no value to allow both initializers because only the last value is preserved.

Example

The following code produces this diagnostic because the field f is being initialized twice:

class C {
  int f;

  C() : f = 0, f = 1;
}

Common fixes

Remove one of the initializers:

class C {
  int f;

  C() : f = 0;
}

field_initialized_in_initializer_and_declaration

Fields can’t be initialized in the constructor if they are final and were already initialized at their declaration.

Description

The analyzer produces this diagnostic when a final field is initialized in both the declaration of the field and in an initializer in a constructor. Final fields can only be assigned once, so it can’t be initialized in both places.

Example

The following code produces this diagnostic because f is :

class C {
  final int f = 0;
  C() : f = 1;
}

Common fixes

If the initialization doesn’t depend on any values passed to the constructor, and if all of the constructors need to initialize the field to the same value, then remove the initializer from the constructor:

class C {
  final int f = 0;
  C();
}

If the initialization depends on a value passed to the constructor, or if different constructors need to initialize the field differently, then remove the initializer in the field’s declaration:

class C {
  final int f;
  C() : f = 1;
}

field_initialized_in_parameter_and_initializer

Fields can’t be initialized in both the parameter list and the initializers.

Description

The analyzer produces this diagnostic when a field is initialized in both the parameter list and in the initializer list of a constructor.

Example

The following code produces this diagnostic because the field f is initialized both by an initializing formal parameter and in the initializer list:

class C {
  int f;

  C(this.f) : f = 0;
}

Common fixes

If the field should be initialized by the parameter, then remove the initialization in the initializer list:

class C {
  int f;

  C(this.f);
}

If the field should be initialized in the initializer list and the parameter isn’t needed, then remove the parameter:

class C {
  int f;

  C() : f = 0;
}

If the field should be initialized in the initializer list and the parameter is needed, then make it a normal parameter:

class C {
  int f;

  C(int g) : f = g * 2;
}

field_initializer_factory_constructor

Initializing formal parameters can’t be used in factory constructors.

Description

The analyzer produces this diagnostic when a factory constructor has an initializing formal parameter. Factory constructors can’t assign values to fields because no instance is created; hence, there is no field to assign.

Example

The following code produces this diagnostic because the factory constructor uses an initializing formal parameter:

class C {
  int? f;

  factory C(this.f) => throw 0;
}

Common fixes

Replace the initializing formal parameter with a normal parameter:

class C {
  int? f;

  factory C(int f) => throw 0;
}

field_initializer_in_struct

Constructors in subclasses of ‘Struct’ and ‘Union’ can’t have field initializers.

Description

The analyzer produces this diagnostic when a constructor in a subclass of either Struct or Union has one or more field initializers.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class C has a constructor with an initializer for the field f:

// @dart = 2.9
import 'dart:ffi';

final class C extends Struct {
  @Int32()
  int f;

  C() : f = 0;
}

Common fixes

Remove the field initializer:

// @dart = 2.9
import 'dart:ffi';

final class C extends Struct {
  @Int32()
  int f;

  C();
}

field_initializer_not_assignable

The initializer type ‘{0}’ can’t be assigned to the field type ‘{1}’ in a const constructor.

The initializer type ‘{0}’ can’t be assigned to the field type ‘{1}’.

Description

The analyzer produces this diagnostic when the initializer list of a constructor initializes a field to a value that isn’t assignable to the field.

Example

The following code produces this diagnostic because 0 has the type int, and an int can’t be assigned to a field of type String:

class C {
  String s;

  C() : s = 0;
}

Common fixes

If the type of the field is correct, then change the value assigned to it so that the value has a valid type:

class C {
  String s;

  C() : s = '0';
}

If the type of the value is correct, then change the type of the field to allow the assignment:

class C {
  int s;

  C() : s = 0;
}

field_initializer_outside_constructor

Field formal parameters can only be used in a constructor.

Initializing formal parameters can only be used in constructors.

Description

The analyzer produces this diagnostic when an initializing formal parameter is used in the parameter list for anything other than a constructor.

Example

The following code produces this diagnostic because the initializing formal parameter this.x is being used in the method m:

class A {
  int x = 0;

  m([this.x = 0]) {}
}

Common fixes

Replace the initializing formal parameter with a normal parameter and assign the field within the body of the method:

class A {
  int x = 0;

  m([int x = 0]) {
    this.x = x;
  }
}

field_initializer_redirecting_constructor

The redirecting constructor can’t have a field initializer.

Description

The analyzer produces this diagnostic when a redirecting constructor initializes a field in the object. This isn’t allowed because the instance that has the field hasn’t been created at the point at which it should be initialized.

Examples

The following code produces this diagnostic because the constructor C.zero, which redirects to the constructor C, has an initializing formal parameter that initializes the field f:

class C {
  int f;

  C(this.f);

  C.zero(this.f) : this(f);
}

The following code produces this diagnostic because the constructor C.zero, which redirects to the constructor C, has an initializer that initializes the field f:

class C {
  int f;

  C(this.f);

  C.zero() : f = 0, this(1);
}

Common fixes

If the initialization is done by an initializing formal parameter, then use a normal parameter:

class C {
  int f;

  C(this.f);

  C.zero(int f) : this(f);
}

If the initialization is done in an initializer, then remove the initializer:

class C {
  int f;

  C(this.f);

  C.zero() : this(0);
}

field_initializing_formal_not_assignable

The parameter type ‘{0}’ is incompatible with the field type ‘{1}’.

Description

The analyzer produces this diagnostic when the type of an initializing formal parameter isn’t assignable to the type of the field being initialized.

Example

The following code produces this diagnostic because the initializing formal parameter has the type String, but the type of the field is int. The parameter must have a type that is a subtype of the field’s type.

class C {
  int f;

  C(String this.f);
}

Common fixes

If the type of the field is incorrect, then change the type of the field to match the type of the parameter, and consider removing the type from the parameter:

class C {
  String f;

  C(this.f);
}

If the type of the parameter is incorrect, then remove the type of the parameter:

class C {
  int f;

  C(this.f);
}

If the types of both the field and the parameter are correct, then use an initializer rather than an initializing formal parameter to convert the parameter value into a value of the correct type:

class C {
  int f;

  C(String s) : f = int.parse(s);
}

field_in_struct_with_initializer

Fields in subclasses of ‘Struct’ and ‘Union’ can’t have initializers.

Description

The analyzer produces this diagnostic when a field in a subclass of Struct has an initializer.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field p has an initializer:

// @dart = 2.9
import 'dart:ffi';

final class C extends Struct {
  Pointer p = nullptr;
}

Common fixes

Remove the initializer:

// @dart = 2.9
import 'dart:ffi';

final class C extends Struct {
  Pointer p;
}

field_must_be_external_in_struct

Fields of ‘Struct’ and ‘Union’ subclasses must be marked external.

Description

The analyzer produces this diagnostic when a field in a subclass of either Struct or Union isn’t marked as being external.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field a isn’t marked as being external:

import 'dart:ffi';

final class C extends Struct {
  @Int16()
  int a;
}

Common fixes

Add the required external modifier:

import 'dart:ffi';

final class C extends Struct {
  @Int16()
  external int a;
}

final_initialized_in_declaration_and_constructor

‘{0}’ is final and was given a value when it was declared, so it can’t be set to a new value.

Description

The analyzer produces this diagnostic when a final field is initialized twice: once where it’s declared and once by a constructor’s parameter.

Example

The following code produces this diagnostic because the field f is initialized twice:

class C {
  final int f = 0;

  C(this.f);
}

Common fixes

If the field should have the same value for all instances, then remove the initialization in the parameter list:

class C {
  final int f = 0;

  C();
}

If the field can have different values in different instances, then remove the initialization in the declaration:

class C {
  final int f;

  C(this.f);
}

final_not_initialized

The final variable ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a final field or variable isn’t initialized.

Example

The following code produces this diagnostic because x doesn’t have an initializer:

final x;

Common fixes

For variables and static fields, you can add an initializer:

final x = 0;

For instance fields, you can add an initializer as shown in the previous example, or you can initialize the field in every constructor. You can initialize the field by using an initializing formal parameter:

class C {
  final int x;
  C(this.x);
}

You can also initialize the field by using an initializer in the constructor:

class C {
  final int x;
  C(int y) : x = y * 2;
}

final_not_initialized_constructor

All final variables must be initialized, but ‘{0}’ and ‘{1}’ aren’t.

All final variables must be initialized, but ‘{0}’ isn’t.

All final variables must be initialized, but ‘{0}’, ‘{1}’, and {2} others aren’t.

Description

The analyzer produces this diagnostic when a class defines one or more final instance fields without initializers and has at least one constructor that doesn’t initialize those fields. All final instance fields must be initialized when the instance is created, either by the field’s initializer or by the constructor.

Example

The following code produces this diagnostic:

class C {
  final String value;

  C();
}

Common fixes

If the value should be passed in to the constructor directly, then use an initializing formal parameter to initialize the field value:

class C {
  final String value;

  C(this.value);
}

If the value should be computed indirectly from a value provided by the caller, then add a parameter and include an initializer:

class C {
  final String value;

  C(Object o) : value = o.toString();
}

If the value of the field doesn’t depend on values that can be passed to the constructor, then add an initializer for the field as part of the field declaration:

class C {
  final String value = '';

  C();
}

If the value of the field doesn’t depend on values that can be passed to the constructor but different constructors need to initialize it to different values, then add an initializer for the field in the initializer list:

class C {
  final String value;

  C() : value = '';

  C.named() : value = 'c';
}

However, if the value is the same for all instances, then consider using a static field instead of an instance field:

class C {
  static const String value = '';

  C();
}

flutter_field_not_map

The value of the ‘flutter’ field is expected to be a map.

Description

The analyzer produces this diagnostic when the value of the flutter key isn’t a map.

Example

The following code produces this diagnostic because the value of the top-level flutter key is a string:

name: example
flutter: true

Common fixes

If you need to specify Flutter-specific options, then change the value to be a map:

name: example
flutter:
  uses-material-design: true

If you don’t need to specify Flutter-specific options, then remove the flutter key:

name: example

for_in_of_invalid_element_type

The type ‘{0}’ used in the ‘for’ loop must implement ‘{1}’ with a type argument that can be assigned to ‘{2}’.

Description

The analyzer produces this diagnostic when the Iterable or Stream in a for-in loop has an element type that can’t be assigned to the loop variable.

Example

The following code produces this diagnostic because <String>[] has an element type of String, and String can’t be assigned to the type of e (int):

void f() {
  for (int e in <String>[]) {
    print(e);
  }
}

Common fixes

If the type of the loop variable is correct, then update the type of the iterable:

void f() {
  for (int e in <int>[]) {
    print(e);
  }
}

If the type of the iterable is correct, then update the type of the loop variable:

void f() {
  for (String e in <String>[]) {
    print(e);
  }
}

for_in_of_invalid_type

The type ‘{0}’ used in the ‘for’ loop must implement ‘{1}’.

Description

The analyzer produces this diagnostic when the expression following in in a for-in loop has a type that isn’t a subclass of Iterable.

Example

The following code produces this diagnostic because m is a Map, and Map isn’t a subclass of Iterable:

void f(Map<String, String> m) {
  for (String s in m) {
    print(s);
  }
}

Common fixes

Replace the expression with one that produces an iterable value:

void f(Map<String, String> m) {
  for (String s in m.values) {
    print(s);
  }
}

for_in_with_const_variable

A for-in loop variable can’t be a ‘const’.

Description

The analyzer produces this diagnostic when the loop variable declared in a for-in loop is declared to be a const. The variable can’t be a const because the value can’t be computed at compile time.

Example

The following code produces this diagnostic because the loop variable x is declared to be a const:

void f() {
  for (const x in [0, 1, 2]) {
    print(x);
  }
}

Common fixes

If there’s a type annotation, then remove the const modifier from the declaration.

If there’s no type, then replace the const modifier with final, var, or a type annotation:

void f() {
  for (final x in [0, 1, 2]) {
    print(x);
  }
}

generic_method_type_instantiation_on_dynamic

A method tear-off on a receiver whose type is ‘dynamic’ can’t have type arguments.

Description

The analyzer produces this diagnostic when an instance method is being torn off from a receiver whose type is dynamic, and the tear-off includes type arguments. Because the analyzer can’t know how many type parameters the method has, or whether it has any type parameters, there’s no way it can validate that the type arguments are correct. As a result, the type arguments aren’t allowed.

Example

The following code produces this diagnostic because the type of p is dynamic and the tear-off of m has type arguments:

void f(dynamic list) {
  list.fold<int>;
}

Common fixes

If you can use a more specific type than dynamic, then change the type of the receiver:

void f(List<Object> list) {
  list.fold<int>;
}

If you can’t use a more specific type, then remove the type arguments:

void f(dynamic list) {
  list.cast;
}

generic_struct_subclass

The class ‘{0}’ can’t extend ‘Struct’ or ‘Union’ because ‘{0}’ is generic.

Description

The analyzer produces this diagnostic when a subclass of either Struct or Union has a type parameter.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class S defines the type parameter T:

import 'dart:ffi';

final class S<T> extends Struct {
  external Pointer notEmpty;
}

Common fixes

Remove the type parameters from the class:

import 'dart:ffi';

final class S extends Struct {
  external Pointer notEmpty;
}

getter_not_subtype_setter_types

The return type of getter ‘{0}’ is ‘{1}’ which isn’t a subtype of the type ‘{2}’ of its setter ‘{3}’.

Description

The analyzer produces this diagnostic when the return type of a getter isn’t a subtype of the type of the parameter of a setter with the same name.

The subtype relationship is a requirement whether the getter and setter are in the same class or whether one of them is in a superclass of the other.

Example

The following code produces this diagnostic because the return type of the getter x is num, the parameter type of the setter x is int, and num isn’t a subtype of int:

class C {
  num get x => 0;

  set x(int y) {}
}

Common fixes

If the type of the getter is correct, then change the type of the setter:

class C {
  num get x => 0;

  set x(num y) {}
}

If the type of the setter is correct, then change the type of the getter:

class C {
  int get x => 0;

  set x(int y) {}
}

illegal_async_generator_return_type

_Functions marked ‘async*’ must have a return type that is a supertype of ‘Stream' for some type 'T'._

Description

The analyzer produces this diagnostic when the body of a function has the async* modifier even though the return type of the function isn’t either Stream or a supertype of Stream.

Example

The following code produces this diagnostic because the body of the function f has the ‘async*’ modifier even though the return type int isn’t a supertype of Stream:

int f() async* {}

Common fixes

If the function should be asynchronous, then change the return type to be either Stream or a supertype of Stream:

Stream<int> f() async* {}

If the function should be synchronous, then remove the async* modifier:

int f() => 0;

illegal_async_return_type

Functions marked ‘async’ must have a return type which is a supertype of ‘Future’.

Description

The analyzer produces this diagnostic when the body of a function has the async modifier even though the return type of the function isn’t assignable to Future.

Example

The following code produces this diagnostic because the body of the function f has the async modifier even though the return type isn’t assignable to Future:

int f() async {
  return 0;
}

Common fixes

If the function should be asynchronous, then change the return type to be assignable to Future:

Future<int> f() async {
  return 0;
}

If the function should be synchronous, then remove the async modifier:

int f() => 0;

illegal_concrete_enum_member

A concrete instance member named ‘{0}’ can’t be declared in a class that implements ‘Enum’.

A concrete instance member named ‘{0}’ can’t be inherited from ‘{1}’ in a class that implements ‘Enum’.

Description

The analyzer produces this diagnostic when either an enum declaration, a class that implements Enum, or a mixin with a superclass constraint of Enum, declares or inherits a concrete instance member named either index, hashCode, or ==.

Examples

The following code produces this diagnostic because the enum E declares an instance getter named index:

enum E {
  v;

  int get index => 0;
}

The following code produces this diagnostic because the class C, which implements Enum, declares an instance field named hashCode:

abstract class C implements Enum {
  int hashCode = 0;
}

The following code produces this diagnostic because the class C, which indirectly implements Enum through the class A, declares an instance getter named hashCode:

abstract class A implements Enum {}

abstract class C implements A {
  int get hashCode => 0;
}

The following code produces this diagnostic because the mixin M, which has Enum in the on clause, declares an explicit operator named ==:

mixin M on Enum {
  bool operator ==(Object? other) => false;
}

Common fixes

Rename the conflicting member:

enum E {
  v;

  int get getIndex => 0;
}

illegal_enum_values

An instance member named ‘values’ can’t be declared in a class that implements ‘Enum’.

An instance member named ‘values’ can’t be inherited from ‘{0}’ in a class that implements ‘Enum’.

Description

The analyzer produces this diagnostic when either a class that implements Enum or a mixin with a superclass constraint of Enum has an instance member named values.

Examples

The following code produces this diagnostic because the class C, which implements Enum, declares an instance field named values:

abstract class C implements Enum {
  int get values => 0;
}

The following code produces this diagnostic because the class B, which implements Enum, inherits an instance method named values from A:

abstract class A {
  int values() => 0;
}

abstract class B extends A implements Enum {}

Common fixes

Change the name of the conflicting member:

abstract class C implements Enum {
  int get value => 0;
}

illegal_sync_generator_return_type

_Functions marked ‘sync*’ must have a return type that is a supertype of ‘Iterable' for some type 'T'._

Description

The analyzer produces this diagnostic when the body of a function has the sync* modifier even though the return type of the function isn’t either Iterable or a supertype of Iterable.

Example

The following code produces this diagnostic because the body of the function f has the ‘sync*’ modifier even though the return type int isn’t a supertype of Iterable:

int f() sync* {}

Common fixes

If the function should return an iterable, then change the return type to be either Iterable or a supertype of Iterable:

Iterable<int> f() sync* {}

If the function should return a single value, then remove the sync* modifier:

int f() => 0;

implements_non_class

Classes and mixins can only implement other classes and mixins.

Description

The analyzer produces this diagnostic when a name used in the implements clause of a class or mixin declaration is defined to be something other than a class or mixin.

Example

The following code produces this diagnostic because x is a variable rather than a class or mixin:

var x;
class C implements x {}

Common fixes

If the name is the name of an existing class or mixin that’s already being imported, then add a prefix to the import so that the local definition of the name doesn’t shadow the imported name.

If the name is the name of an existing class or mixin that isn’t being imported, then add an import, with a prefix, for the library in which it’s declared.

Otherwise, either replace the name in the implements clause with the name of an existing class or mixin, or remove the name from the implements clause.

implements_repeated

‘{0}’ can only be implemented once.

Description

The analyzer produces this diagnostic when a single class is specified more than once in an implements clause.

Example

The following code produces this diagnostic because A is in the list twice:

class A {}
class B implements A, A {}

Common fixes

Remove all except one occurrence of the class name:

class A {}
class B implements A {}

implements_super_class

‘{0}’ can’t be used in both the ‘extends’ and ‘implements’ clauses.

‘{0}’ can’t be used in both the ‘extends’ and ‘with’ clauses.

Description

The analyzer produces this diagnostic when a class is listed in the extends clause of a class declaration and also in either the implements or with clause of the same declaration.

Example

The following code produces this diagnostic because the class A is used in both the extends and implements clauses for the class B:

class A {}

class B extends A implements A {}

The following code produces this diagnostic because the class A is used in both the extends and with clauses for the class B:

mixin class A {}

class B extends A with A {}

Common fixes

If you want to inherit the implementation from the class, then remove the class from the implements clause:

class A {}

class B extends A {}

If you don’t want to inherit the implementation from the class, then remove the extends clause:

class A {}

class B implements A {}

implicit_super_initializer_missing_arguments

The implicitly invoked unnamed constructor from ‘{0}’ has required parameters.

Description

The analyzer produces this diagnostic when a constructor implicitly invokes the unnamed constructor from the superclass, the unnamed constructor of the superclass has a required parameter, and there’s no super parameter corresponding to the required parameter.

Examples

The following code produces this diagnostic because the unnamed constructor in the class B implicitly invokes the unnamed constructor in the class A, but the constructor in A has a required positional parameter named x:

class A {
  A(int x);
}

class B extends A {
  B();
}

The following code produces this diagnostic because the unnamed constructor in the class B implicitly invokes the unnamed constructor in the class A, but the constructor in A has a required named parameter named x:

class A {
  A({required int x});
}

class B extends A {
  B();
}

Common fixes

If you can add a parameter to the constructor in the subclass, then add a super parameter corresponding to the required parameter in the superclass’ constructor. The new parameter can either be required:

class A {
  A({required int x});
}

class B extends A {
  B({required super.x});
}

or it can be optional:

class A {
  A({required int x});
}

class B extends A {
  B({super.x = 0});
}

If you can’t add a parameter to the constructor in the subclass, then add an explicit super constructor invocation with the required argument:

class A {
  A(int x);
}

class B extends A {
  B() : super(0);
}

implicit_this_reference_in_initializer

The instance member ‘{0}’ can’t be accessed in an initializer.

Description

The analyzer produces this diagnostic when it finds a reference to an instance member in a constructor’s initializer list.

Example

The following code produces this diagnostic because defaultX is an instance member:

class C {
  int x;

  C() : x = defaultX;

  int get defaultX => 0;
}

Common fixes

If the member can be made static, then do so:

class C {
  int x;

  C() : x = defaultX;

  static int get defaultX => 0;
}

If not, then replace the reference in the initializer with a different expression that doesn’t use an instance member:

class C {
  int x;

  C() : x = 0;

  int get defaultX => 0;
}

import_deferred_library_with_load_function

The imported library defines a top-level function named ‘loadLibrary’ that is hidden by deferring this library.

Description

The analyzer produces this diagnostic when a library that declares a function named loadLibrary is imported using a deferred import. A deferred import introduces an implicit function named loadLibrary. This function is used to load the contents of the deferred library, and the implicit function hides the explicit declaration in the deferred library.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines a function named loadLibrary:

void loadLibrary(Library library) {}

class Library {}

The following code produces this diagnostic because the implicit declaration of a.loadLibrary is hiding the explicit declaration of loadLibrary in a.dart:

import 'a.dart' deferred as a;

void f() {
  a.Library();
}

Common fixes

If the imported library isn’t required to be deferred, then remove the keyword deferred:

import 'a.dart' as a;

void f() {
  a.Library();
}

If the imported library is required to be deferred and you need to reference the imported function, then rename the function in the imported library:

void populateLibrary(Library library) {}

class Library {}

If the imported library is required to be deferred and you don’t need to reference the imported function, then add a hide clause:

import 'a.dart' deferred as a hide loadLibrary;

void f() {
  a.Library();
}

import_internal_library

The library ‘{0}’ is internal and can’t be imported.

Description

The analyzer produces this diagnostic when it finds an import whose dart: URI references an internal library.

Example

The following code produces this diagnostic because _interceptors is an internal library:

import 'dart:_interceptors';

Common fixes

Remove the import directive.

import_of_legacy_library_into_null_safe

The library ‘{0}’ is legacy, and shouldn’t be imported into a null safe library.

Description

The analyzer produces this diagnostic when a library that is null safe imports a library that isn’t null safe.

Example

Given a file a.dart that contains the following:

// @dart = 2.9

class A {}

The following code produces this diagnostic because a library that null safe is importing a library that isn’t null safe:

import 'a.dart';

A? f() => null;

Common fixes

If you can migrate the imported library to be null safe, then migrate it and update or remove the migrated library’s language version.

If you can’t migrate the imported library, then the importing library needs to have a language version that is before 2.12, when null safety was enabled by default.

import_of_non_library

The imported library ‘{0}’ can’t have a part-of directive.

Description

The analyzer produces this diagnostic when a part file is imported into a library.

Example

Given a part file named part.dart containing the following:

part of lib;

The following code produces this diagnostic because imported files can’t have a part-of directive:

library lib;

import 'part.dart';

Common fixes

Import the library that contains the part file rather than the part file itself.

inconsistent_inheritance

Superinterfaces don’t have a valid override for ‘{0}’: {1}.

Description

The analyzer produces this diagnostic when a class inherits two or more conflicting signatures for a member and doesn’t provide an implementation that satisfies all the inherited signatures.

Example

The following code produces this diagnostic because C is inheriting the declaration of m from A, and that implementation isn’t consistent with the signature of m that’s inherited from B:

class A {
  void m({int a}) {}
}

class B {
  void m({int b}) {}
}

class C extends A implements B {
}

Common fixes

Add an implementation of the method that satisfies all the inherited signatures:

class A {
  void m({int a}) {}
}

class B {
  void m({int b}) {}
}

class C extends A implements B {
  void m({int a, int b}) {}
}

inconsistent_language_version_override

Parts must have exactly the same language version override as the library.

Description

The analyzer produces this diagnostic when a part file has a language version override comment that specifies a different language version than the one being used for the library to which the part belongs.

Example

Given a part file named part.dart that contains the following:

// @dart = 2.14
part of 'test.dart';

The following code produces this diagnostic because the parts of a library must have the same language version as the defining compilation unit:

// @dart = 2.15
part 'part.dart';

Common fixes

Remove the language version override from the part file, so that it implicitly uses the same version as the defining compilation unit:

part of 'test.dart';

If necessary, either adjust the language version override in the defining compilation unit to be appropriate for the code in the part, or migrate the code in the part file to be consistent with the new language version.

inconsistent_pattern_variable_logical_or

The variable ‘{0}’ has a different type and/or finality in this branch of the logical-or pattern.

Description

The analyzer produces this diagnostic when a pattern variable that is declared on all branches of a logical-or pattern doesn’t have the same type on every branch. It is also produced when the variable has a different finality on different branches. A pattern variable declared on multiple branches of a logical-or pattern is required to have the same type and finality in each branch, so that the type and finality of the variable can be known in code that’s guarded by the logical-or pattern.

Examples

The following code produces this diagnostic because the variable a is defined to be an int on one branch and a double on the other:

void f(Object? x) {
  if (x case (int a) || (double a)) {
    print(a);
  }
}

The following code produces this diagnostic because the variable a is final in the first branch and isn’t final in the second branch:

void f(Object? x) {
  if (x case (final int a) || (int a)) {
    print(a);
  }
}

Common fixes

If the finality of the variable is different, decide whether it should be final or not final and make the cases consistent:

void f(Object? x) {
  if (x case (int a) || (int a)) {
    print(a);
  }
}

If the type of the variable is different and the type isn’t critical to the condition being matched, then ensure that the variable has the same type on both branches:

void f(Object? x) {
  if (x case (num a) || (num a)) {
    print(a);
  }
}

If the type of the variable is different and the type is critical to the condition being matched, then consider breaking the condition into multiple if statements or case clauses:

void f(Object? x) {
  if (x case int a) {
    print(a);
  } else if (x case double a) {
    print(a);
  }
}

initializer_for_non_existent_field

‘{0}’ isn’t a field in the enclosing class.

Description

The analyzer produces this diagnostic when a constructor initializes a field that isn’t declared in the class containing the constructor. Constructors can’t initialize fields that aren’t declared and fields that are inherited from superclasses.

Example

The following code produces this diagnostic because the initializer is initializing x, but x isn’t a field in the class:

class C {
  int y;

  C() : x = 0;
}

Common fixes

If a different field should be initialized, then change the name to the name of the field:

class C {
  int y;

  C() : y = 0;
}

If the field must be declared, then add a declaration:

class C {
  int x;
  int y;

  C() : x = 0;
}

initializer_for_static_field

‘{0}’ is a static field in the enclosing class. Fields initialized in a constructor can’t be static.

Description

The analyzer produces this diagnostic when a static field is initialized in a constructor using either an initializing formal parameter or an assignment in the initializer list.

Example

The following code produces this diagnostic because the static field a is being initialized by the initializing formal parameter this.a:

class C {
  static int? a;
  C(this.a);
}

Common fixes

If the field should be an instance field, then remove the keyword static:

class C {
  int? a;
  C(this.a);
}

If you intended to initialize an instance field and typed the wrong name, then correct the name of the field being initialized:

class C {
  static int? a;
  int? b;
  C(this.b);
}

If you really want to initialize the static field, then move the initialization into the constructor body:

class C {
  static int? a;
  C(int? c) {
    a = c;
  }
}

initializing_formal_for_non_existent_field

‘{0}’ isn’t a field in the enclosing class.

Description

The analyzer produces this diagnostic when an initializing formal parameter is found in a constructor in a class that doesn’t declare the field being initialized. Constructors can’t initialize fields that aren’t declared and fields that are inherited from superclasses.

Example

The following code produces this diagnostic because the field x isn’t defined:

class C {
  int y;

  C(this.x);
}

Common fixes

If the field name was wrong, then change it to the name of an existing field:

class C {
  int y;

  C(this.y);
}

If the field name is correct but hasn’t yet been defined, then declare the field:

class C {
  int x;
  int y;

  C(this.x);
}

If the parameter is needed but shouldn’t initialize a field, then convert it to a normal parameter and use it:

class C {
  int y;

  C(int x) : y = x * 2;
}

If the parameter isn’t needed, then remove it:

class C {
  int y;

  C();
}

instance_access_to_static_member

The static {1} ‘{0}’ can’t be accessed through an instance.

Description

The analyzer produces this diagnostic when an access operator is used to access a static member through an instance of the class.

Example

The following code produces this diagnostic because zero is a static field, but it’s being accessed as if it were an instance field:

void f(C c) {
  c.zero;
}

class C {
  static int zero = 0;
}

Common fixes

Use the class to access the static member:

void f(C c) {
  C.zero;
}

class C {
  static int zero = 0;
}

instance_member_access_from_factory

Instance members can’t be accessed from a factory constructor.

Description

The analyzer produces this diagnostic when a factory constructor contains an unqualified reference to an instance member. In a generative constructor, the instance of the class is created and initialized before the body of the constructor is executed, so the instance can be bound to this and accessed just like it would be in an instance method. But, in a factory constructor, the instance isn’t created before executing the body, so this can’t be used to reference it.

Example

The following code produces this diagnostic because x isn’t in scope in the factory constructor:

class C {
  int x;
  factory C() {
    return C._(x);
  }
  C._(this.x);
}

Common fixes

Rewrite the code so that it doesn’t reference the instance member:

class C {
  int x;
  factory C() {
    return C._(0);
  }
  C._(this.x);
}

instance_member_access_from_static

Instance members can’t be accessed from a static method.

Description

The analyzer produces this diagnostic when a static method contains an unqualified reference to an instance member.

Example

The following code produces this diagnostic because the instance field x is being referenced in a static method:

class C {
  int x;

  static int m() {
    return x;
  }
}

Common fixes

If the method must reference the instance member, then it can’t be static, so remove the keyword:

class C {
  int x;

  int m() {
    return x;
  }
}

If the method can’t be made an instance method, then add a parameter so that an instance of the class can be passed in:

class C {
  int x;

  static int m(C c) {
    return c.x;
  }
}

instantiate_abstract_class

Abstract classes can’t be instantiated.

Description

The analyzer produces this diagnostic when it finds a constructor invocation and the constructor is declared in an abstract class. Even though you can’t create an instance of an abstract class, abstract classes can declare constructors that can be invoked by subclasses.

Example

The following code produces this diagnostic because C is an abstract class:

abstract class C {}

var c = new C();

Common fixes

If there’s a concrete subclass of the abstract class that can be used, then create an instance of the concrete subclass.

instantiate_enum

Enums can’t be instantiated.

Description

The analyzer produces this diagnostic when an enum is instantiated. It’s invalid to create an instance of an enum by invoking a constructor; only the instances named in the declaration of the enum can exist.

Example

The following code produces this diagnostic because the enum E is being instantiated:

// @dart = 2.16
enum E {a}

var e = E();

Common fixes

If you intend to use an instance of the enum, then reference one of the constants defined in the enum:

// @dart = 2.16
enum E {a}

var e = E.a;

If you intend to use an instance of a class, then use the name of that class in place of the name of the enum.

instantiate_type_alias_expands_to_type_parameter

Type aliases that expand to a type parameter can’t be instantiated.

Description

The analyzer produces this diagnostic when a constructor invocation is found where the type being instantiated is a type alias for one of the type parameters of the type alias. This isn’t allowed because the value of the type parameter is a type rather than a class.

Example

The following code produces this diagnostic because it creates an instance of A, even though A is a type alias that is defined to be equivalent to a type parameter:

typedef A<T> = T;

void f() {
  const A<int>();
}

Common fixes

Use either a class name or a type alias defined to be a class, rather than a type alias defined to be a type parameter:

typedef A<T> = C<T>;

void f() {
  const A<int>();
}

class C<T> {
  const C();
}

integer_literal_imprecise_as_double

The integer literal is being used as a double, but can’t be represented as a 64-bit double without overflow or loss of precision: ‘{0}’.

Description

The analyzer produces this diagnostic when an integer literal is being implicitly converted to a double, but can’t be represented as a 64-bit double without overflow or loss of precision. Integer literals are implicitly converted to a double if the context requires the type double.

Example

The following code produces this diagnostic because the integer value 9223372036854775807 can’t be represented exactly as a double:

double x = 9223372036854775807;

Common fixes

If you need to use the exact value, then use the class BigInt to represent the value:

var x = BigInt.parse('9223372036854775807');

If you need to use a double, then change the value to one that can be represented exactly:

double x = 9223372036854775808;

integer_literal_out_of_range

The integer literal {0} can’t be represented in 64 bits.

Description

The analyzer produces this diagnostic when an integer literal has a value that is too large (positive) or too small (negative) to be represented in a 64-bit word.

Example

The following code produces this diagnostic because the value can’t be represented in 64 bits:

var x = 9223372036854775810;

Common fixes

If you need to represent the current value, then wrap it in an instance of the class BigInt:

var x = BigInt.parse('9223372036854775810');

invalid_annotation

Annotation must be either a const variable reference or const constructor invocation.

Description

The analyzer produces this diagnostic when an annotation is found that is using something that is neither a variable marked as const or the invocation of a const constructor.

Getters can’t be used as annotations.

Examples

The following code produces this diagnostic because the variable v isn’t a const variable:

var v = 0;

@v
void f() {
}

The following code produces this diagnostic because f isn’t a variable:

@f
void f() {
}

The following code produces this diagnostic because f isn’t a constructor:

@f()
void f() {
}

The following code produces this diagnostic because g is a getter:

@g
int get g => 0;

Common fixes

If the annotation is referencing a variable that isn’t a const constructor, add the keyword const to the variable’s declaration:

const v = 0;

@v
void f() {
}

If the annotation isn’t referencing a variable, then remove it:

int v = 0;

void f() {
}

invalid_annotation_constant_value_from_deferred_library

Constant values from a deferred library can’t be used in annotations.

Description

The analyzer produces this diagnostic when a constant defined in a library that is imported as a deferred library is referenced in the argument list of an annotation. Annotations are evaluated at compile time, and values from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

The following code produces this diagnostic because the constant pi is being referenced in the argument list of an annotation, even though the library that defines it is being imported as a deferred library:

import 'dart:math' deferred as math;

class C {
  const C(double d);
}

@C(math.pi)
void f () {}

Common fixes

If you need to reference the imported constant, then remove the deferred keyword:

import 'dart:math' as math;

class C {
  const C(double d);
}

@C(math.pi)
void f () {}

If the import is required to be deferred and there’s another constant that is appropriate, then use that constant in place of the constant from the deferred library.

invalid_annotation_from_deferred_library

Constant values from a deferred library can’t be used as annotations.

Description

The analyzer produces this diagnostic when a constant from a library that is imported using a deferred import is used as an annotation. Annotations are evaluated at compile time, and constants from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

The following code produces this diagnostic because the constant pi is being used as an annotation when the library dart:math is imported as deferred:

import 'dart:math' deferred as math;

@math.pi
void f() {}

Common fixes

If you need to reference the constant as an annotation, then remove the keyword deferred from the import:

import 'dart:math' as math;

@math.pi
void f() {}

If you can use a different constant as an annotation, then replace the annotation with a different constant:

@deprecated
void f() {}

invalid_annotation_target

The annotation ‘{0}’ can only be used on {1}.

Description

The analyzer produces this diagnostic when an annotation is applied to a kind of declaration that it doesn’t support.

Example

The following code produces this diagnostic because the optionalTypeArgs annotation isn’t defined to be valid for top-level variables:

import 'package:meta/meta.dart';

@optionalTypeArgs
int x = 0;

Common fixes

Remove the annotation from the declaration.

invalid_assignment

A value of type ‘{0}’ can’t be assigned to a variable of type ‘{1}’.

Description

The analyzer produces this diagnostic when the static type of an expression that is assigned to a variable isn’t assignable to the type of the variable.

Example

The following code produces this diagnostic because the type of the initializer (int) isn’t assignable to the type of the variable (String):

int i = 0;
String s = i;

Common fixes

If the value being assigned is always assignable at runtime, even though the static types don’t reflect that, then add an explicit cast.

Otherwise, change the value being assigned so that it has the expected type. In the previous example, this might look like:

int i = 0;
String s = i.toString();

If you can’t change the value, then change the type of the variable to be compatible with the type of the value being assigned:

int i = 0;
int s = i;

invalid_dependency

Publishable packages can’t have ‘{0}’ dependencies.

Description

The analyzer produces this diagnostic when a publishable package includes a package in the dependencies list of its pubspec.yaml file that isn’t a pub-hosted dependency.

To learn more about the different types of dependency sources, check out Package dependencies.

Example

The following code produces this diagnostic because the dependency on the package transmogrify isn’t a pub-hosted dependency.

name: example
dependencies:
  transmogrify:
    path: ../transmogrify

Common fixes

If you want to publish the package to pub.dev, then change the dependency to a hosted package that is published on pub.dev.

If the package isn’t intended to be published on pub.dev, then add a publish_to: none entry to its pubspec.yaml file to mark it as not intended to be published:

name: example
publish_to: none
dependencies:
  transmogrify:
    path: ../transmogrify

invalid_exception_value

The method {0} can’t have an exceptional return value (the second argument) when the return type of the function is either ‘void’, ‘Handle’ or ‘Pointer’.

Description

The analyzer produces this diagnostic when an invocation of the method Pointer.fromFunction or NativeCallable.isolateLocal has a second argument (the exceptional return value) and the type to be returned from the invocation is either void, Handle or Pointer.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because a second argument is provided when the return type of f is void:

import 'dart:ffi';

typedef T = Void Function(Int8);

void f(int i) {}

void g() {
  Pointer.fromFunction<T>(f, 42);
}

Common fixes

Remove the exception value:

import 'dart:ffi';

typedef T = Void Function(Int8);

void f(int i) {}

void g() {
  Pointer.fromFunction<T>(f);
}

invalid_export_of_internal_element

The member ‘{0}’ can’t be exported as a part of a package’s public API.

Description

The analyzer produces this diagnostic when a public library exports a declaration that is marked with the internal annotation.

Example

Given a file a.dart in the src directory that contains:

import 'package:meta/meta.dart';

@internal class One {}

The following code, when found in a public library produces this diagnostic because the export directive is exporting a name that is only intended to be used internally:

export 'src/a.dart';

Common fixes

If the export is needed, then add a hide clause to hide the internal names:

export 'src/a.dart' hide One;

If the export isn’t needed, then remove it.

invalid_export_of_internal_element_indirectly

The member ‘{0}’ can’t be exported as a part of a package’s public API, but is indirectly exported as part of the signature of ‘{1}’.

Description

The analyzer produces this diagnostic when a public library exports a top-level function with a return type or at least one parameter type that is marked with the internal annotation.

Example

Given a file a.dart in the src directory that contains the following:

import 'package:meta/meta.dart';

@internal
typedef IntFunction = int Function();

int f(IntFunction g) => g();

The following code produces this diagnostic because the function f has a parameter of type IntFunction, and IntFunction is only intended to be used internally:

export 'src/a.dart' show f;

Common fixes

If the function must be public, then make all the types in the function’s signature public types.

If the function doesn’t need to be exported, then stop exporting it, either by removing it from the show clause, adding it to the hide clause, or by removing the export.

invalid_extension_argument_count

Extension overrides must have exactly one argument: the value of ‘this’ in the extension method.

Description

The analyzer produces this diagnostic when an extension override doesn’t have exactly one argument. The argument is the expression used to compute the value of this within the extension method, so there must be one argument.

Examples

The following code produces this diagnostic because there are no arguments:

extension E on String {
  String join(String other) => '$this $other';
}

void f() {
  E().join('b');
}

And, the following code produces this diagnostic because there’s more than one argument:

extension E on String {
  String join(String other) => '$this $other';
}

void f() {
  E('a', 'b').join('c');
}

Common fixes

Provide one argument for the extension override:

extension E on String {
  String join(String other) => '$this $other';
}

void f() {
  E('a').join('b');
}

invalid_factory_method_decl

Factory method ‘{0}’ must have a return type.

Description

The analyzer produces this diagnostic when a method that is annotated with the factory annotation has a return type of void.

Example

The following code produces this diagnostic because the method createC is annotated with the factory annotation but doesn’t return any value:

import 'package:meta/meta.dart';

class Factory {
  @factory
  void createC() {}
}

class C {}

Common fixes

Change the return type to something other than void:

import 'package:meta/meta.dart';

class Factory {
  @factory
  C createC() => C();
}

class C {}

invalid_factory_method_impl

Factory method ‘{0}’ doesn’t return a newly allocated object.

Description

The analyzer produces this diagnostic when a method that is annotated with the factory annotation doesn’t return a newly allocated object.

Example

The following code produces this diagnostic because the method createC returns the value of a field rather than a newly created instance of C:

import 'package:meta/meta.dart';

class Factory {
  C c = C();

  @factory
  C createC() => c;
}

class C {}

Common fixes

Change the method to return a newly created instance of the return type:

import 'package:meta/meta.dart';

class Factory {
  @factory
  C createC() => C();
}

class C {}

invalid_factory_name_not_a_class

The name of a factory constructor must be the same as the name of the immediately enclosing class.

Description

The analyzer produces this diagnostic when the name of a factory constructor isn’t the same as the name of the surrounding class.

Example

The following code produces this diagnostic because the name of the factory constructor (A) isn’t the same as the surrounding class (C):

class A {}

class C {
  factory A() => throw 0;
}

Common fixes

If the factory returns an instance of the surrounding class, and you intend it to be an unnamed factory constructor, then rename the factory:

class A {}

class C {
  factory C() => throw 0;
}

If the factory returns an instance of the surrounding class, and you intend it to be a named factory constructor, then prefix the name of the factory constructor with the name of the surrounding class:

class A {}

class C {
  factory C.a() => throw 0;
}

If the factory returns an instance of a different class, then move the factory to that class:

class A {
  factory A() => throw 0;
}

class C {}

If the factory returns an instance of a different class, but you can’t modify that class or don’t want to move the factory, then convert it to be a static method:

class A {}

class C {
  static A a() => throw 0;
}

invalid_field_name

Record field names can’t be a dollar sign followed by an integer when the integer is the index of a positional field.

Record field names can’t be private.

Record field names can’t be the same as a member from ‘Object’.

Description

The analyzer produces this diagnostic when either a record literal or a record type annotation has a field whose name is invalid. The name is invalid if it is:

• private (starts with _)
• the same as one of the members defined on Object
• the same as the name of a positional field (an exception is made if the field is a positional field with the specified name)

Examples

The following code produces this diagnostic because the record literal has a field named toString, which is a method defined on Object:

var r = (a: 1, toString: 4);

The following code produces this diagnostic because the record type annotation has a field named hashCode, which is a getter defined on Object:

void f(({int a, int hashCode}) r) {}

The following code produces this diagnostic because the record literal has a private field named _a:

var r = (_a: 1, b: 2);

The following code produces this diagnostic because the record type annotation has a private field named _a:

void f(({int _a, int b}) r) {}

The following code produces this diagnostic because the record literal has a field named $1, which is also the name of a different positional parameter:

var r = (2, $1: 1);

The following code produces this diagnostic because the record type annotation has a field named $1, which is also the name of a different positional parameter:

void f((int, String, {int $1}) r) {}

Common fixes

Rename the field:

var r = (a: 1, d: 4);

invalid_field_type_in_struct

Fields in struct classes can’t have the type ‘{0}’. They can only be declared as ‘int’, ‘double’, ‘Array’, ‘Pointer’, or subtype of ‘Struct’ or ‘Union’.

Description

The analyzer produces this diagnostic when a field in a subclass of Struct has a type other than int, double, Array, Pointer, or subtype of Struct or Union.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field str has the type String, which isn’t one of the allowed types for fields in a subclass of Struct:

import 'dart:ffi';

final class C extends Struct {
  external String s;

  @Int32()
  external int i;
}

Common fixes

Use one of the allowed types for the field:

import 'dart:ffi';
import 'package:ffi/ffi.dart';

final class C extends Struct {
  external Pointer<Utf8> s;

  @Int32()
  external int i;
}

invalid_implementation_override

‘{1}.{0}’ (‘{2}’) isn’t a valid concrete implementation of ‘{3}.{0}’ (‘{4}’).

The setter ‘{1}.{0}’ (‘{2}’) isn’t a valid concrete implementation of ‘{3}.{0}’ (‘{4}’).

Description

The analyzer produces this diagnostic when all of the following are true:

• A class defines an abstract member.
• There is a concrete implementation of that member in a superclass.
• The concrete implementation isn’t a valid implementation of the abstract method.

The concrete implementation can be invalid because of incompatibilities in either the return type, the types of parameters, or the type variables.

Example

The following code produces this diagnostic because the method A.add has a parameter of type int, and the overriding method B.add has a corresponding parameter of type num:

class A {
  int add(int a) => a;
}
class B extends A {
  int add(num a);
}

This is a problem because in an invocation of B.add like the following:

void f(B b) {
  b.add(3.4);
}

B.add is expecting to be able to take, for example, a double, but when the method A.add is executed (because it’s the only concrete implementation of add), a runtime exception will be thrown because a double can’t be assigned to a parameter of type int.

Common fixes

If the method in the subclass can conform to the implementation in the superclass, then change the declaration in the subclass (or remove it if it’s the same):

class A {
  int add(int a) => a;
}
class B	extends A {
  int add(int a);
}

If the method in the superclass can be generalized to be a valid implementation of the method in the subclass, then change the superclass method:

class A {
  int add(num a) => a.floor();
}
class B	extends A {
  int add(num a);
}

If neither the method in the superclass nor the method in the subclass can be changed, then provide a concrete implementation of the method in the subclass:

class A {
  int add(int a) => a;
}
class B	extends A {
  int add(num a) => a.floor();
}

invalid_inline_function_type

Inline function types can’t be used for parameters in a generic function type.

Description

The analyzer produces this diagnostic when a generic function type has a function-valued parameter that is written using the older inline function type syntax.

Example

The following code produces this diagnostic because the parameter f, in the generic function type used to define F, uses the inline function type syntax:

typedef F = int Function(int f(String s));

Common fixes

Use the generic function syntax for the parameter’s type:

typedef F = int Function(int Function(String));

invalid_internal_annotation

Only public elements in a package’s private API can be annotated as being internal.

Description

The analyzer produces this diagnostic when a declaration is annotated with the internal annotation and that declaration is either in a public library or has a private name.

Example

The following code, when in a public library, produces this diagnostic because the internal annotation can’t be applied to declarations in a public library:

import 'package:meta/meta.dart';

@internal
class C {}

The following code, whether in a public or internal library, produces this diagnostic because the internal annotation can’t be applied to declarations with private names:

import 'package:meta/meta.dart';

@internal
class _C {}

void f(_C c) {}

Common fixes

If the declaration has a private name, then remove the annotation:

class _C {}

void f(_C c) {}

If the declaration has a public name and is intended to be internal to the package, then move the annotated declaration into an internal library (in other words, a library inside the src directory).

Otherwise, remove the use of the annotation:

class C {}

invalid_language_version_override

The Dart language version override comment can’t be followed by any non-whitespace characters.

The Dart language version override comment must be specified with a version number, like ‘2.0’, after the ‘=’ character.

The Dart language version override comment must be specified with an ‘=’ character.

The Dart language version override comment must be specified with exactly two slashes.

The Dart language version override comment must be specified with the word ‘dart’ in all lower case.

The Dart language version override number can’t be prefixed with a letter.

The Dart language version override number must begin with ‘@dart’.

The language version override can’t specify a version greater than the latest known language version: {0}.{1}.

The language version override must be specified before any declaration or directive.

Description

The analyzer produces this diagnostic when a comment that appears to be an attempt to specify a language version override doesn’t conform to the requirements for such a comment. For more information, see Per-library language version selection.

Example

The following code produces this diagnostic because the word dart must be lowercase in such a comment and because there’s no equal sign between the word dart and the version number:

// @Dart 2.13

Common fixes

If the comment is intended to be a language version override, then change the comment to follow the correct format:

// @dart = 2.13

invalid_literal_annotation

Only const constructors can have the @literal annotation.

Description

The analyzer produces this diagnostic when the literal annotation is applied to anything other than a const constructor.

Examples

The following code produces this diagnostic because the constructor isn’t a const constructor:

import 'package:meta/meta.dart';

class C {
  @literal
  C();
}

The following code produces this diagnostic because x isn’t a constructor:

import 'package:meta/meta.dart';

@literal
var x;

Common fixes

If the annotation is on a constructor and the constructor should always be invoked with const, when possible, then mark the constructor with the const keyword:

import 'package:meta/meta.dart';

class C {
  @literal
  const C();
}

If the constructor can’t be marked as const, then remove the annotation.

If the annotation is on anything other than a constructor, then remove the annotation:

var x;

invalid_modifier_on_constructor

The modifier ‘{0}’ can’t be applied to the body of a constructor.

Description

The analyzer produces this diagnostic when the body of a constructor is prefixed by one of the following modifiers: async, async*, or sync*. Constructor bodies must be synchronous.

Example

The following code produces this diagnostic because the body of the constructor for C is marked as being async:

class C {
  C() async {}
}

Common fixes

If the constructor can be synchronous, then remove the modifier:

class C {
  C();
}

If the constructor can’t be synchronous, then use a static method to create the instance instead:

class C {
  C();
  static Future<C> c() async {
    return C();
  }
}

invalid_modifier_on_setter

Setters can’t use ‘async’, ‘async*’, or ‘sync*‘.

Description

The analyzer produces this diagnostic when the body of a setter is prefixed by one of the following modifiers: async, async*, or sync*. Setter bodies must be synchronous.

Example

The following code produces this diagnostic because the body of the setter x is marked as being async:

class C {
  set x(int i) async {}
}

Common fixes

If the setter can be synchronous, then remove the modifier:

class C {
  set x(int i) {}
}

If the setter can’t be synchronous, then use a method to set the value instead:

class C {
  void x(int i) async {}
}

invalid_non_virtual_annotation

The annotation ‘@nonVirtual’ can only be applied to a concrete instance member.

Description

The analyzer produces this diagnostic when the nonVirtual annotation is found on a declaration other than a member of a class, mixin, or enum, or if the member isn’t a concrete instance member.

Examples

The following code produces this diagnostic because the annotation is on a class declaration rather than a member inside the class:

import 'package:meta/meta.dart';

@nonVirtual
class C {}

The following code produces this diagnostic because the method m is an abstract method:

import 'package:meta/meta.dart';

abstract class C {
  @nonVirtual
  void m();
}

The following code produces this diagnostic because the method m is a static method:

import 'package:meta/meta.dart';

abstract class C {
  @nonVirtual
  static void m() {}
}

Common fixes

If the declaration isn’t a member of a class, mixin, or enum, then remove the annotation:

class C {}

If the member is intended to be a concrete instance member, then make it so:

import 'package:meta/meta.dart';

abstract class C {
  @nonVirtual
  void m() {}
}

If the member is not intended to be a concrete instance member, then remove the annotation:

abstract class C {
  static void m() {}
}

invalid_null_aware_operator

The receiver can’t be null because of short-circuiting, so the null-aware operator ‘{0}’ can’t be used.

The receiver can’t be null, so the null-aware operator ‘{0}’ is unnecessary.

Description

The analyzer produces this diagnostic when a null-aware operator (?., ?.., ?[, ?..[, or ...?) is used on a receiver that’s known to be non-nullable.

Examples

The following code produces this diagnostic because s can’t be null:

int? getLength(String s) {
  return s?.length;
}

The following code produces this diagnostic because a can’t be null:

var a = [];
var b = [...?a];

The following code produces this diagnostic because s?.length can’t return null:

void f(String? s) {
  s?.length?.isEven;
}

The reason s?.length can’t return null is because the null-aware operator following s short-circuits the evaluation of both length and isEven if s is null. In other words, if s is null, then neither length nor isEven will be invoked, and if s is non-null, then length can’t return a null value. Either way, isEven can’t be invoked on a null value, so the null-aware operator isn’t necessary. See Understanding null safety for more details.

The following code produces this diagnostic because s can’t be null.

void f(Object? o) {
  var s = o as String;
  s?.length;
}

The reason s can’t be null, despite the fact that o can be null, is because of the cast to String, which is a non-nullable type. If o ever has the value null, the cast will fail and the invocation of length will not happen.

Common fixes

Replace the null-aware operator with a non-null-aware equivalent; for example, change ?. to .:

int getLength(String s) {
  return s.length;
}

(Note that the return type was also changed to be non-nullable, which might not be appropriate in some cases.)

invalid_override

‘{1}.{0}’ (‘{2}’) isn’t a valid override of ‘{3}.{0}’ (‘{4}’).

The setter ‘{1}.{0}’ (‘{2}’) isn’t a valid override of ‘{3}.{0}’ (‘{4}’).

Description

The analyzer produces this diagnostic when a member of a class is found that overrides a member from a supertype and the override isn’t valid. An override is valid if all of these are true:

• It allows all of the arguments allowed by the overridden member.
• It doesn’t require any arguments that aren’t required by the overridden member.
• The type of every parameter of the overridden member is assignable to the corresponding parameter of the override.
• The return type of the override is assignable to the return type of the overridden member.

Example

The following code produces this diagnostic because the type of the parameter s (String) isn’t assignable to the type of the parameter i (int):

class A {
  void m(int i) {}
}

class B extends A {
  void m(String s) {}
}

Common fixes

If the invalid method is intended to override the method from the superclass, then change it to conform:

class A {
  void m(int i) {}
}

class B extends A {
  void m(int i) {}
}

If it isn’t intended to override the method from the superclass, then rename it:

class A {
  void m(int i) {}
}

class B extends A {
  void m2(String s) {}
}

invalid_override_of_non_virtual_member

The member ‘{0}’ is declared non-virtual in ‘{1}’ and can’t be overridden in subclasses.

Description

The analyzer produces this diagnostic when a member of a class, mixin, or enum overrides a member that has the @nonVirtual annotation on it.

Example

The following code produces this diagnostic because the method m in B overrides the method m in A, and the method m in A is annotated with the @nonVirtual annotation:

import 'package:meta/meta.dart';

class A {
  @nonVirtual
  void m() {}
}

class B extends A {
  @override
  void m() {}
}

Common fixes

If the annotation on the method in the superclass is correct (the method in the superclass is not intended to be overridden), then remove or rename the overriding method:

import 'package:meta/meta.dart';

class A {
  @nonVirtual
  void m() {}
}

class B extends A {}

If the method in the superclass is intended to be overridden, then remove the @nonVirtual annotation:

class A {
  void m() {}
}

class B extends A {
  @override
  void m() {}
}

invalid_pattern_variable_in_shared_case_scope

The variable ‘{0}’ doesn’t have the same type and/or finality in all cases that share this body.

The variable ‘{0}’ is available in some, but not all cases that share this body.

The variable ‘{0}’ is not available because there is a label or ‘default’ case.

Description

The analyzer produces this diagnostic when multiple case clauses in a switch statement share a body, and at least one of them declares a variable that is referenced in the shared statements, but the variable is either not declared in all of the case clauses or it is declared in inconsistent ways.

If the variable isn’t declared in all of the case clauses, then it won’t have a value if one of the clauses that doesn’t declare the variable is the one that matches and executes the body. This includes the situation where one of the case clauses is the default clause.

If the variable is declared in inconsistent ways, either being final in some cases and not final in others or having a different type in different cases, then the semantics of what the type or finality of the variable should be are not defined.

Examples

The following code produces this diagnostic because the variable a is only declared in one of the case clauses, and won’t have a value if the second clause is the one that matched x:

void f(Object? x) {
  switch (x) {
    case int a when a > 0:
    case 0:
      a;
  }
}

The following code produces this diagnostic because the variable a isn’t declared in the default clause, and won’t have a value if the body is executed because none of the other clauses matched x:

void f(Object? x) {
  switch (x) {
    case int a when a > 0:
    default:
      a;
  }
}

The following code produces this diagnostic because the variable a won’t have a value if the body is executed because a different group of cases caused control to continue at the label:

void f(Object? x) {
  switch (x) {
    someLabel:
    case int a when a > 0:
      a;
    case int b when b < 0:
      continue someLabel;
  }
}

The following code produces this diagnostic because the variable a, while being assigned in all of the case clauses, doesn’t have then same type associated with it in every clause:

void f(Object? x) {
  switch (x) {
    case int a when a < 0:
    case num a when a > 0:
      a;
  }
}

The following code produces this diagnostic because the variable a is final in the first case clause and isn’t final in the second case clause:

void f(Object? x) {
  switch (x) {
    case final int a when a < 0:
    case int a when a > 0:
      a;
  }
}

Common fixes

If the variable isn’t declared in all of the cases, and you need to reference it in the statements, then declare it in the other cases:

void f(Object? x) {
  switch (x) {
    case int a when a > 0:
    case int a when a == 0:
      a;
  }
}

If the variable isn’t declared in all of the cases, and you don’t need to reference it in the statements, then remove the references to it and remove the declarations from the other cases:

void f(int x) {
  switch (x) {
    case > 0:
    case 0:
  }
}

If the type of the variable is different, decide the type the variable should have and make the cases consistent:

void f(Object? x) {
  switch (x) {
    case num a when a < 0:
    case num a when a > 0:
      a;
  }
}

If the finality of the variable is different, decide whether it should be final or not final and make the cases consistent:

void f(Object? x) {
  switch (x) {
    case final int a when a < 0:
    case final int a when a > 0:
      a;
  }
}

invalid_platforms_field

The ‘platforms’ field must be a map with platforms as keys.

Description

The analyzer produces this diagnostic when a top-level platforms field is specified, but its value is not a map with keys. To learn more about specifying your package’s supported platforms, check out the documentation on platform declarations.

Example

The following pubspec.yaml produces this diagnostic because platforms should be a map.

name: example
platforms:
  - android
  - web
  - ios

Common fixes

If you can rely on automatic platform detection, then omit the top-level platforms field.

name: example

If you need to manually specify the list of supported platforms, then write the platforms field as a map with platform names as keys.

name: example
platforms:
  android:
  web:
  ios:

invalid_reference_to_generative_enum_constructor

Generative enum constructors can only be used as targets of redirection.

Description

The analyzer produces this diagnostic when a generative constructor defined on an enum is used anywhere other than to create one of the enum constants or as the target of a redirection from another constructor in the same enum.

Example

The following code produces this diagnostic because the constructor for E is being used to create an instance in the function f:

enum E {
  a(0);

  const E(int x);
}

E f() => const E(2);

Common fixes

If there’s an enum constant with the same value, or if you add such a constant, then reference the constant directly:

enum E {
  a(0), b(2);

  const E(int x);
}

E f() => E.b;

If you need to use a constructor invocation, then use a factory constructor:

enum E {
  a(0);

  const E(int x);

  factory E.c(int x) => a;
}

E f() => E.c(2);

invalid_reference_to_this

Invalid reference to ‘this’ expression.

Description

The analyzer produces this diagnostic when this is used outside of an instance method or a generative constructor. The reserved word this is only defined in the context of an instance method, a generative constructor, or the initializer of a late instance field declaration.

Example

The following code produces this diagnostic because v is a top-level variable:

C f() => this;

class C {}

Common fixes

Use a variable of the appropriate type in place of this, declaring it if necessary:

C f(C c) => c;

class C {}

invalid_return_type_for_catch_error

A value of type ‘{0}’ can’t be returned by the ‘onError’ handler because it must be assignable to ‘{1}’.

The return type ‘{0}’ isn’t assignable to ‘{1}’, as required by ‘Future.catchError’.

Description

The analyzer produces this diagnostic when an invocation of Future.catchError has an argument whose return type isn’t compatible with the type returned by the instance of Future. At runtime, the method catchError attempts to return the value from the callback as the result of the future, which results in another exception being thrown.

Examples

The following code produces this diagnostic because future is declared to return an int while callback is declared to return a String, and String isn’t a subtype of int:

void f(Future<int> future, String Function(dynamic, StackTrace) callback) {
  future.catchError(callback);
}

The following code produces this diagnostic because the closure being passed to catchError returns an int while future is declared to return a String:

void f(Future<String> future) {
  future.catchError((error, stackTrace) => 3);
}

Common fixes

If the instance of Future is declared correctly, then change the callback to match:

void f(Future<int> future, int Function(dynamic, StackTrace) callback) {
  future.catchError(callback);
}

If the declaration of the instance of Future is wrong, then change it to match the callback:

void f(Future<String> future, String Function(dynamic, StackTrace) callback) {
  future.catchError(callback);
}

invalid_sealed_annotation

The annotation ‘@sealed’ can only be applied to classes.

Description

The analyzer produces this diagnostic when a declaration other than a class declaration has the @sealed annotation on it.

Example

The following code produces this diagnostic because the @sealed annotation is on a method declaration:

import 'package:meta/meta.dart';

class A {
  @sealed
  void m() {}
}

Common fixes

Remove the annotation:

class A {
  void m() {}
}

invalid_super_formal_parameter_location

Super parameters can only be used in non-redirecting generative constructors.

Description

The analyzer produces this diagnostic when a super parameter is used anywhere other than a non-redirecting generative constructor.

Examples

The following code produces this diagnostic because the super parameter x is in a redirecting generative constructor:

class A {
  A(int x);
}

class B extends A {
  B.b(super.x) : this._();
  B._() : super(0);
}

The following code produces this diagnostic because the super parameter x isn’t in a generative constructor:

class A {
  A(int x);
}

class C extends A {
  factory C.c(super.x) => C._();
  C._() : super(0);
}

The following code produces this diagnostic because the super parameter x is in a method:

class A {
  A(int x);
}

class D extends A {
  D() : super(0);

  void m(super.x) {}
}

Common fixes

If the function containing the super parameter can be changed to be a non-redirecting generative constructor, then do so:

class A {
  A(int x);
}

class B extends A {
  B.b(super.x);
}

If the function containing the super parameter can’t be changed to be a non-redirecting generative constructor, then remove the super:

class A {
  A(int x);
}

class D extends A {
  D() : super(0);

  void m(int x) {}
}

invalid_type_argument_in_const_literal

Constant list literals can’t include a type parameter as a type argument, such as ‘{0}’.

Constant map literals can’t include a type parameter as a type argument, such as ‘{0}’.

Constant set literals can’t include a type parameter as a type argument, such as ‘{0}’.

Description

The analyzer produces this diagnostic when a type parameter is used as a type argument in a list, map, or set literal that is prefixed by const. This isn’t allowed because the value of the type parameter (the actual type that will be used at runtime) can’t be known at compile time.

Examples

The following code produces this diagnostic because the type parameter T is being used as a type argument when creating a constant list:

List<T> newList<T>() => const <T>[];

The following code produces this diagnostic because the type parameter T is being used as a type argument when creating a constant map:

Map<String, T> newSet<T>() => const <String, T>{};

The following code produces this diagnostic because the type parameter T is being used as a type argument when creating a constant set:

Set<T> newSet<T>() => const <T>{};

Common fixes

If the type that will be used for the type parameter can be known at compile time, then remove the type parameter:

List<int> newList() => const <int>[];

If the type that will be used for the type parameter can’t be known until runtime, then remove the keyword const:

List<T> newList<T>() => <T>[];

invalid_uri

Invalid URI syntax: ‘{0}’.

Description

The analyzer produces this diagnostic when a URI in a directive doesn’t conform to the syntax of a valid URI.

Example

The following code produces this diagnostic because '#' isn’t a valid URI:

import '#';

Common fixes

Replace the invalid URI with a valid URI.

invalid_use_of_covariant_in_extension

Can’t have modifier ‘{0}’ in an extension.

Description

The analyzer produces this diagnostic when a member declared inside an extension uses the keyword covariant in the declaration of a parameter. Extensions aren’t classes and don’t have subclasses, so the keyword serves no purpose.

Example

The following code produces this diagnostic because i is marked as being covariant:

extension E on String {
  void a(covariant int i) {}
}

Common fixes

Remove the covariant keyword:

extension E on String {
  void a(int i) {}
}

invalid_use_of_internal_member

The member ‘{0}’ can only be used within its package.

Description

The analyzer produces this diagnostic when a reference to a declaration that is annotated with the internal annotation is found outside the package containing the declaration.

Example

Given a package p that defines a library containing a declaration marked with the internal annotation:

import 'package:meta/meta.dart';

@internal
class C {}

The following code produces this diagnostic because it’s referencing the class C, which isn’t intended to be used outside the package p:

import 'package:p/src/p.dart';

void f(C c) {}

Common fixes

Remove the reference to the internal declaration.

invalid_use_of_null_value

An expression whose value is always ‘null’ can’t be dereferenced.

Description

The analyzer produces this diagnostic when an expression whose value will always be null is dereferenced.

Example

The following code produces this diagnostic because x will always be null:

int f(Null x) {
  return x.length;
}

Common fixes

If the value is allowed to be something other than null, then change the type of the expression:

int f(String? x) {
  return x!.length;
}

invalid_use_of_type_outside_library

The class ‘{0}’ can’t be extended outside of its library because it’s a final class.

The class ‘{0}’ can’t be extended outside of its library because it’s an interface class.

The class ‘{0}’ can’t be extended, implemented, or mixed in outside of its library because it’s a sealed class.

The class ‘{0}’ can’t be implemented outside of its library because it’s a base class.

The class ‘{0}’ can’t be implemented outside of its library because it’s a final class.

The class ‘{0}’ can’t be used as a mixin superclass constraint outside of its library because it’s a final class.

The mixin ‘{0}’ can’t be implemented outside of its library because it’s a base mixin.

Description

The analyzer produces this diagnostic when an extends, implements, with, or on clause uses a class or mixin in a way that isn’t allowed given the modifiers on that class or mixin’s declaration.

The message specifies how the declaration is being used and why it isn’t allowed.

Example

Given a file a.dart that defines a base class A:

base class A {}

The following code produces this diagnostic because the class B implements the class A, but the base modifier prevents A from being implemented outside of the library where it’s defined:

import 'a.dart';

final class B implements A {}

Common fixes

Use of this type is restricted outside of its declaring library. If a different, unrestricted type is available that can provide similar functionality, then replace the type:

class B implements C {}
class C {}

If there isn’t a different type that would be appropriate, then remove the type, and possibly the whole clause:

class B {}

invalid_use_of_visible_for_overriding_member

The member ‘{0}’ can only be used for overriding.

Description

The analyzer produces this diagnostic when an instance member that is annotated with visibleForOverriding is referenced outside the library in which it’s declared for any reason other than to override it.

Example

Given a file a.dart containing the following declaration:

import 'package:meta/meta.dart';

class A {
  @visibleForOverriding
  void a() {}
}

The following code produces this diagnostic because the method m is being invoked even though the only reason it’s public is to allow it to be overridden:

import 'a.dart';

class B extends A {
  void b() {
    a();
  }
}

Common fixes

Remove the invalid use of the member.

invalid_use_of_visible_for_testing_member

The member ‘{0}’ can only be used within ‘{1}’ or a test.

Description

The analyzer produces this diagnostic when a member annotated with @visibleForTesting is referenced anywhere other than the library in which it is declared or in a library in the test directory.

Example

Given a file c.dart that contains the following:

import 'package:meta/meta.dart';

class C {
  @visibleForTesting
  void m() {}
}

The following code, when not inside the test directory, produces this diagnostic because the method m is marked as being visible only for tests:

import 'c.dart';

void f(C c) {
  c.m();
}

Common fixes

If the annotated member should not be referenced outside of tests, then remove the reference:

import 'c.dart';

void f(C c) {}

If it’s OK to reference the annotated member outside of tests, then remove the annotation:

class C {
  void m() {}
}

invalid_visibility_annotation

The member ‘{0}’ is annotated with ‘{1}’, but this annotation is only meaningful on declarations of public members.

Description

The analyzer produces this diagnostic when either the visibleForTemplate or visibleForTesting annotation is applied to a non-public declaration.

Example

The following code produces this diagnostic:

import 'package:meta/meta.dart';

@visibleForTesting
void _someFunction() {}

void f() => _someFunction();

Common fixes

If the declaration doesn’t need to be used by test code, then remove the annotation:

void _someFunction() {}

void f() => _someFunction();

If it does, then make it public:

import 'package:meta/meta.dart';

@visibleForTesting
void someFunction() {}

void f() => someFunction();

invalid_visible_for_overriding_annotation

The annotation ‘visibleForOverriding’ can only be applied to a public instance member that can be overridden.

Description

The analyzer produces this diagnostic when anything other than a public instance member of a class is annotated with visibleForOverriding. Because only public instance members can be overridden outside the defining library, there’s no value to annotating any other declarations.

Example

The following code produces this diagnostic because the annotation is on a class, and classes can’t be overridden:

import 'package:meta/meta.dart';

@visibleForOverriding
class C {}

Common fixes

Remove the annotation:

class C {}

invalid_visible_outside_template_annotation

The annotation ‘visibleOutsideTemplate’ can only be applied to a member of a class, enum, or mixin that is annotated with ‘visibleForTemplate’.

Description

The analyzer produces this diagnostic when the @visibleOutsideTemplate annotation is used incorrectly. This annotation is only meant to annotate members of a class, enum, or mixin that has the @visibleForTemplate annotation, to opt those members out of the visibility restrictions that @visibleForTemplate imposes.

Examples

The following code produces this diagnostic because there is no @visibleForTemplate annotation at the class level:

import 'package:angular_meta/angular_meta.dart';

class C {
  @visibleOutsideTemplate
  int m() {
    return 1;
  }
}

The following code produces this diagnostic because the annotation is on a class declaration, not a member of a class, enum, or mixin:

import 'package:angular_meta/angular_meta.dart';

@visibleOutsideTemplate
class C {}

Common fixes

If the class is only visible so that templates can reference it, then add the @visibleForTemplate annotation to the class:

import 'package:angular_meta/angular_meta.dart';

@visibleForTemplate
class C {
  @visibleOutsideTemplate
  int m() {
    return 1;
  }
}

If the @visibleOutsideTemplate annotation is on anything other than a member of a class, enum, or mixin with the @visibleForTemplate annotation, remove the annotation:

class C {}

invocation_of_extension_without_call

The extension ‘{0}’ doesn’t define a ‘call’ method so the override can’t be used in an invocation.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a function but the extension doesn’t declare a call method.

Example

The following code produces this diagnostic because the extension E doesn’t define a call method:

extension E on String {}

void f() {
  E('')();
}

Common fixes

If the extension is intended to define a call method, then declare it:

extension E on String {
  int call() => 0;
}

void f() {
  E('')();
}

If the extended type defines a call method, then remove the extension override.

If the call method isn’t defined, then rewrite the code so that it doesn’t invoke the call method.

invocation_of_non_function

‘{0}’ isn’t a function.

Description

The analyzer produces this diagnostic when it finds a function invocation, but the name of the function being invoked is defined to be something other than a function.

Example

The following code produces this diagnostic because Binary is the name of a function type, not a function:

typedef Binary = int Function(int, int);

int f() {
  return Binary(1, 2);
}

Common fixes

Replace the name with the name of a function.

invocation_of_non_function_expression

The expression doesn’t evaluate to a function, so it can’t be invoked.

Description

The analyzer produces this diagnostic when a function invocation is found, but the name being referenced isn’t the name of a function, or when the expression computing the function doesn’t compute a function.

Examples

The following code produces this diagnostic because x isn’t a function:

int x = 0;

int f() => x;

var y = x();

The following code produces this diagnostic because f() doesn’t return a function:

int x = 0;

int f() => x;

var y = f()();

Common fixes

If you need to invoke a function, then replace the code before the argument list with the name of a function or with an expression that computes a function:

int x = 0;

int f() => x;

var y = f();

label_in_outer_scope

Can’t reference label ‘{0}’ declared in an outer method.

Description

The analyzer produces this diagnostic when a break or continue statement references a label that is declared in a method or function containing the function in which the break or continue statement appears. The break and continue statements can’t be used to transfer control outside the function that contains them.

Example

The following code produces this diagnostic because the label loop is declared outside the local function g:

void f() {
  loop:
  while (true) {
    void g() {
      break loop;
    }

    g();
  }
}

Common fixes

Try rewriting the code so that it isn’t necessary to transfer control outside the local function, possibly by inlining the local function:

void f() {
  loop:
  while (true) {
    break loop;
  }
}

If that isn’t possible, then try rewriting the local function so that a value returned by the function can be used to determine whether control is transferred:

void f() {
  loop:
  while (true) {
    bool g() {
      return true;
    }

    if (g()) {
      break loop;
    }
  }
}

label_undefined

Can’t reference an undefined label ‘{0}’.

Description

The analyzer produces this diagnostic when it finds a reference to a label that isn’t defined in the scope of the break or continue statement that is referencing it.

Example

The following code produces this diagnostic because the label loop isn’t defined anywhere:

void f() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      if (j != 0) {
        break loop;
      }
    }
  }
}

Common fixes

If the label should be on the innermost enclosing do, for, switch, or while statement, then remove the label:

void f() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      if (j != 0) {
        break;
      }
    }
  }
}

If the label should be on some other statement, then add the label:

void f() {
  loop: for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      if (j != 0) {
        break loop;
      }
    }
  }
}

late_final_field_with_const_constructor

Can’t have a late final field in a class with a generative const constructor.

Description

The analyzer produces this diagnostic when a class that has at least one const constructor also has a field marked both late and final.

Example

The following code produces this diagnostic because the class A has a const constructor and the final field f is marked as late:

class A {
  late final int f;

  const A();
}

Common fixes

If the field doesn’t need to be marked late, then remove the late modifier from the field:

class A {
  final int f = 0;

  const A();
}

If the field must be marked late, then remove the const modifier from the constructors:

class A {
  late final int f;

  A();
}

late_final_local_already_assigned

The late final local variable is already assigned.

Description

The analyzer produces this diagnostic when the analyzer can prove that a local variable marked as both late and final was already assigned a value at the point where another assignment occurs.

Because final variables can only be assigned once, subsequent assignments are guaranteed to fail, so they’re flagged.

Example

The following code produces this diagnostic because the final variable v is assigned a value in two places:

int f() {
  late final int v;
  v = 0;
  v += 1;
  return v;
}

Common fixes

If you need to be able to reassign the variable, then remove the final keyword:

int f() {
  late int v;
  v = 0;
  v += 1;
  return v;
}

If you don’t need to reassign the variable, then remove all except the first of the assignments:

int f() {
  late final int v;
  v = 0;
  return v;
}

leaf_call_must_not_return_handle

FFI leaf call can’t return a ‘Handle’.

Description

The analyzer produces this diagnostic when the value of the isLeaf argument in an invocation of either Pointer.asFunction or DynamicLibrary.lookupFunction is true and the function that would be returned would have a return type of Handle.

The analyzer also produces this diagnostic when the value of the isLeaf argument in an Native annotation is true and the type argument on the annotation is a function type whose return type is Handle.

In all of these cases, leaf calls are only supported for the types bool, int, float, double, and, as a return type void.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the function p returns a Handle, but the isLeaf argument is true:

import 'dart:ffi';

void f(Pointer<NativeFunction<Handle Function()>> p) {
  p.asFunction<Object Function()>(isLeaf: true);
}

Common fixes

If the function returns a handle, then remove the isLeaf argument:

import 'dart:ffi';

void f(Pointer<NativeFunction<Handle Function()>> p) {
  p.asFunction<Object Function()>();
}

If the function returns one of the supported types, then correct the type information:

import 'dart:ffi';

void f(Pointer<NativeFunction<Int32 Function()>> p) {
  p.asFunction<int Function()>(isLeaf: true);
}

leaf_call_must_not_take_handle

FFI leaf call can’t take arguments of type ‘Handle’.

Description

The analyzer produces this diagnostic when the value of the isLeaf argument in an invocation of either Pointer.asFunction or DynamicLibrary.lookupFunction is true and the function that would be returned would have a parameter of type Handle.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the function p has a parameter of type Handle, but the isLeaf argument is true:

import 'dart:ffi';

void f(Pointer<NativeFunction<Void Function(Handle)>> p) {
  p.asFunction<void Function(Object)>(isLeaf: true);
}

Common fixes

If the function has at least one parameter of type Handle, then remove the isLeaf argument:

import 'dart:ffi';

void f(Pointer<NativeFunction<Void Function(Handle)>> p) {
  p.asFunction<void Function(Object)>();
}

If none of the function’s parameters are Handles, then correct the type information:

import 'dart:ffi';

void f(Pointer<NativeFunction<Void Function(Int8)>> p) {
  p.asFunction<void Function(int)>(isLeaf: true);
}

list_element_type_not_assignable

The element type ‘{0}’ can’t be assigned to the list type ‘{1}’.

Description

The analyzer produces this diagnostic when the type of an element in a list literal isn’t assignable to the element type of the list.

Example

The following code produces this diagnostic because 2.5 is a double, and the list can hold only integers:

List<int> x = [1, 2.5, 3];

Common fixes

If you intended to add a different object to the list, then replace the element with an expression that computes the intended object:

List<int> x = [1, 2, 3];

If the object shouldn’t be in the list, then remove the element:

List<int> x = [1, 3];

If the object being computed is correct, then widen the element type of the list to allow all of the different types of objects it needs to contain:

List<num> x = [1, 2.5, 3];

main_first_positional_parameter_type

_The type of the first positional parameter of the ‘main’ function must be a supertype of ‘List'._

Description

The analyzer produces this diagnostic when the first positional parameter of a function named main isn’t a supertype of List<String>.

Example

The following code produces this diagnostic because List<int> isn’t a supertype of List<String>:

void main(List<int> args) {}

Common fixes

If the function is an entry point, then change the type of the first positional parameter to be a supertype of List<String>:

void main(List<String> args) {}

If the function isn’t an entry point, then change the name of the function:

void f(List<int> args) {}

main_has_required_named_parameters

The function ‘main’ can’t have any required named parameters.

Description

The analyzer produces this diagnostic when a function named main has one or more required named parameters.

Example

The following code produces this diagnostic because the function named main has a required named parameter (x):

void main({required int x}) {}

Common fixes

If the function is an entry point, then remove the required keyword:

void main({int? x}) {}

If the function isn’t an entry point, then change the name of the function:

void f({required int x}) {}

main_has_too_many_required_positional_parameters

The function ‘main’ can’t have more than two required positional parameters.

Description

The analyzer produces this diagnostic when a function named main has more than two required positional parameters.

Example

The following code produces this diagnostic because the function main has three required positional parameters:

void main(List<String> args, int x, int y) {}

Common fixes

If the function is an entry point and the extra parameters aren’t used, then remove them:

void main(List<String> args, int x) {}

If the function is an entry point, but the extra parameters used are for when the function isn’t being used as an entry point, then make the extra parameters optional:

void main(List<String> args, int x, [int y = 0]) {}

If the function isn’t an entry point, then change the name of the function:

void f(List<String> args, int x, int y) {}

main_is_not_function

The declaration named ‘main’ must be a function.

Description

The analyzer produces this diagnostic when a library contains a declaration of the name main that isn’t the declaration of a top-level function.

Example

The following code produces this diagnostic because the name main is being used to declare a top-level variable:

var main = 3;

Common fixes

Use a different name for the declaration:

var mainIndex = 3;

map_entry_not_in_map

Map entries can only be used in a map literal.

Description

The analyzer produces this diagnostic when a map entry (a key/value pair) is found in a set literal.

Example

The following code produces this diagnostic because the literal has a map entry even though it’s a set literal:

var collection = <String>{'a' : 'b'};

Common fixes

If you intended for the collection to be a map, then change the code so that it is a map. In the previous example, you could do this by adding another type argument:

var collection = <String, String>{'a' : 'b'};

In other cases, you might need to change the explicit type from Set to Map.

If you intended for the collection to be a set, then remove the map entry, possibly by replacing the colon with a comma if both values should be included in the set:

var collection = <String>{'a', 'b'};

map_key_type_not_assignable

The element type ‘{0}’ can’t be assigned to the map key type ‘{1}’.

Description

The analyzer produces this diagnostic when a key of a key-value pair in a map literal has a type that isn’t assignable to the key type of the map.

Example

The following code produces this diagnostic because 2 is an int, but the keys of the map are required to be Strings:

var m = <String, String>{2 : 'a'};

Common fixes

If the type of the map is correct, then change the key to have the correct type:

var m = <String, String>{'2' : 'a'};

If the type of the key is correct, then change the key type of the map:

var m = <int, String>{2 : 'a'};

map_value_type_not_assignable

The element type ‘{0}’ can’t be assigned to the map value type ‘{1}’.

Description

The analyzer produces this diagnostic when a value of a key-value pair in a map literal has a type that isn’t assignable to the value type of the map.

Example

The following code produces this diagnostic because 2 is an int, but/ the values of the map are required to be Strings:

var m = <String, String>{'a' : 2};

Common fixes

If the type of the map is correct, then change the value to have the correct type:

var m = <String, String>{'a' : '2'};

If the type of the value is correct, then change the value type of the map:

var m = <String, int>{'a' : 2};

mismatched_annotation_on_struct_field

The annotation doesn’t match the declared type of the field.

Description

The analyzer produces this diagnostic when the annotation on a field in a subclass of Struct or Union doesn’t match the Dart type of the field.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the annotation Double doesn’t match the Dart type int:

import 'dart:ffi';

final class C extends Struct {
  @Double()
  external int x;
}

Common fixes

If the type of the field is correct, then change the annotation to match:

import 'dart:ffi';

final class C extends Struct {
  @Int32()
  external int x;
}

If the annotation is correct, then change the type of the field to match:

import 'dart:ffi';

final class C extends Struct {
  @Double()
  external double x;
}

missing_annotation_on_struct_field

Fields of type ‘{0}’ in a subclass of ‘{1}’ must have an annotation indicating the native type.

Description

The analyzer produces this diagnostic when a field in a subclass of Struct or Union whose type requires an annotation doesn’t have one. The Dart types int, double, and Array are used to represent multiple C types, and the annotation specifies which of the compatible C types the field represents.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field x doesn’t have an annotation indicating the underlying width of the integer value:

import 'dart:ffi';

final class C extends Struct {
  external int x;
}

Common fixes

Add an appropriate annotation to the field:

import 'dart:ffi';

final class C extends Struct {
  @Int64()
  external int x;
}

missing_dart_library

Required library ‘{0}’ is missing.

Description

The analyzer produces this diagnostic when either the Dart or Flutter SDK isn’t installed correctly, and, as a result, one of the dart: libraries can’t be found.

Common fixes

Reinstall the Dart or Flutter SDK.

missing_default_value_for_parameter

The parameter ‘{0}’ can’t have a value of ‘null’ because of its type, but the implicit default value is ‘null’.

With null safety, use the ‘required’ keyword, not the ‘@required’ annotation.

Description

The analyzer produces this diagnostic when an optional parameter, whether positional or named, has a potentially non-nullable type and doesn’t specify a default value. Optional parameters that have no explicit default value have an implicit default value of null. If the type of the parameter doesn’t allow the parameter to have a value of null, then the implicit default value isn’t valid.

Examples

The following code produces this diagnostic because x can’t be null, and no non-null default value is specified:

void f([int x]) {}

As does this:

void g({int x}) {}

Common fixes

If you want to use null to indicate that no value was provided, then you need to make the type nullable:

void f([int? x]) {}
void g({int? x}) {}

If the parameter can’t be null, then either provide a default value:

void f([int x = 1]) {}
void g({int x = 2}) {}

or make the parameter a required parameter:

void f(int x) {}
void g({required int x}) {}

missing_dependency

Missing a dependency on imported package ‘{0}’.

Description

The analyzer produces this diagnostic when there’s a package that has been imported in the source but is not listed as a dependency of the importing package.

Example

The following code produces this diagnostic because the package path is not listed as a dependency, while there is an import statement with package path in the source code of package example:

name: example
dependencies:
  meta: ^1.0.2

Common fixes

Add the missing package path to the dependencies field:

name: example
dependencies:
  meta: ^1.0.2
  path: any

missing_enum_constant_in_switch

Missing case clause for ‘{0}’.

Description

The analyzer produces this diagnostic when a switch statement for an enum doesn’t include an option for one of the values in the enum.

Note that null is always a possible value for an enum and therefore also must be handled.

Example

The following code produces this diagnostic because the enum constant e2 isn’t handled:

enum E { e1, e2 }

void f(E e) {
  switch (e) {
    case E.e1:
      break;
  }
}

Common fixes

If there’s special handling for the missing values, then add a case clause for each of the missing values:

enum E { e1, e2 }

void f(E e) {
  switch (e) {
    case E.e1:
      break;
    case E.e2:
      break;
  }
}

If the missing values should be handled the same way, then add a default clause:

enum E { e1, e2 }

void f(E e) {
  switch (e) {
    case E.e1:
      break;
    default:
      break;
  }
}

missing_exception_value

The method {0} must have an exceptional return value (the second argument) when the return type of the function is neither ‘void’, ‘Handle’, nor ‘Pointer’.

Description

The analyzer produces this diagnostic when an invocation of the method Pointer.fromFunction or NativeCallable.isolateLocal doesn’t have a second argument (the exceptional return value) when the type to be returned from the invocation is neither void, Handle, nor Pointer.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the type returned by f is expected to be an 8-bit integer but the call to fromFunction doesn’t include an exceptional return argument:

import 'dart:ffi';

int f(int i) => i * 2;

void g() {
  Pointer.fromFunction<Int8 Function(Int8)>(f);
}

Common fixes

Add an exceptional return type:

import 'dart:ffi';

int f(int i) => i * 2;

void g() {
  Pointer.fromFunction<Int8 Function(Int8)>(f, 0);
}

missing_field_type_in_struct

Fields in struct classes must have an explicitly declared type of ‘int’, ‘double’ or ‘Pointer’.

Description

The analyzer produces this diagnostic when a field in a subclass of Struct or Union doesn’t have a type annotation. Every field must have an explicit type, and the type must either be int, double, Pointer, or a subclass of either Struct or Union.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field str doesn’t have a type annotation:

import 'dart:ffi';

final class C extends Struct {
  external var str;

  @Int32()
  external int i;
}

Common fixes

Explicitly specify the type of the field:

import 'dart:ffi';
import 'package:ffi/ffi.dart';

final class C extends Struct {
  external Pointer<Utf8> str;

  @Int32()
  external int i;
}

missing_name

The ‘name’ field is required but missing.

Description

The analyzer produces this diagnostic when there’s no top-level name key. The name key provides the name of the package, which is required.

Example

The following code produces this diagnostic because the package doesn’t have a name:

dependencies:
  meta: ^1.0.2

Common fixes

Add the top-level key name with a value that’s the name of the package:

name: example
dependencies:
  meta: ^1.0.2

missing_named_pattern_field_name

The getter name is not specified explicitly, and the pattern is not a variable.

Description

The analyzer produces this diagnostic when, within an object pattern, the specification of a property and the pattern used to match the property’s value doesn’t have either:

• a getter name before the colon
• a variable pattern from which the getter name can be inferred

Example

The following code produces this diagnostic because there is no getter name before the colon and no variable pattern after the colon in the object pattern (C(:0)):

abstract class C {
  int get f;
}

void f(C c) {
  switch (c) {
    case C(:0):
      break;
  }
}

Common fixes

If you need to use the actual value of the property within the pattern’s scope, then add a variable pattern where the name of the variable is the same as the name of the property being matched:

abstract class C {
  int get f;
}

void f(C c) {
  switch (c) {
    case C(:var f) when f == 0:
      print(f);
  }
}

If you don’t need to use the actual value of the property within the pattern’s scope, then add the name of the property being matched before the colon:

abstract class C {
  int get f;
}

void f(C c) {
  switch (c) {
    case C(f: 0):
      break;
  }
}

missing_override_of_must_be_overridden

Missing concrete implementation of ‘{0}’.

Missing concrete implementations of ‘{0}’ and ‘{1}’.

Missing concrete implementations of ‘{0}’, ‘{1}’, and {2} more.

Description

The analyzer produces this diagnostic when an instance member that has the @mustBeOverridden annotation isn’t overridden in a subclass.

Example

The following code produces this diagnostic because the class B doesn’t have an override of the inherited method A.m when A.m is annotated with @mustBeOverridden:

import 'package:meta/meta.dart';

class A {
  @mustBeOverridden
  void m() {}
}

class B extends A {}

Common fixes

If the annotation is appropriate for the member, then override the member in the subclass:

import 'package:meta/meta.dart';

class A {
  @mustBeOverridden
  void m() {}
}

class B extends A {
  @override
  void m() {}
}

If the annotation isn’t appropriate for the member, then remove the annotation:

class A {
  void m() {}
}

class B extends A {}

missing_required_argument

The named parameter ‘{0}’ is required, but there’s no corresponding argument.

Description

The analyzer produces this diagnostic when an invocation of a function is missing a required named parameter.

Example

The following code produces this diagnostic because the invocation of f doesn’t include a value for the required named parameter end:

void f(int start, {required int end}) {}
void g() {
  f(3);
}

Common fixes

Add a named argument corresponding to the missing required parameter:

void f(int start, {required int end}) {}
void g() {
  f(3, end: 5);
}

missing_required_param

The parameter ‘{0}’ is required.

The parameter ‘{0}’ is required. {1}.

Description

The analyzer produces this diagnostic when a method or function with a named parameter that is annotated as being required is invoked without providing a value for the parameter.

Example

The following code produces this diagnostic because the named parameter x is required:

import 'package:meta/meta.dart';

void f({@required int x}) {}

void g() {
  f();
}

Common fixes

Provide the required value:

import 'package:meta/meta.dart';

void f({@required int x}) {}

void g() {
  f(x: 2);
}

missing_return

This function has a return type of ‘{0}’, but doesn’t end with a return statement.

Description

Any function or method that doesn’t end with either an explicit return or a throw implicitly returns null. This is rarely the desired behavior. The analyzer produces this diagnostic when it finds an implicit return.

Example

The following code produces this diagnostic because f doesn’t end with a return:

int f(int x) {
  if (x < 0) {
    return 0;
  }
}

Common fixes

Add a return statement that makes the return value explicit, even if null is the appropriate value.

missing_size_annotation_carray

Fields of type ‘Array’ must have exactly one ‘Array’ annotation.

Description

The analyzer produces this diagnostic when a field in a subclass of either Struct or Union has a type of Array but doesn’t have a single Array annotation indicating the dimensions of the array.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field a0 doesn’t have an Array annotation:

import 'dart:ffi';

final class C extends Struct {
  external Array<Uint8> a0;
}

Common fixes

Ensure that there’s exactly one Array annotation on the field:

import 'dart:ffi';

final class C extends Struct {
  @Array(8)
  external Array<Uint8> a0;
}

missing_variable_pattern

Variable pattern ‘{0}’ is missing in this branch of the logical-or pattern.

Description

The analyzer produces this diagnostic when one branch of a logical-or pattern doesn’t declare a variable that is declared on the other branch of the same pattern.

Example

The following code produces this diagnostic because the right-hand side of the logical-or pattern doesn’t declare the variable a:

void f((int, int) r) {
  if (r case (var a, 0) || (0, _)) {
    print(a);
  }
}

Common fixes

If the variable needs to be referenced in the controlled statements, then add a declaration of the variable to every branch of the logical-or pattern:

void f((int, int) r) {
  if (r case (var a, 0) || (0, var a)) {
    print(a);
  }
}

If the variable doesn’t need to be referenced in the controlled statements, then remove the declaration of the variable from every branch of the logical-or pattern:

void f((int, int) r) {
  if (r case (_, 0) || (0, _)) {
    print('found a zero');
  }
}

If the variable needs to be referenced if one branch of the pattern matches but not when the other matches, then break the pattern into two pieces:

void f((int, int) r) {
  switch (r) {
    case (var a, 0):
      print(a);
    case (0, _):
      print('found a zero');
  }
}

mixin_application_concrete_super_invoked_member_type

The super-invoked member ‘{0}’ has the type ‘{1}’, and the concrete member in the class has the type ‘{2}’.

Description

The analyzer produces this diagnostic when a mixin that invokes a method using super is used in a class where the concrete implementation of that method has a different signature than the signature defined for that method by the mixin’s on type. The reason this is an error is because the invocation in the mixin might invoke the method in a way that’s incompatible with the method that will actually be executed.

Example

The following code produces this diagnostic because the class C uses the mixin M, the mixin M invokes foo using super, and the abstract version of foo declared in I (the mixin’s on type) doesn’t have the same signature as the concrete version of foo declared in A:

class I {
  void foo([int? p]) {}
}

class A {
  void foo(int p) {}
}

abstract class B extends A implements I {
  @override
  void foo([int? p]);
}

mixin M on I {
  void bar() {
    super.foo(42);
  }
}

abstract class C extends B with M {}

Common fixes

If the class doesn’t need to use the mixin, then remove it from the with clause:

class I {
  void foo([int? p]) {}
}

class A {
  void foo(int? p) {}
}

abstract class B extends A implements I {
  @override
  void foo([int? p]);
}

mixin M on I {
  void bar() {
    super.foo(42);
  }
}

abstract class C extends B {}

If the class needs to use the mixin, then ensure that there’s a concrete implementation of the method that conforms to the signature expected by the mixin:

class I {
  void foo([int? p]) {}
}

class A {
  void foo(int? p) {}
}

abstract class B extends A implements I {
  @override
  void foo([int? p]) {
    super.foo(p);
  }
}

mixin M on I {
  void bar() {
    super.foo(42);
  }
}

abstract class C extends B with M {}

mixin_application_not_implemented_interface

‘{0}’ can’t be mixed onto ‘{1}’ because ‘{1}’ doesn’t implement ‘{2}’.

Description

The analyzer produces this diagnostic when a mixin that has a superclass constraint is used in a mixin application with a superclass that doesn’t implement the required constraint.

Example

The following code produces this diagnostic because the mixin M requires that the class to which it’s applied be a subclass of A, but Object isn’t a subclass of A:

class A {}

mixin M on A {}

class X = Object with M;

Common fixes

If you need to use the mixin, then change the superclass to be either the same as or a subclass of the superclass constraint:

class A {}

mixin M on A {}

class X = A with M;

mixin_application_no_concrete_super_invoked_member

The class doesn’t have a concrete implementation of the super-invoked member ‘{0}’.

The class doesn’t have a concrete implementation of the super-invoked setter ‘{0}’.

Description

The analyzer produces this diagnostic when a mixin application contains an invocation of a member from its superclass, and there’s no concrete member of that name in the mixin application’s superclass.

Example

The following code produces this diagnostic because the mixin M contains the invocation super.m(), and the class A, which is the superclass of the mixin application A+M, doesn’t define a concrete implementation of m:

abstract class A {
  void m();
}

mixin M on A {
  void bar() {
    super.m();
  }
}

abstract class B extends A with M {}

Common fixes

If you intended to apply the mixin M to a different class, one that has a concrete implementation of m, then change the superclass of B to that class:

abstract class A {
  void m();
}

mixin M on A {
  void bar() {
    super.m();
  }
}

class C implements A {
  void m() {}
}

abstract class B extends C with M {}

If you need to make B a subclass of A, then add a concrete implementation of m in A:

abstract class A {
  void m() {}
}

mixin M on A {
  void bar() {
    super.m();
  }
}

abstract class B extends A with M {}

mixin_class_declaration_extends_not_object

The class ‘{0}’ can’t be declared a mixin because it extends a class other than ‘Object’.

Description

The analyzer produces this diagnostic when a class that is marked with the mixin modifier extends a class other than Object. A mixin class can’t have a superclass other than Object.

Example

The following code produces this diagnostic because the class B, which has the modifier mixin, extends A:

class A {}

mixin class B extends A {}

Common fixes

If you want the class to be used as a mixin, then change the superclass to Object, either explicitly or by removing the extends clause:

class A {}

mixin class B {}

If the class needs to have a superclass other than Object, then remove the mixin modifier:

class A {}

class B extends A {}

If you need both a mixin and a subclass of a class other than Object, then move the members of the subclass to a new mixin, remove the mixin modifier from the subclass, and apply the new mixin to the subclass:

class A {}

class B extends A with M {}

mixin M {}

Depending on the members of the subclass this might require adding an on clause to the mixin.

mixin_class_declares_constructor

The class ‘{0}’ can’t be used as a mixin because it declares a constructor.

Description

The analyzer produces this diagnostic when a class is used as a mixin and the mixed-in class defines a constructor.

Example

The following code produces this diagnostic because the class A, which defines a constructor, is being used as a mixin:

//@dart=2.19
class A {
  A();
}

class B with A {}

Common fixes

If it’s possible to convert the class to a mixin, then do so:

mixin A {
}

class B with A {}

If the class can’t be a mixin and it’s possible to remove the constructor, then do so:

//@dart=2.19
class A {
}

class B with A {}

If the class can’t be a mixin and you can’t remove the constructor, then try extending or implementing the class rather than mixing it in:

class A {
  A();
}

class B extends A {}

mixin_inherits_from_not_object

The class ‘{0}’ can’t be used as a mixin because it extends a class other than ‘Object’.

Description

The analyzer produces this diagnostic when a class that extends a class other than Object is used as a mixin.

Example

The following code produces this diagnostic because the class B, which extends A, is being used as a mixin by C:

//@dart=2.19
class A {}

class B extends A {}

class C with B {}

Common fixes

If the class being used as a mixin can be changed to extend Object, then change it:

//@dart=2.19
class A {}

class B {}

class C with B {}

If the class being used as a mixin can’t be changed and the class that’s using it extends Object, then extend the class being used as a mixin:

class A {}

class B extends A {}

class C extends B {}

If the class doesn’t extend Object or if you want to be able to mix in the behavior from B in other places, then create a real mixin:

class A {}

mixin M on A {}

class B extends A with M {}

class C extends A with M {}

mixin_instantiate

Mixins can’t be instantiated.

Description

The analyzer produces this diagnostic when a mixin is instantiated.

Example

The following code produces this diagnostic because the mixin M is being instantiated:

mixin M {}

var m = M();

Common fixes

If you intend to use an instance of a class, then use the name of that class in place of the name of the mixin.

mixin_of_non_class

Classes can only mix in mixins and classes.

Description

The analyzer produces this diagnostic when a name in a with clause is defined to be something other than a mixin or a class.

Example

The following code produces this diagnostic because F is defined to be a function type:

typedef F = int Function(String);

class C with F {}

Common fixes

Remove the invalid name from the list, possibly replacing it with the name of the intended mixin or class:

typedef F = int Function(String);

class C {}

mixin_on_sealed_class

The class ‘{0}’ shouldn’t be used as a mixin constraint because it is sealed, and any class mixing in this mixin must have ‘{0}’ as a superclass.

Description

The analyzer produces this diagnostic when the superclass constraint of a mixin is a class from a different package that was marked as sealed. Classes that are sealed can’t be extended, implemented, mixed in, or used as a superclass constraint.

Example

If the package p defines a sealed class:

import 'package:meta/meta.dart';

@sealed
class C {}

Then, the following code, when in a package other than p, produces this diagnostic:

import 'package:p/p.dart';

mixin M on C {}

Common fixes

If the classes that use the mixin don’t need to be subclasses of the sealed class, then consider adding a field and delegating to the wrapped instance of the sealed class.

mixin_super_class_constraint_deferred_class

Deferred classes can’t be used as superclass constraints.

Description

The analyzer produces this diagnostic when a superclass constraint of a mixin is imported from a deferred library.

Example

The following code produces this diagnostic because the superclass constraint of math.Random is imported from a deferred library:

import 'dart:async' deferred as async;

mixin M<T> on async.Stream<T> {}

Common fixes

If the import doesn’t need to be deferred, then remove the deferred keyword:

import 'dart:async' as async;

mixin M<T> on async.Stream<T> {}

If the import does need to be deferred, then remove the superclass constraint:

mixin M<T> {}

mixin_super_class_constraint_non_interface

Only classes and mixins can be used as superclass constraints.

Description

The analyzer produces this diagnostic when a type following the on keyword in a mixin declaration is neither a class nor a mixin.

Example

The following code produces this diagnostic because F is neither a class nor a mixin:

typedef F = void Function();

mixin M on F {}

Common fixes

If the type was intended to be a class but was mistyped, then replace the name.

Otherwise, remove the type from the on clause.

multiple_redirecting_constructor_invocations

Constructors can have only one ‘this’ redirection, at most.

Description

The analyzer produces this diagnostic when a constructor redirects to more than one other constructor in the same class (using this).

Example

The following code produces this diagnostic because the unnamed constructor in C is redirecting to both this.a and this.b:

class C {
  C() : this.a(), this.b();
  C.a();
  C.b();
}

Common fixes

Remove all but one of the redirections:

class C {
  C() : this.a();
  C.a();
  C.b();
}

multiple_super_initializers

A constructor can have at most one ‘super’ initializer.

Description

The analyzer produces this diagnostic when the initializer list of a constructor contains more than one invocation of a constructor from the superclass. The initializer list is required to have exactly one such call, which can either be explicit or implicit.

Example

The following code produces this diagnostic because the initializer list for B’s constructor invokes both the constructor one and the constructor two from the superclass A:

class A {
  int? x;
  String? s;
  A.one(this.x);
  A.two(this.s);
}

class B extends A {
  B() : super.one(0), super.two('');
}

Common fixes

If one of the super constructors will initialize the instance fully, then remove the other:

class A {
  int? x;
  String? s;
  A.one(this.x);
  A.two(this.s);
}

class B extends A {
  B() : super.one(0);
}

If the initialization achieved by one of the super constructors can be performed in the body of the constructor, then remove its super invocation and perform the initialization in the body:

class A {
  int? x;
  String? s;
  A.one(this.x);
  A.two(this.s);
}

class B extends A {
  B() : super.one(0) {
    s = '';
  }
}

If the initialization can only be performed in a constructor in the superclass, then either add a new constructor or modify one of the existing constructors so there’s a constructor that allows all the required initialization to occur in a single call:

class A {
  int? x;
  String? s;
  A.one(this.x);
  A.two(this.s);
  A.three(this.x, this.s);
}

class B extends A {
  B() : super.three(0, '');
}

must_be_a_native_function_type

The type ‘{0}’ given to ‘{1}’ must be a valid ‘dart:ffi’ native function type.

Description

The analyzer produces this diagnostic when an invocation of either Pointer.fromFunction, DynamicLibrary.lookupFunction, or a NativeCallable constructor, has a type argument(whether explicit or inferred) that isn’t a native function type.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the type T can be any subclass of Function but the type argument for fromFunction is required to be a native function type:

import 'dart:ffi';

int f(int i) => i * 2;

class C<T extends Function> {
  void g() {
    Pointer.fromFunction<T>(f, 0);
  }
}

Common fixes

Use a native function type as the type argument to the invocation:

import 'dart:ffi';

int f(int i) => i * 2;

class C<T extends Function> {
  void g() {
    Pointer.fromFunction<Int32 Function(Int32)>(f, 0);
  }
}

must_be_a_subtype

The type ‘{0}’ must be a subtype of ‘{1}’ for ‘{2}’.

Description

The analyzer produces this diagnostic in two cases:

• In an invocation of Pointer.fromFunction, or a NativeCallable constructor where the type argument (whether explicit or inferred) isn’t a supertype of the type of the function passed as the first argument to the method.
• In an invocation of DynamicLibrary.lookupFunction where the first type argument isn’t a supertype of the second type argument.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the type of the function f (String Function(int)) isn’t a subtype of the type argument T (Int8 Function(Int8)):

import 'dart:ffi';

typedef T = Int8 Function(Int8);

double f(double i) => i;

void g() {
  Pointer.fromFunction<T>(f, 5.0);
}

Common fixes

If the function is correct, then change the type argument to match:

import 'dart:ffi';

typedef T = Float Function(Float);

double f(double i) => i;

void g() {
  Pointer.fromFunction<T>(f, 5.0);
}

If the type argument is correct, then change the function to match:

import 'dart:ffi';

typedef T = Int8 Function(Int8);

int f(int i) => i;

void g() {
  Pointer.fromFunction<T>(f, 5);
}

must_be_immutable

This class (or a class that this class inherits from) is marked as ‘@immutable’, but one or more of its instance fields aren’t final: {0}

Description

The analyzer produces this diagnostic when an immutable class defines one or more instance fields that aren’t final. A class is immutable if it’s marked as being immutable using the annotation immutable or if it’s a subclass of an immutable class.

Example

The following code produces this diagnostic because the field x isn’t final:

import 'package:meta/meta.dart';

@immutable
class C {
  int x;

  C(this.x);
}

Common fixes

If instances of the class should be immutable, then add the keyword final to all non-final field declarations:

import 'package:meta/meta.dart';

@immutable
class C {
  final int x;

  C(this.x);
}

If the instances of the class should be mutable, then remove the annotation, or choose a different superclass if the annotation is inherited:

class C {
  int x;

  C(this.x);
}

must_call_super

This method overrides a method annotated as ‘@mustCallSuper’ in ‘{0}’, but doesn’t invoke the overridden method.

Description

The analyzer produces this diagnostic when a method that overrides a method that is annotated as mustCallSuper doesn’t invoke the overridden method as required.

Example

The following code produces this diagnostic because the method m in B doesn’t invoke the overridden method m in A:

import 'package:meta/meta.dart';

class A {
  @mustCallSuper
  m() {}
}

class B extends A {
  @override
  m() {}
}

Common fixes

Add an invocation of the overridden method in the overriding method:

import 'package:meta/meta.dart';

class A {
  @mustCallSuper
  m() {}
}

class B extends A {
  @override
  m() {
    super.m();
  }
}

must_return_void

The return type of the function passed to ‘NativeCallable.listener’ must be ‘void’ rather than ‘{0}’.

Description

The analyzer produces this diagnostic when you pass a function that doesn’t return void to the NativeCallable.listener constructor.

NativeCallable.listener creates a native callable that can be invoked from any thread. The native code that invokes the callable sends a message back to the isolate that created the callable, and doesn’t wait for a response. So it isn’t possible to return a result from the callable.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the function f returns int rather than void.

import 'dart:ffi';

int f(int i) => i * 2;

void g() {
  NativeCallable<Int32 Function(Int32)>.listener(f);
}

Common fixes

Change the return type of the function to void.

import 'dart:ffi';

void f(int i) => print(i * 2);

void g() {
  NativeCallable<Void Function(Int32)>.listener(f);
}

name_not_string

The value of the ‘name’ field is required to be a string.

Description

The analyzer produces this diagnostic when the top-level name key has a value that isn’t a string.

Example

The following code produces this diagnostic because the value following the name key is a list:

name:
  - example

Common fixes

Replace the value with a string:

name: example

new_with_undefined_constructor_default

The class ‘{0}’ doesn’t have an unnamed constructor.

Description

The analyzer produces this diagnostic when an unnamed constructor is invoked on a class that defines named constructors but the class doesn’t have an unnamed constructor.

Example

The following code produces this diagnostic because A doesn’t define an unnamed constructor:

class A {
  A.a();
}

A f() => A();

Common fixes

If one of the named constructors does what you need, then use it:

class A {
  A.a();
}

A f() => A.a();

If none of the named constructors does what you need, and you’re able to add an unnamed constructor, then add the constructor:

class A {
  A();
  A.a();
}

A f() => A();

non_abstract_class_inherits_abstract_member

Missing concrete implementation of ‘{0}’.

Missing concrete implementations of ‘{0}’ and ‘{1}’.

Missing concrete implementations of ‘{0}’, ‘{1}’, ‘{2}’, ‘{3}’, and {4} more.

Missing concrete implementations of ‘{0}’, ‘{1}’, ‘{2}’, and ‘{3}’.

Missing concrete implementations of ‘{0}’, ‘{1}’, and ‘{2}’.

Description

The analyzer produces this diagnostic when a concrete class inherits one or more abstract members, and doesn’t provide or inherit an implementation for at least one of those abstract members.

Example

The following code produces this diagnostic because the class B doesn’t have a concrete implementation of m:

abstract class A {
  void m();
}

class B extends A {}

Common fixes

If the subclass can provide a concrete implementation for some or all of the abstract inherited members, then add the concrete implementations:

abstract class A {
  void m();
}

class B extends A {
  void m() {}
}

If there is a mixin that provides an implementation of the inherited methods, then apply the mixin to the subclass:

abstract class A {
  void m();
}

class B extends A with M {}

mixin M {
  void m() {}
}

If the subclass can’t provide a concrete implementation for all of the abstract inherited members, then mark the subclass as being abstract:

abstract class A {
  void m();
}

abstract class B extends A {}

non_bool_condition

Conditions must have a static type of ‘bool’.

Description

The analyzer produces this diagnostic when a condition, such as an if or while loop, doesn’t have the static type bool.

Example

The following code produces this diagnostic because x has the static type int:

void f(int x) {
  if (x) {
    // ...
  }
}

Common fixes

Change the condition so that it produces a Boolean value:

void f(int x) {
  if (x == 0) {
    // ...
  }
}

non_bool_expression

The expression in an assert must be of type ‘bool’.

Description

The analyzer produces this diagnostic when the first expression in an assert has a type other than bool.

Example

The following code produces this diagnostic because the type of p is int, but a bool is required:

void f(int p) {
  assert(p);
}

Common fixes

Change the expression so that it has the type bool:

void f(int p) {
  assert(p > 0);
}

non_bool_negation_expression

A negation operand must have a static type of ‘bool’.

Description

The analyzer produces this diagnostic when the operand of the unary negation operator (!) doesn’t have the type bool.

Example

The following code produces this diagnostic because x is an int when it must be a bool:

int x = 0;
bool y = !x;

Common fixes

Replace the operand with an expression that has the type bool:

int x = 0;
bool y = !(x > 0);

non_bool_operand

The operands of the operator ‘{0}’ must be assignable to ‘bool’.

Description

The analyzer produces this diagnostic when one of the operands of either the && or || operator doesn’t have the type bool.

Example

The following code produces this diagnostic because a isn’t a Boolean value:

int a = 3;
bool b = a || a > 1;

Common fixes

Change the operand to a Boolean value:

int a = 3;
bool b = a == 0 || a > 1;

non_constant_annotation_constructor

Annotation creation can only call a const constructor.

Description

The analyzer produces this diagnostic when an annotation is the invocation of an existing constructor even though the invoked constructor isn’t a const constructor.

Example

The following code produces this diagnostic because the constructor for C isn’t a const constructor:

@C()
void f() {
}

class C {
  C();
}

Common fixes

If it’s valid for the class to have a const constructor, then create a const constructor that can be used for the annotation:

@C()
void f() {
}

class C {
  const C();
}

If it isn’t valid for the class to have a const constructor, then either remove the annotation or use a different class for the annotation.

non_constant_case_expression

Case expressions must be constant.

Description

The analyzer produces this diagnostic when the expression in a case clause isn’t a constant expression.

Example

The following code produces this diagnostic because j isn’t a constant:

void f(int i, int j) {
  switch (i) {
    case j:
      // ...
      break;
  }
}

Common fixes

Either make the expression a constant expression, or rewrite the switch statement as a sequence of if statements:

void f(int i, int j) {
  if (i == j) {
    // ...
  }
}

non_constant_case_expression_from_deferred_library

Constant values from a deferred library can’t be used as a case expression.

Description

The analyzer produces this diagnostic when the expression in a case clause references a constant from a library that is imported using a deferred import. In order for switch statements to be compiled efficiently, the constants referenced in case clauses need to be available at compile time, and constants from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines the constant zero:

const zero = 0;

The following code produces this diagnostic because the library a.dart is imported using a deferred import, and the constant a.zero, declared in the imported library, is used in a case clause:

import 'a.dart' deferred as a;

void f(int x) {
  switch (x) {
    case a.zero:
      // ...
      break;
  }
}

Common fixes

If you need to reference the constant from the imported library, then remove the deferred keyword:

import 'a.dart' as a;

void f(int x) {
  switch (x) {
    case a.zero:
      // ...
      break;
  }
}

If you need to reference the constant from the imported library and also need the imported library to be deferred, then rewrite the switch statement as a sequence of if statements:

import 'a.dart' deferred as a;

void f(int x) {
  if (x == a.zero) {
    // ...
  }
}

If you don’t need to reference the constant, then replace the case expression:

void f(int x) {
  switch (x) {
    case 0:
      // ...
      break;
  }
}

non_constant_default_value

The default value of an optional parameter must be constant.

Description

The analyzer produces this diagnostic when an optional parameter, either named or positional, has a default value that isn’t a compile-time constant.

Example

The following code produces this diagnostic:

var defaultValue = 3;

void f([int value = defaultValue]) {}

Common fixes

If the default value can be converted to be a constant, then convert it:

const defaultValue = 3;

void f([int value = defaultValue]) {}

If the default value needs to change over time, then apply the default value inside the function:

var defaultValue = 3;

void f([int value]) {
  value ??= defaultValue;
}

non_constant_default_value_from_deferred_library

Constant values from a deferred library can’t be used as a default parameter value.

Description

The analyzer produces this diagnostic when the default value of an optional parameter uses a constant from a library imported using a deferred import. Default values need to be available at compile time, and constants from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines the constant zero:

const zero = 0;

The following code produces this diagnostic because zero is declared in a library imported using a deferred import:

import 'a.dart' deferred as a;

void f({int x = a.zero}) {}

Common fixes

If you need to reference the constant from the imported library, then remove the deferred keyword:

import 'a.dart' as a;

void f({int x = a.zero}) {}

If you don’t need to reference the constant, then replace the default value:

void f({int x = 0}) {}

non_constant_list_element

The values in a const list literal must be constants.

Description

The analyzer produces this diagnostic when an element in a constant list literal isn’t a constant value. The list literal can be constant either explicitly (because it’s prefixed by the const keyword) or implicitly (because it appears in a constant context).

Example

The following code produces this diagnostic because x isn’t a constant, even though it appears in an implicitly constant list literal:

var x = 2;
var y = const <int>[0, 1, x];

Common fixes

If the list needs to be a constant list, then convert the element to be a constant. In the example above, you might add the const keyword to the declaration of x:

const x = 2;
var y = const <int>[0, 1, x];

If the expression can’t be made a constant, then the list can’t be a constant either, so you must change the code so that the list isn’t a constant. In the example above this means removing the const keyword before the list literal:

var x = 2;
var y = <int>[0, 1, x];

non_constant_map_element

The elements in a const map literal must be constant.

Description

The analyzer produces this diagnostic when an if element or a spread element in a constant map isn’t a constant element.

Examples

The following code produces this diagnostic because it’s attempting to spread a non-constant map:

var notConst = <int, int>{};
var map = const <int, int>{...notConst};

Similarly, the following code produces this diagnostic because the condition in the if element isn’t a constant expression:

bool notConst = true;
var map = const <int, int>{if (notConst) 1 : 2};

Common fixes

If the map needs to be a constant map, then make the elements constants. In the spread example, you might do that by making the collection being spread a constant:

const notConst = <int, int>{};
var map = const <int, int>{...notConst};

If the map doesn’t need to be a constant map, then remove the const keyword:

bool notConst = true;
var map = <int, int>{if (notConst) 1 : 2};

non_constant_map_key

The keys in a const map literal must be constant.

Description

The analyzer produces this diagnostic when a key in a constant map literal isn’t a constant value.

Example

The following code produces this diagnostic because a isn’t a constant:

var a = 'a';
var m = const {a: 0};

Common fixes

If the map needs to be a constant map, then make the key a constant:

const a = 'a';
var m = const {a: 0};

If the map doesn’t need to be a constant map, then remove the const keyword:

var a = 'a';
var m = {a: 0};

non_constant_map_pattern_key

Key expressions in map patterns must be constants.

Description

The analyzer produces this diagnostic when a key in a map pattern isn’t a constant expression.

Example

The following code produces this diagnostic because the key A() isn’t a constant:

void f(Object x) {
  if (x case {A(): 0}) {}
}

class A {
  const A();
}

Common fixes

Use a constant for the key:

void f(Object x) {
  if (x case {const A(): 0}) {}
}

class A {
  const A();
}

non_constant_map_value

The values in a const map literal must be constant.

Description

The analyzer produces this diagnostic when a value in a constant map literal isn’t a constant value.

Example

The following code produces this diagnostic because a isn’t a constant:

var a = 'a';
var m = const {0: a};

Common fixes

If the map needs to be a constant map, then make the key a constant:

const a = 'a';
var m = const {0: a};

If the map doesn’t need to be a constant map, then remove the const keyword:

var a = 'a';
var m = {0: a};

non_constant_relational_pattern_expression

The relational pattern expression must be a constant.

Description

The analyzer produces this diagnostic when the value in a relational pattern expression isn’t a constant expression.

Example

The following code produces this diagnostic because the operand of the > operator, a, isn’t a constant:

final a = 0;

void f(int x) {
  if (x case > a) {}
}

Common fixes

Replace the value with a constant expression:

const a = 0;

void f(int x) {
  if (x case > a) {}
}

non_constant_set_element

The values in a const set literal must be constants.

Description

The analyzer produces this diagnostic when a constant set literal contains an element that isn’t a compile-time constant.

Example

The following code produces this diagnostic because i isn’t a constant:

var i = 0;

var s = const {i};

Common fixes

If the element can be changed to be a constant, then change it:

const i = 0;

var s = const {i};

If the element can’t be a constant, then remove the keyword const:

var i = 0;

var s = {i};

non_constant_type_argument

The type arguments to ‘{0}’ must be known at compile time, so they can’t be type parameters.

Description

The analyzer produces this diagnostic when the type arguments to a method are required to be known at compile time, but a type parameter, whose value can’t be known at compile time, is used as a type argument.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the type argument to Pointer.asFunction must be known at compile time, but the type parameter R, which isn’t known at compile time, is being used as the type argument:

import 'dart:ffi';

typedef T = int Function(int);

class C<R extends T> {
  void m(Pointer<NativeFunction<T>> p) {
    p.asFunction<R>();
  }
}

Common fixes

Remove any uses of type parameters:

import 'dart:ffi';

class C {
  void m(Pointer<NativeFunction<Int64 Function(Int64)>> p) {
    p.asFunction<int Function(int)>();
  }
}

non_const_call_to_literal_constructor

This instance creation must be ‘const’, because the {0} constructor is marked as ‘@literal’.

Description

The analyzer produces this diagnostic when a constructor that has the literal annotation is invoked without using the const keyword, but all of the arguments to the constructor are constants. The annotation indicates that the constructor should be used to create a constant value whenever possible.

Example

The following code produces this diagnostic:

import 'package:meta/meta.dart';

class C {
  @literal
  const C();
}

C f() => C();

Common fixes

Add the keyword const before the constructor invocation:

import 'package:meta/meta.dart';

class C {
  @literal
  const C();
}

void f() => const C();

non_const_generative_enum_constructor

Generative enum constructors must be ‘const’.

Description

The analyzer produces this diagnostic when an enum declaration contains a generative constructor that isn’t marked as const.

Example

The following code produces this diagnostic because the constructor in E isn’t marked as being const:

enum E {
  e;

  E();
}

Common fixes

Add the const keyword before the constructor:

enum E {
  e;

  const E();
}

non_covariant_type_parameter_position_in_representation_type

An extension type parameter can’t be used in a non-covariant position of its representation type.

Description

The analyzer produces this diagnostic when a type parameter of an extension type is used in a non-covariant position in the representation type of that extension type.

Example

The following code produces this diagnostic because the type parameter T is used as a parameter type in the function type void Function(T), and parameters are not covariant:

extension type A<T>(void Function(T) f) {}

Common fixes

Remove the use of the type parameter:

extension type A(void Function(String) f) {}

non_exhaustive_switch_expression

The type ‘{0}’ is not exhaustively matched by the switch cases since it doesn’t match ‘{1}’.

Description

The analyzer produces this diagnostic when a switch expression is missing a case for one or more of the possible values that could flow through it.

Example

The following code produces this diagnostic because the switch expression doesn’t have a case for the value E.three:

enum E { one, two, three }

String f(E e) => switch (e) {
    E.one => 'one',
    E.two => 'two',
  };

Common fixes

If the missing values are distinctly meaningful to the switch expression, then add a case for each of the values missing a match:

enum E { one, two, three }

String f(E e) => switch (e) {
    E.one => 'one',
    E.two => 'two',
    E.three => 'three',
  };

If the missing values don’t need to be matched, then add a wildcard pattern that returns a simple default:

enum E { one, two, three }

String f(E e) => switch (e) {
    E.one => 'one',
    E.two => 'two',
    _ => 'unknown',
  };

Be aware that a wildcard pattern will handle any values added to the type in the future. You will lose the ability to have the compiler warn you if the switch needs to be updated to account for newly added types.

non_exhaustive_switch_statement

The type ‘{0}’ is not exhaustively matched by the switch cases since it doesn’t match ‘{1}’.

Description

The analyzer produces this diagnostic when a switch statement switching over an exhaustive type is missing a case for one or more of the possible values that could flow through it.

Example

The following code produces this diagnostic because the switch statement doesn’t have a case for the value E.three, and E is an exhaustive type:

enum E { one, two, three }

void f(E e) {
  switch (e) {
    case E.one:
    case E.two:
  }
}

Common fixes

Add a case for each of the constants that aren’t currently being matched:

enum E { one, two, three }

void f(E e) {
  switch (e) {
    case E.one:
    case E.two:
      break;
    case E.three:
  }
}

If the missing values don’t need to be matched, then add a default clause or a wildcard pattern:

enum E { one, two, three }

void f(E e) {
  switch (e) {
    case E.one:
    case E.two:
      break;
    default:
  }
}

But be aware that adding a default clause or wildcard pattern will cause any future values of the exhaustive type to also be handled, so you will have lost the ability for the compiler to warn you if the switch needs to be updated.

non_final_field_in_enum

Enums can only declare final fields.

Description

The analyzer produces this diagnostic when an instance field in an enum isn’t marked as final.

Example

The following code produces this diagnostic because the field f isn’t a final field:

enum E {
  c;

  int f = 0;
}

Common fixes

If the field must be defined for the enum, then mark the field as being final:

enum E {
  c;

  final int f = 0;
}

If the field can be removed, then remove it:

enum E {
  c
}

non_generative_constructor

The generative constructor ‘{0}’ is expected, but a factory was found.

Description

The analyzer produces this diagnostic when the initializer list of a constructor invokes a constructor from the superclass, and the invoked constructor is a factory constructor. Only a generative constructor can be invoked in the initializer list.

Example

The following code produces this diagnostic because the invocation of the constructor super.one() is invoking a factory constructor:

class A {
  factory A.one() = B;
  A.two();
}

class B extends A {
  B() : super.one();
}

Common fixes

Change the super invocation to invoke a generative constructor:

class A {
  factory A.one() = B;
  A.two();
}

class B extends A {
  B() : super.two();
}

If the generative constructor is the unnamed constructor, and if there are no arguments being passed to it, then you can remove the super invocation.

non_generative_implicit_constructor

The unnamed constructor of superclass ‘{0}’ (called by the default constructor of ‘{1}’) must be a generative constructor, but factory found.

Description

The analyzer produces this diagnostic when a class has an implicit generative constructor and the superclass has an explicit unnamed factory constructor. The implicit constructor in the subclass implicitly invokes the unnamed constructor in the superclass, but generative constructors can only invoke another generative constructor, not a factory constructor.

Example

The following code produces this diagnostic because the implicit constructor in B invokes the unnamed constructor in A, but the constructor in A is a factory constructor, when a generative constructor is required:

class A {
  factory A() => throw 0;
  A.named();
}

class B extends A {}

Common fixes

If the unnamed constructor in the superclass can be a generative constructor, then change it to be a generative constructor:

class A {
  A();
  A.named();
}

class B extends A { }

If the unnamed constructor can’t be a generative constructor and there are other generative constructors in the superclass, then explicitly invoke one of them:

class A {
  factory A() => throw 0;
  A.named();
}

class B extends A {
  B() : super.named();
}

If there are no generative constructors that can be used and none can be added, then implement the superclass rather than extending it:

class A {
  factory A() => throw 0;
  A.named();
}

class B implements A {}

non_native_function_type_argument_to_pointer

Can’t invoke ‘asFunction’ because the function signature ‘{0}’ for the pointer isn’t a valid C function signature.

Description

The analyzer produces this diagnostic when the method asFunction is invoked on a pointer to a native function, but the signature of the native function isn’t a valid C function signature.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because function signature associated with the pointer p (FNative) isn’t a valid C function signature:

import 'dart:ffi';

typedef FNative = int Function(int);
typedef F = int Function(int);

class C {
  void f(Pointer<NativeFunction<FNative>> p) {
    p.asFunction<F>();
  }
}

Common fixes

Make the NativeFunction signature a valid C signature:

import 'dart:ffi';

typedef FNative = Int8 Function(Int8);
typedef F = int Function(int);

class C {
  void f(Pointer<NativeFunction<FNative>> p) {
    p.asFunction<F>();
  }
}

non_positive_array_dimension

Array dimensions must be positive numbers.

Description

The analyzer produces this diagnostic when a dimension given in an Array annotation is less than or equal to zero (0).

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because an array dimension of -1 was provided:

import 'dart:ffi';

final class MyStruct extends Struct {
  @Array(-8)
  external Array<Uint8> a0;
}

Common fixes

Change the dimension to be a positive integer:

import 'dart:ffi';

final class MyStruct extends Struct {
  @Array(8)
  external Array<Uint8> a0;
}

non_sized_type_argument

The type ‘{1}’ isn’t a valid type argument for ‘{0}’. The type argument must be a native integer, ‘Float’, ‘Double’, ‘Pointer’, or subtype of ‘Struct’, ‘Union’, or ‘AbiSpecificInteger’.

Description

The analyzer produces this diagnostic when the type argument for the class Array isn’t one of the valid types: either a native integer, Float, Double, Pointer, or subtype of Struct, Union, or AbiSpecificInteger.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the type argument to Array is Void, and Void isn’t one of the valid types:

import 'dart:ffi';

final class C extends Struct {
  @Array(8)
  external Array<Void> a0;
}

Common fixes

Change the type argument to one of the valid types:

import 'dart:ffi';

final class C extends Struct {
  @Array(8)
  external Array<Uint8> a0;
}

non_sync_factory

Factory bodies can’t use ‘async’, ‘async*’, or ‘sync*‘.

Description

The analyzer produces this diagnostic when the body of a factory constructor is marked with async, async*, or sync*. All constructors, including factory constructors, are required to return an instance of the class in which they’re declared, not a Future, Stream, or Iterator.

Example

The following code produces this diagnostic because the body of the factory constructor is marked with async:

class C {
  factory C() async {
    return C._();
  }
  C._();
}

Common fixes

If the member must be declared as a factory constructor, then remove the keyword appearing before the body:

class C {
  factory C() {
    return C._();
  }
  C._();
}

If the member must return something other than an instance of the enclosing class, then make the member a static method:

class C {
  static Future<C> m() async {
    return C._();
  }
  C._();
}

non_type_as_type_argument

The name ‘{0}’ isn’t a type, so it can’t be used as a type argument.

Description

The analyzer produces this diagnostic when an identifier that isn’t a type is used as a type argument.

Example

The following code produces this diagnostic because x is a variable, not a type:

var x = 0;
List<x> xList = [];

Common fixes

Change the type argument to be a type:

var x = 0;
List<int> xList = [];

non_type_in_catch_clause

The name ‘{0}’ isn’t a type and can’t be used in an on-catch clause.

Description

The analyzer produces this diagnostic when the identifier following the on in a catch clause is defined to be something other than a type.

Example

The following code produces this diagnostic because f is a function, not a type:

void f() {
  try {
    // ...
  } on f {
    // ...
  }
}

Common fixes

Change the name to the type of object that should be caught:

void f() {
  try {
    // ...
  } on FormatException {
    // ...
  }
}

non_void_return_for_operator

The return type of the operator []= must be ‘void’.

Description

The analyzer produces this diagnostic when a declaration of the operator []= has a return type other than void.

Example

The following code produces this diagnostic because the declaration of the operator []= has a return type of int:

class C {
  int operator []=(int index, int value) => 0;
}

Common fixes

Change the return type to void:

class C {
  void operator []=(int index, int value) => 0;
}

non_void_return_for_setter

The return type of the setter must be ‘void’ or absent.

Description

The analyzer produces this diagnostic when a setter is defined with a return type other than void.

Example

The following code produces this diagnostic because the setter p has a return type of int:

class C {
  int set p(int i) => 0;
}

Common fixes

Change the return type to void or omit the return type:

class C {
  set p(int i) => 0;
}

not_assigned_potentially_non_nullable_local_variable

The non-nullable local variable ‘{0}’ must be assigned before it can be used.

Description

The analyzer produces this diagnostic when a local variable is referenced and has all these characteristics:

• Has a type that’s potentially non-nullable.
• Doesn’t have an initializer.
• Isn’t marked as late.
• The analyzer can’t prove that the local variable will be assigned before the reference based on the specification of definite assignment.

Examples

The following code produces this diagnostic because x can’t have a value of null, but is referenced before a value was assigned to it:

String f() {
  int x;
  return x.toString();
}

The following code produces this diagnostic because the assignment to x might not be executed, so it might have a value of null:

int g(bool b) {
  int x;
  if (b) {
    x = 1;
  }
  return x * 2;
}

The following code produces this diagnostic because the analyzer can’t prove, based on definite assignment analysis, that x won’t be referenced without having a value assigned to it:

int h(bool b) {
  int x;
  if (b) {
    x = 1;
  }
  if (b) {
    return x * 2;
  }
  return 0;
}

Common fixes

If null is a valid value, then make the variable nullable:

String f() {
  int? x;
  return x!.toString();
}

If null isn’t a valid value, and there’s a reasonable default value, then add an initializer:

int g(bool b) {
  int x = 2;
  if (b) {
    x = 1;
  }
  return x * 2;
}

Otherwise, ensure that a value was assigned on every possible code path before the value is accessed:

int g(bool b) {
  int x;
  if (b) {
    x = 1;
  } else {
    x = 2;
  }
  return x * 2;
}

You can also mark the variable as late, which removes the diagnostic, but if the variable isn’t assigned a value before it’s accessed, then it results in an exception being thrown at runtime. This approach should only be used if you’re sure that the variable will always be assigned, even though the analyzer can’t prove it based on definite assignment analysis.

int h(bool b) {
  late int x;
  if (b) {
    x = 1;
  }
  if (b) {
    return x * 2;
  }
  return 0;
}

not_a_type

{0} isn’t a type.

Description

The analyzer produces this diagnostic when a name is used as a type but declared to be something other than a type.

Example

The following code produces this diagnostic because f is a function:

f() {}
g(f v) {}

Common fixes

Replace the name with the name of a type.

not_binary_operator

‘{0}’ isn’t a binary operator.

Description

The analyzer produces this diagnostic when an operator that can only be used as a unary operator is used as a binary operator.

Example

The following code produces this diagnostic because the operator ~ can only be used as a unary operator:

var a = 5 ~ 3;

Common fixes

Replace the operator with the correct binary operator:

var a = 5 - 3;

not_enough_positional_arguments

1 positional argument expected by ‘{0}’, but 0 found.

1 positional argument expected, but 0 found.

{0} positional arguments expected by ‘{2}’, but {1} found.

{0} positional arguments expected, but {1} found.

Description

The analyzer produces this diagnostic when a method or function invocation has fewer positional arguments than the number of required positional parameters.

Example

The following code produces this diagnostic because f declares two required parameters, but only one argument is provided:

void f(int a, int b) {}
void g() {
  f(0);
}

Common fixes

Add arguments corresponding to the remaining parameters:

void f(int a, int b) {}
void g() {
  f(0, 1);
}

not_initialized_non_nullable_instance_field

Non-nullable instance field ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a field is declared and has all these characteristics:

• Has a type that’s potentially non-nullable
• Doesn’t have an initializer
• Isn’t marked as late

Examples

The following code produces this diagnostic because x is implicitly initialized to null when it isn’t allowed to be null:

class C {
  int x;
}

Similarly, the following code produces this diagnostic because x is implicitly initialized to null, when it isn’t allowed to be null, by one of the constructors, even though it’s initialized by other constructors:

class C {
  int x;

  C(this.x);

  C.n();
}

Common fixes

If there’s a reasonable default value for the field that’s the same for all instances, then add an initializer expression:

class C {
  int x = 0;
}

If the value of the field should be provided when an instance is created, then add a constructor that sets the value of the field or update an existing constructor:

class C {
  int x;

  C(this.x);
}

You can also mark the field as late, which removes the diagnostic, but if the field isn’t assigned a value before it’s accessed, then it results in an exception being thrown at runtime. This approach should only be used if you’re sure that the field will always be assigned before it’s referenced.

class C {
  late int x;
}

not_initialized_non_nullable_variable

The non-nullable variable ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a static field or top-level variable has a type that’s non-nullable and doesn’t have an initializer. Fields and variables that don’t have an initializer are normally initialized to null, but the type of the field or variable doesn’t allow it to be set to null, so an explicit initializer must be provided.

Examples

The following code produces this diagnostic because the field f can’t be initialized to null:

class C {
  static int f;
}

Similarly, the following code produces this diagnostic because the top-level variable v can’t be initialized to null:

int v;

Common fixes

If the field or variable can’t be initialized to null, then add an initializer that sets it to a non-null value:

class C {
  static int f = 0;
}

If the field or variable should be initialized to null, then change the type to be nullable:

int? v;

If the field or variable can’t be initialized in the declaration but will always be initialized before it’s referenced, then mark it as being late:

class C {
  static late int f;
}

not_iterable_spread

Spread elements in list or set literals must implement ‘Iterable’.

Description

The analyzer produces this diagnostic when the static type of the expression of a spread element that appears in either a list literal or a set literal doesn’t implement the type Iterable.

Example

The following code produces this diagnostic:

var m = <String, int>{'a': 0, 'b': 1};
var s = <String>{...m};

Common fixes

The most common fix is to replace the expression with one that produces an iterable object:

var m = <String, int>{'a': 0, 'b': 1};
var s = <String>{...m.keys};

not_map_spread

Spread elements in map literals must implement ‘Map’.

Description

The analyzer produces this diagnostic when the static type of the expression of a spread element that appears in a map literal doesn’t implement the type Map.

Example

The following code produces this diagnostic because l isn’t a Map:

var l =  <String>['a', 'b'];
var m = <int, String>{...l};

Common fixes

The most common fix is to replace the expression with one that produces a map:

var l =  <String>['a', 'b'];
var m = <int, String>{...l.asMap()};

no_annotation_constructor_arguments

Annotation creation must have arguments.

Description

The analyzer produces this diagnostic when an annotation consists of a single identifier, but that identifier is the name of a class rather than a variable. To create an instance of the class, the identifier must be followed by an argument list.

Example

The following code produces this diagnostic because C is a class, and a class can’t be used as an annotation without invoking a const constructor from the class:

class C {
  const C();
}

@C
var x;

Common fixes

Add the missing argument list:

class C {
  const C();
}

@C()
var x;

no_combined_super_signature

Can’t infer missing types in ‘{0}’ from overridden methods: {1}.

Description

The analyzer produces this diagnostic when there is a method declaration for which one or more types needs to be inferred, and those types can’t be inferred because none of the overridden methods has a function type that is a supertype of all the other overridden methods, as specified by override inference.

Example

The following code produces this diagnostic because the method m declared in the class C is missing both the return type and the type of the parameter a, and neither of the missing types can be inferred for it:

abstract class A {
  A m(String a);
}

abstract class B {
  B m(int a);
}

abstract class C implements A, B {
  m(a);
}

In this example, override inference can’t be performed because the overridden methods are incompatible in these ways:

• Neither parameter type (String and int) is a supertype of the other.
• Neither return type is a subtype of the other.

Common fixes

If possible, add types to the method in the subclass that are consistent with the types from all the overridden methods:

abstract class A {
  A m(String a);
}

abstract class B {
  B m(int a);
}

abstract class C implements A, B {
  C m(Object a);
}

no_generative_constructors_in_superclass

The class ‘{0}’ can’t extend ‘{1}’ because ‘{1}’ only has factory constructors (no generative constructors), and ‘{0}’ has at least one generative constructor.

Description

The analyzer produces this diagnostic when a class that has at least one generative constructor (whether explicit or implicit) has a superclass that doesn’t have any generative constructors. Every generative constructor, except the one defined in Object, invokes, either explicitly or implicitly, one of the generative constructors from its superclass.

Example

The following code produces this diagnostic because the class B has an implicit generative constructor that can’t invoke a generative constructor from A because A doesn’t have any generative constructors:

class A {
  factory A.none() => throw '';
}

class B extends A {}

Common fixes

If the superclass should have a generative constructor, then add one:

class A {
  A();
  factory A.none() => throw '';
}

class B extends A {}

If the subclass shouldn’t have a generative constructor, then remove it by adding a factory constructor:

class A {
  factory A.none() => throw '';
}

class B extends A {
  factory B.none() => throw '';
}

If the subclass must have a generative constructor but the superclass can’t have one, then implement the superclass instead:

class A {
  factory A.none() => throw '';
}

class B implements A {}

nullable_type_in_catch_clause

A potentially nullable type can’t be used in an ‘on’ clause because it isn’t valid to throw a nullable expression.

Description

The analyzer produces this diagnostic when the type following on in a catch clause is a nullable type. It isn’t valid to specify a nullable type because it isn’t possible to catch null (because it’s a runtime error to throw null).

Example

The following code produces this diagnostic because the exception type is specified to allow null when null can’t be thrown:

void f() {
  try {
    // ...
  } on FormatException? {
  }
}

Common fixes

Remove the question mark from the type:

void f() {
  try {
    // ...
  } on FormatException {
  }
}

nullable_type_in_extends_clause

A class can’t extend a nullable type.

Description

The analyzer produces this diagnostic when a class declaration uses an extends clause to specify a superclass, and the superclass is followed by a ?.

It isn’t valid to specify a nullable superclass because doing so would have no meaning; it wouldn’t change either the interface or implementation being inherited by the class containing the extends clause.

Note, however, that it is valid to use a nullable type as a type argument to the superclass, such as class A extends B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type, and nullable types can’t be used in an extends clause:

class A {}
class B extends A? {}

Common fixes

Remove the question mark from the type:

class A {}
class B extends A {}

nullable_type_in_implements_clause

A class or mixin can’t implement a nullable type.

Description

The analyzer produces this diagnostic when a class or mixin declaration has an implements clause, and an interface is followed by a ?.

It isn’t valid to specify a nullable interface because doing so would have no meaning; it wouldn’t change the interface being inherited by the class containing the implements clause.

Note, however, that it is valid to use a nullable type as a type argument to the interface, such as class A implements B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type, and nullable types can’t be used in an implements clause:

class A {}
class B implements A? {}

Common fixes

Remove the question mark from the type:

class A {}
class B implements A {}

nullable_type_in_on_clause

A mixin can’t have a nullable type as a superclass constraint.

Description

The analyzer produces this diagnostic when a mixin declaration uses an on clause to specify a superclass constraint, and the class that’s specified is followed by a ?.

It isn’t valid to specify a nullable superclass constraint because doing so would have no meaning; it wouldn’t change the interface being depended on by the mixin containing the on clause.

Note, however, that it is valid to use a nullable type as a type argument to the superclass constraint, such as mixin A on B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type and nullable types can’t be used in an on clause:

class C {}
mixin M on C? {}

Common fixes

Remove the question mark from the type:

class C {}
mixin M on C {}

nullable_type_in_with_clause

A class or mixin can’t mix in a nullable type.

Description

The analyzer produces this diagnostic when a class or mixin declaration has a with clause, and a mixin is followed by a ?.

It isn’t valid to specify a nullable mixin because doing so would have no meaning; it wouldn’t change either the interface or implementation being inherited by the class containing the with clause.

Note, however, that it is valid to use a nullable type as a type argument to the mixin, such as class A with B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type, and nullable types can’t be used in a with clause:

mixin M {}
class C with M? {}

Common fixes

Remove the question mark from the type:

mixin M {}
class C with M {}

null_argument_to_non_null_type

‘{0}’ shouldn’t be called with a null argument for the non-nullable type argument ‘{1}’.

Description

The analyzer produces this diagnostic when null is passed to either the constructor Future.value or the method Completer.complete when the type argument used to create the instance was non-nullable. Even though the type system can’t express this restriction, passing in a null results in a runtime exception.

Example

The following code produces this diagnostic because null is being passed to the constructor Future.value even though the type argument is the non-nullable type String:

Future<String> f() {
  return Future.value(null);
}

Common fixes

Pass in a non-null value:

Future<String> f() {
  return Future.value('');
}

null_check_always_fails

This null-check will always throw an exception because the expression will always evaluate to ‘null’.

Description

The analyzer produces this diagnostic when the null check operator (!) is used on an expression whose value can only be null. In such a case the operator always throws an exception, which likely isn’t the intended behavior.

Example

The following code produces this diagnostic because the function g will always return null, which means that the null check in f will always throw:

void f() {
  g()!;
}

Null g() => null;

Common fixes

If you intend to always throw an exception, then replace the null check with an explicit throw expression to make the intent more clear:

void f() {
  g();
  throw TypeError();
}

Null g() => null;

obsolete_colon_for_default_value

Using a colon as the separator before a default value is no longer supported.

Description

The analyzer produces this diagnostic when a colon (:) is used as the separator before the default value of an optional named parameter. While this syntax used to be allowed, it was removed in favor of using an equal sign (=).

Example

The following code produces this diagnostic because a colon is being used before the default value of the optional parameter i:

void f({int i : 0}) {}

Common fixes

Replace the colon with an equal sign:

void f({int i = 0}) {}

on_repeated

The type ‘{0}’ can be included in the superclass constraints only once.

Description

The analyzer produces this diagnostic when the same type is listed in the superclass constraints of a mixin multiple times.

Example

The following code produces this diagnostic because A is included twice in the superclass constraints for M:

mixin M on A, A {
}

class A {}
class B {}

Common fixes

If a different type should be included in the superclass constraints, then replace one of the occurrences with the other type:

mixin M on A, B {
}

class A {}
class B {}

If no other type was intended, then remove the repeated type name:

mixin M on A {
}

class A {}
class B {}

optional_parameter_in_operator

Optional parameters aren’t allowed when defining an operator.

Description

The analyzer produces this diagnostic when one or more of the parameters in an operator declaration are optional.

Example

The following code produces this diagnostic because the parameter other is an optional parameter:

class C {
  C operator +([C? other]) => this;
}

Common fixes

Make all of the parameters be required parameters:

class C {
  C operator +(C other) => this;
}

override_on_non_overriding_member

The field doesn’t override an inherited getter or setter.

The getter doesn’t override an inherited getter.

The method doesn’t override an inherited method.

The setter doesn’t override an inherited setter.

Description

The analyzer produces this diagnostic when a class member is annotated with the @override annotation, but the member isn’t declared in any of the supertypes of the class.

Example

The following code produces this diagnostic because m isn’t declared in any of the supertypes of C:

class C {
  @override
  String m() => '';
}

Common fixes

If the member is intended to override a member with a different name, then update the member to have the same name:

class C {
  @override
  String toString() => '';
}

If the member is intended to override a member that was removed from the superclass, then consider removing the member from the subclass.

If the member can’t be removed, then remove the annotation.

packed_annotation

Structs must have at most one ‘Packed’ annotation.

Description

The analyzer produces this diagnostic when a subclass of Struct has more than one Packed annotation.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class C, which is a subclass of Struct, has two Packed annotations:

import 'dart:ffi';

@Packed(1)
@Packed(1)
final class C extends Struct {
  external Pointer<Uint8> notEmpty;
}

Common fixes

Remove all but one of the annotations:

import 'dart:ffi';

@Packed(1)
final class C extends Struct {
  external Pointer<Uint8> notEmpty;
}

packed_annotation_alignment

Only packing to 1, 2, 4, 8, and 16 bytes is supported.

Description

The analyzer produces this diagnostic when the argument to the Packed annotation isn’t one of the allowed values: 1, 2, 4, 8, or 16.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the argument to the Packed annotation (3) isn’t one of the allowed values:

import 'dart:ffi';

@Packed(3)
final class C extends Struct {
  external Pointer<Uint8> notEmpty;
}

Common fixes

Change the alignment to be one of the allowed values:

import 'dart:ffi';

@Packed(4)
final class C extends Struct {
  external Pointer<Uint8> notEmpty;
}

part_of_different_library

Expected this library to be part of ‘{0}’, not ‘{1}’.

Description

The analyzer produces this diagnostic when a library attempts to include a file as a part of itself when the other file is a part of a different library.

Example

Given a file part.dart containing

part of 'library.dart';

The following code, in any file other than library.dart, produces this diagnostic because it attempts to include part.dart as a part of itself when part.dart is a part of a different library:

part 'package:a/part.dart';

Common fixes

If the library should be using a different file as a part, then change the URI in the part directive to be the URI of the other file.

If the part file should be a part of this library, then update the URI (or library name) in the part-of directive to be the URI (or name) of the correct library.

part_of_non_part

The included part ‘{0}’ must have a part-of directive.

Description

The analyzer produces this diagnostic when a part directive is found and the referenced file doesn’t have a part-of directive.

Example

Given a file a.dart containing:

class A {}

The following code produces this diagnostic because a.dart doesn’t contain a part-of directive:

part 'a.dart';

Common fixes

If the referenced file is intended to be a part of another library, then add a part-of directive to the file:

part of 'test.dart';

class A {}

If the referenced file is intended to be a library, then replace the part directive with an import directive:

import 'a.dart';

part_of_unnamed_library

The library is unnamed. A URI is expected, not a library name ‘{0}’, in the part-of directive.

Description

The analyzer produces this diagnostic when a library that doesn’t have a library directive (and hence has no name) contains a part directive and the part of directive in the part file uses a name to specify the library that it’s a part of.

Example

Given a part file named part_file.dart containing the following code:

part of lib;

The following code produces this diagnostic because the library including the part file doesn’t have a name even though the part file uses a name to specify which library it’s a part of:

part 'part_file.dart';

Common fixes

Change the part of directive in the part file to specify its library by URI:

part of 'test.dart';

path_does_not_exist

The path ‘{0}’ doesn’t exist.

Description

The analyzer produces this diagnostic when a dependency has a path key referencing a directory that doesn’t exist.

Example

Assuming that the directory doesNotExist doesn’t exist, the following code produces this diagnostic because it’s listed as the path of a package:

name: example
dependencies:
  local_package:
    path: doesNotExist

Common fixes

If the path is correct, then create a directory at that path.

If the path isn’t correct, then change the path to match the path to the root of the package.

path_not_posix

The path ‘{0}’ isn’t a POSIX-style path.

Description

The analyzer produces this diagnostic when a dependency has a path key whose value is a string, but isn’t a POSIX-style path.

Example

The following code produces this diagnostic because the path following the path key is a Windows path:

name: example
dependencies:
  local_package:
    path: E:\local_package

Common fixes

Convert the path to a POSIX path.

path_pubspec_does_not_exist

The directory ‘{0}’ doesn’t contain a pubspec.

Description

The analyzer produces this diagnostic when a dependency has a path key that references a directory that doesn’t contain a pubspec.yaml file.

Example

Assuming that the directory local_package doesn’t contain a file pubspec.yaml, the following code produces this diagnostic because it’s listed as the path of a package:

name: example
dependencies:
  local_package:
    path: local_package

Common fixes

If the path is intended to be the root of a package, then add a pubspec.yaml file in the directory:

name: local_package

If the path is wrong, then replace it with the correct path.

pattern_assignment_not_local_variable

Only local variables can be assigned in pattern assignments.

Description

The analyzer produces this diagnostic when a pattern assignment assigns a value to anything other than a local variable. Patterns can’t assign to fields or top-level variables.

Example

If the code is cleaner when destructuring with a pattern, then rewrite the code to assign the value to a local variable in a pattern declaration, assigning the non-local variable separately:

class C {
  var x = 0;

  void f((int, int) r) {
    (x, _) = r;
  }
}

Common fixes

If the code is cleaner when using a pattern assignment, then rewrite the code to assign the value to a local variable, assigning the non-local variable separately:

class C {
  var x = 0;

  void f((int, int) r) {
    var (a, _) = r;
    x = a;
  }
}

If the code is cleaner without using a pattern assignment, then rewrite the code to not use a pattern assignment:

class C {
  var x = 0;

  void f((int, int) r) {
    x = r.$1;
  }
}

pattern_constant_from_deferred_library

Constant values from a deferred library can’t be used in patterns.

Description

The analyzer produces this diagnostic when a pattern contains a value declared in a different library, and that library is imported using a deferred import. Constants are evaluated at compile time, but values from deferred libraries aren’t available at compile time.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines the constant zero:

const zero = 0;

The following code produces this diagnostic because the constant pattern a.zero is imported using a deferred import:

import 'a.dart' deferred as a;

void f(int x) {
  switch (x) {
    case a.zero:
      // ...
      break;
  }
}

Common fixes

If you need to reference the constant from the imported library, then remove the deferred keyword:

import 'a.dart' as a;

void f(int x) {
  switch (x) {
    case a.zero:
      // ...
      break;
  }
}

If you need to reference the constant from the imported library and also need the imported library to be deferred, then rewrite the switch statement as a sequence of if statements:

import 'a.dart' deferred as a;

void f(int x) {
  if (x == a.zero) {
    // ...
  }
}

If you don’t need to reference the constant, then replace the case expression:

void f(int x) {
  switch (x) {
    case 0:
      // ...
      break;
  }
}

pattern_type_mismatch_in_irrefutable_context

The matched value of type ‘{0}’ isn’t assignable to the required type ‘{1}’.

Description

The analyzer produces this diagnostic when the type of the value on the right-hand side of a pattern assignment or pattern declaration doesn’t match the type required by the pattern being used to match it.

Example

The following code produces this diagnostic because x might not be a String and hence might not match the object pattern:

void f(Object x) {
  var String(length: a) = x;
  print(a);
}

Common fixes

Change the code so that the type of the expression on the right-hand side matches the type required by the pattern:

void f(String x) {
  var String(length: a) = x;
  print(a);
}

pattern_variable_assignment_inside_guard

Pattern variables can’t be assigned inside the guard of the enclosing guarded pattern.

Description

The analyzer produces this diagnostic when a pattern variable is assigned a value inside a guard (when) clause.

Example

The following code produces this diagnostic because the variable a is assigned a value inside the guard clause:

void f(int x) {
  if (x case var a when (a = 1) > 0) {
    print(a);
  }
}

Common fixes

If there’s a value you need to capture, then assign it to a different variable:

void f(int x) {
  var b;
  if (x case var a when (b = 1) > 0) {
    print(a + b);
  }
}

If there isn’t a value you need to capture, then remove the assignment:

void f(int x) {
  if (x case var a when 1 > 0) {
    print(a);
  }
}

platform_value_disallowed

Keys in the platforms field can’t have values.

Description

The analyzer produces this diagnostic when a key in the platforms map has a value. To learn more about specifying your package’s supported platforms, check out the documentation on platform declarations.

Example

The following pubspec.yaml produces this diagnostic because the key web has a value.

name: example
platforms:
  web: "chrome"

Common fixes

Omit the value and leave the key without a value:

name: example
platforms:
  web:

Values for keys in the platforms field are currently reserved for potential future behavior.

positional_field_in_object_pattern

Object patterns can only use named fields.

Description

The analyzer produces this diagnostic when an object pattern contains a field without specifying the getter name. Object pattern fields match against values that the object’s getters return. Without a getter name specified, the pattern field can’t access a value to attempt to match against.

Example

The following code produces this diagnostic because the object pattern String(1) doesn’t specify which getter of String to access and compare with the value 1:

void f(Object o) {
  if (o case String(1)) {}
}

Common fixes

Add the getter name to access the value, followed by a colon before the pattern to match against:

void f(Object o) {
  if (o case String(length: 1)) {}
}

positional_super_formal_parameter_with_positional_argument

Positional super parameters can’t be used when the super constructor invocation has a positional argument.

Description

The analyzer produces this diagnostic when some, but not all, of the positional parameters provided to the constructor of the superclass are using a super parameter.

Positional super parameters are associated with positional parameters in the super constructor by their index. That is, the first super parameter is associated with the first positional parameter in the super constructor, the second with the second, and so on. The same is true for positional arguments. Having both positional super parameters and positional arguments means that there are two values associated with the same parameter in the superclass’s constructor, and hence isn’t allowed.

Example

The following code produces this diagnostic because the constructor B.new is using a super parameter to pass one of the required positional parameters to the super constructor in A, but is explicitly passing the other in the super constructor invocation:

class A {
  A(int x, int y);
}

class B extends A {
  B(int x, super.y) : super(x);
}

Common fixes

If all the positional parameters can be super parameters, then convert the normal positional parameters to be super parameters:

class A {
  A(int x, int y);
}

class B extends A {
  B(super.x, super.y);
}

If some positional parameters can’t be super parameters, then convert the super parameters to be normal parameters:

class A {
  A(int x, int y);
}

class B extends A {
  B(int x, int y) : super(x, y);
}

prefix_collides_with_top_level_member

The name ‘{0}’ is already used as an import prefix and can’t be used to name a top-level element.

Description

The analyzer produces this diagnostic when a name is used as both an import prefix and the name of a top-level declaration in the same library.

Example

The following code produces this diagnostic because f is used as both an import prefix and the name of a function:

import 'dart:math' as f;

int f() => f.min(0, 1);

Common fixes

If you want to use the name for the import prefix, then rename the top-level declaration:

import 'dart:math' as f;

int g() => f.min(0, 1);

If you want to use the name for the top-level declaration, then rename the import prefix:

import 'dart:math' as math;

int f() => math.min(0, 1);

prefix_identifier_not_followed_by_dot

The name ‘{0}’ refers to an import prefix, so it must be followed by ‘.’.

Description

The analyzer produces this diagnostic when an import prefix is used by itself, without accessing any of the names declared in the libraries associated with the prefix. Prefixes aren’t variables, and therefore can’t be used as a value.

Example

The following code produces this diagnostic because the prefix math is being used as if it were a variable:

import 'dart:math' as math;

void f() {
  print(math);
}

Common fixes

If the code is incomplete, then reference something in one of the libraries associated with the prefix:

import 'dart:math' as math;

void f() {
  print(math.pi);
}

If the name is wrong, then correct the name.

prefix_shadowed_by_local_declaration

The prefix ‘{0}’ can’t be used here because it’s shadowed by a local declaration.

Description

The analyzer produces this diagnostic when an import prefix is used in a context where it isn’t visible because it was shadowed by a local declaration.

Example

The following code produces this diagnostic because the prefix a is being used to access the class Future, but isn’t visible because it’s shadowed by the parameter a:

import 'dart:async' as a;

a.Future? f(int a) {
  a.Future? x;
  return x;
}

Common fixes

Rename either the prefix:

import 'dart:async' as p;

p.Future? f(int a) {
  p.Future? x;
  return x;
}

Or rename the local variable:

import 'dart:async' as a;

a.Future? f(int p) {
  a.Future? x;
  return x;
}

private_collision_in_mixin_application

The private name ‘{0}’, defined by ‘{1}’, conflicts with the same name defined by ‘{2}’.

Description

The analyzer produces this diagnostic when two mixins that define the same private member are used together in a single class in a library other than the one that defines the mixins.

Example

Given a file a.dart containing the following code:

mixin A {
  void _foo() {}
}

mixin B {
  void _foo() {}
}

The following code produces this diagnostic because the mixins A and B both define the method _foo:

import 'a.dart';

class C extends Object with A, B {}

Common fixes

If you don’t need both of the mixins, then remove one of them from the with clause:

import 'a.dart';

class C extends Object with A, B {}

If you need both of the mixins, then rename the conflicting member in one of the two mixins.

private_optional_parameter

Named parameters can’t start with an underscore.

Description

The analyzer produces this diagnostic when the name of a named parameter starts with an underscore.

Example

The following code produces this diagnostic because the named parameter _x starts with an underscore:

class C {
  void m({int _x = 0}) {}
}

Common fixes

Rename the parameter so that it doesn’t start with an underscore:

class C {
  void m({int x = 0}) {}
}

private_setter

The setter ‘{0}’ is private and can’t be accessed outside the library that declares it.

Description

The analyzer produces this diagnostic when a private setter is used in a library where it isn’t visible.

Example

Given a file a.dart that contains the following:

class A {
  static int _f = 0;
}

The following code produces this diagnostic because it references the private setter _f even though the setter isn’t visible:

import 'a.dart';

void f() {
  A._f = 0;
}

Common fixes

If you’re able to make the setter public, then do so:

class A {
  static int f = 0;
}

If you aren’t able to make the setter public, then find a different way to implement the code.

read_potentially_unassigned_final

The final variable ‘{0}’ can’t be read because it’s potentially unassigned at this point.

Description

The analyzer produces this diagnostic when a final local variable that isn’t initialized at the declaration site is read at a point where the compiler can’t prove that the variable is always initialized before it’s referenced.

Example

The following code produces this diagnostic because the final local variable x is read (on line 3) when it’s possible that it hasn’t yet been initialized:

int f() {
  final int x;
  return x;
}

Common fixes

Ensure that the variable has been initialized before it’s read:

int f(bool b) {
  final int x;
  if (b) {
    x = 0;
  } else {
    x = 1;
  }
  return x;
}

record_literal_one_positional_no_trailing_comma

A record literal with exactly one positional field requires a trailing comma.

Description

The analyzer produces this diagnostic when a record literal with a single positional field doesn’t have a trailing comma after the field.

In some locations a record literal with a single positional field could also be a parenthesized expression. A trailing comma is required to disambiguate these two valid interpretations.

Example

The following code produces this diagnostic because the record literal has one positional field but doesn’t have a trailing comma:

var r = const (1);

Common fixes

Add a trailing comma:

var r = const (1,);

record_type_one_positional_no_trailing_comma

A record type with exactly one positional field requires a trailing comma.

Description

The analyzer produces this diagnostic when a record type annotation with a single positional field doesn’t have a trailing comma after the field.

In some locations a record type with a single positional field could also be a parenthesized expression. A trailing comma is required to disambiguate these two valid interpretations.

Example

The following code produces this diagnostic because the record type has one positional field, but doesn’t have a trailing comma:

void f((int) r) {}

Common fixes

Add a trailing comma:

void f((int,) r) {}

recursive_compile_time_constant

The compile-time constant expression depends on itself.

Description

The analyzer produces this diagnostic when the value of a compile-time constant is defined in terms of itself, either directly or indirectly, creating an infinite loop.

Example

The following code produces this diagnostic twice because both of the constants are defined in terms of the other:

const secondsPerHour = minutesPerHour * 60;
const minutesPerHour = secondsPerHour / 60;

Common fixes

Break the cycle by finding an alternative way of defining at least one of the constants:

const secondsPerHour = minutesPerHour * 60;
const minutesPerHour = 60;

recursive_constructor_redirect

Constructors can’t redirect to themselves either directly or indirectly.

Description

The analyzer produces this diagnostic when a constructor redirects to itself, either directly or indirectly, creating an infinite loop.

Examples

The following code produces this diagnostic because the generative constructors C.a and C.b each redirect to the other:

class C {
  C.a() : this.b();
  C.b() : this.a();
}

The following code produces this diagnostic because the factory constructors A and B each redirect to the other:

abstract class A {
  factory A() = B;
}
class B implements A {
  factory B() = A;
  B.named();
}

Common fixes

In the case of generative constructors, break the cycle by finding defining at least one of the constructors to not redirect to another constructor:

class C {
  C.a() : this.b();
  C.b();
}

In the case of factory constructors, break the cycle by defining at least one of the factory constructors to do one of the following:

• Redirect to a generative constructor:

abstract class A {
  factory A() = B;
}
class B implements A {
  factory B() = B.named;
  B.named();
}

• Not redirect to another constructor:

abstract class A {
  factory A() = B;
}
class B implements A {
  factory B() {
    return B.named();
  }

  B.named();
}

• Not be a factory constructor:

abstract class A {
  factory A() = B;
}
class B implements A {
  B();
  B.named();
}

recursive_interface_inheritance

‘{0}’ can’t be a superinterface of itself: {1}.

‘{0}’ can’t extend itself.

‘{0}’ can’t implement itself.

‘{0}’ can’t use itself as a mixin.

‘{0}’ can’t use itself as a superclass constraint.

Description

The analyzer produces this diagnostic when there’s a circularity in the type hierarchy. This happens when a type, either directly or indirectly, is declared to be a subtype of itself.

Example

The following code produces this diagnostic because the class A is declared to be a subtype of B, and B is a subtype of A:

class A extends B {}
class B implements A {}

Common fixes

Change the type hierarchy so that there’s no circularity.

redeclare_on_non_redeclaring_member

The {0} doesn’t redeclare a {0} declared in a superinterface.

Description

The analyzer produces this diagnostic when a member of an extension type is annotated with @redeclare, but none of the implemented interfaces has a member with the same name.

Example

The following code produces this diagnostic because the member n declared by the extension type E is annotated with @redeclare, but C doesn’t have a member named n:

import 'package:meta/meta.dart';

class C {
  void m() {}
}

extension type E(C c) implements C {
  @redeclare
  void n() {}
}

Common fixes

If the annotated member has the right name, then remove the annotation:

class C {
  void m() {}
}

extension type E(C c) implements C {
  void n() {}
}

If the annotated member is suppose to replace a member from the implemented interfaces, then change the name of the annotated member to match the member being replaced:

import 'package:meta/meta.dart';

class C {
  void m() {}
}

extension type E(C c) implements C {
  @redeclare
  void m() {}
}

redirect_generative_to_missing_constructor

The constructor ‘{0}’ couldn’t be found in ‘{1}’.

Description

The analyzer produces this diagnostic when a generative constructor redirects to a constructor that isn’t defined.

Example

The following code produces this diagnostic because the constructor C.a redirects to the constructor C.b, but C.b isn’t defined:

class C {
  C.a() : this.b();
}

Common fixes

If the missing constructor must be called, then define it:

class C {
  C.a() : this.b();
  C.b();
}

If the missing constructor doesn’t need to be called, then remove the redirect:

class C {
  C.a();
}

redirect_generative_to_non_generative_constructor

Generative constructors can’t redirect to a factory constructor.

Description

The analyzer produces this diagnostic when a generative constructor redirects to a factory constructor.

Example

The following code produces this diagnostic because the generative constructor C.a redirects to the factory constructor C.b:

class C {
  C.a() : this.b();
  factory C.b() => C.a();
}

Common fixes

If the generative constructor doesn’t need to redirect to another constructor, then remove the redirect.

class C {
  C.a();
  factory C.b() => C.a();
}

If the generative constructor must redirect to another constructor, then make the other constructor be a generative (non-factory) constructor:

class C {
  C.a() : this.b();
  C.b();
}

redirect_to_abstract_class_constructor

The redirecting constructor ‘{0}’ can’t redirect to a constructor of the abstract class ‘{1}’.

Description

The analyzer produces this diagnostic when a constructor redirects to a constructor in an abstract class.

Example

The following code produces this diagnostic because the factory constructor in A redirects to a constructor in B, but B is an abstract class:

class A {
  factory A() = B;
}

abstract class B implements A {}

Common fixes

If the code redirects to the correct constructor, then change the class so that it isn’t abstract:

class A {
  factory A() = B;
}

class B implements A {}

Otherwise, change the factory constructor so that it either redirects to a constructor in a concrete class, or has a concrete implementation.

redirect_to_invalid_function_type

The redirected constructor ‘{0}’ has incompatible parameters with ‘{1}’.

Description

The analyzer produces this diagnostic when a factory constructor attempts to redirect to another constructor, but the two have incompatible parameters. The parameters are compatible if all of the parameters of the redirecting constructor can be passed to the other constructor and if the other constructor doesn’t require any parameters that aren’t declared by the redirecting constructor.

Examples

The following code produces this diagnostic because the constructor for A doesn’t declare a parameter that the constructor for B requires:

abstract class A {
  factory A() = B;
}

class B implements A {
  B(int x);
  B.zero();
}

The following code produces this diagnostic because the constructor for A declares a named parameter (y) that the constructor for B doesn’t allow:

abstract class A {
  factory A(int x, {int y}) = B;
}

class B implements A {
  B(int x);
}

Common fixes

If there’s a different constructor that is compatible with the redirecting constructor, then redirect to that constructor:

abstract class A {
  factory A() = B.zero;
}

class B implements A {
  B(int x);
  B.zero();
}

Otherwise, update the redirecting constructor to be compatible:

abstract class A {
  factory A(int x) = B;
}

class B implements A {
  B(int x);
}

redirect_to_invalid_return_type

The return type ‘{0}’ of the redirected constructor isn’t a subtype of ‘{1}’.

Description

The analyzer produces this diagnostic when a factory constructor redirects to a constructor whose return type isn’t a subtype of the type that the factory constructor is declared to produce.

Example

The following code produces this diagnostic because A isn’t a subclass of C, which means that the value returned by the constructor A() couldn’t be returned from the constructor C():

class A {}

class B implements C {}

class C {
  factory C() = A;
}

Common fixes

If the factory constructor is redirecting to a constructor in the wrong class, then update the factory constructor to redirect to the correct constructor:

class A {}

class B implements C {}

class C {
  factory C() = B;
}

If the class defining the constructor being redirected to is the class that should be returned, then make it a subtype of the factory’s return type:

class A implements C {}

class B implements C {}

class C {
  factory C() = A;
}

redirect_to_missing_constructor

The constructor ‘{0}’ couldn’t be found in ‘{1}’.

Description

The analyzer produces this diagnostic when a constructor redirects to a constructor that doesn’t exist.

Example

The following code produces this diagnostic because the factory constructor in A redirects to a constructor in B that doesn’t exist:

class A {
  factory A() = B.name;
}

class B implements A {
  B();
}

Common fixes

If the constructor being redirected to is correct, then define the constructor:

class A {
  factory A() = B.name;
}

class B implements A {
  B();
  B.name();
}

If a different constructor should be invoked, then update the redirect:

class A {
  factory A() = B;
}

class B implements A {
  B();
}

redirect_to_non_class

The name ‘{0}’ isn’t a type and can’t be used in a redirected constructor.

Description

One way to implement a factory constructor is to redirect to another constructor by referencing the name of the constructor. The analyzer produces this diagnostic when the redirect is to something other than a constructor.

Example

The following code produces this diagnostic because f is a function:

C f() => throw 0;

class C {
  factory C() = f;
}

Common fixes

If the constructor isn’t defined, then either define it or replace it with a constructor that is defined.

If the constructor is defined but the class that defines it isn’t visible, then you probably need to add an import.

If you’re trying to return the value returned by a function, then rewrite the constructor to return the value from the constructor’s body:

C f() => throw 0;

class C {
  factory C() => f();
}

redirect_to_non_const_constructor

A constant redirecting constructor can’t redirect to a non-constant constructor.

Description

The analyzer produces this diagnostic when a constructor marked as const redirects to a constructor that isn’t marked as const.

Example

The following code produces this diagnostic because the constructor C.a is marked as const but redirects to the constructor C.b, which isn’t:

class C {
  const C.a() : this.b();
  C.b();
}

Common fixes

If the non-constant constructor can be marked as const, then mark it as const:

class C {
  const C.a() : this.b();
  const C.b();
}

If the non-constant constructor can’t be marked as const, then either remove the redirect or remove const from the redirecting constructor:

class C {
  C.a() : this.b();
  C.b();
}

redirect_to_type_alias_expands_to_type_parameter

A redirecting constructor can’t redirect to a type alias that expands to a type parameter.

Description

The analyzer produces this diagnostic when a redirecting factory constructor redirects to a type alias, and the type alias expands to one of the type parameters of the type alias. This isn’t allowed because the value of the type parameter is a type rather than a class.

Example

The following code produces this diagnostic because the redirect to B<A> is to a type alias whose value is T, even though it looks like the value should be A:

class A implements C {}

typedef B<T> = T;

abstract class C {
  factory C() = B<A>;
}

Common fixes

Use either a class name or a type alias that is defined to be a class rather than a type alias defined to be a type parameter:

class A implements C {}

abstract class C {
  factory C() = A;
}

referenced_before_declaration

Local variable ‘{0}’ can’t be referenced before it is declared.

Description

The analyzer produces this diagnostic when a variable is referenced before it’s declared. In Dart, variables are visible everywhere in the block in which they are declared, but can only be referenced after they are declared.

The analyzer also produces a context message that indicates where the declaration is located.

Example

The following code produces this diagnostic because i is used before it is declared:

void f() {
  print(i);
  int i = 5;
}

Common fixes

If you intended to reference the local variable, move the declaration before the first reference:

void f() {
  int i = 5;
  print(i);
}

If you intended to reference a name from an outer scope, such as a parameter, instance field or top-level variable, then rename the local declaration so that it doesn’t hide the outer variable.

void f(int i) {
  print(i);
  int x = 5;
  print(x);
}

refutable_pattern_in_irrefutable_context

Refutable patterns can’t be used in an irrefutable context.

Description

The analyzer produces this diagnostic when a refutable pattern is used in a context where only an irrefutable pattern is allowed.

The refutable patterns that are disallowed are:

• logical-or
• relational
• null-check
• constant

The contexts that are checked are:

• pattern-based variable declarations
• pattern-based for loops
• assignments with a pattern on the left-hand side

Example

The following code produces this diagnostic because the null-check pattern, which is a refutable pattern, is in a pattern-based variable declaration, which doesn’t allow refutable patterns:

void f(int? x) {
  var (_?) = x;
}

Common fixes

Rewrite the code to not use a refutable pattern in an irrefutable context.

relational_pattern_operand_type_not_assignable

The constant expression type ‘{0}’ is not assignable to the parameter type ‘{1}’ of the ‘{2}’ operator.

Description

The analyzer produces this diagnostic when the operand of a relational pattern has a type that isn’t assignable to the parameter of the operator that will be invoked.

Example

The following code produces this diagnostic because the operand in the relational pattern (0) is an int, but the > operator defined in C expects an object of type C:

class C {
  const C();

  bool operator >(C other) => true;
}

void f(C c) {
  switch (c) {
    case > 0:
      print('positive');
  }
}

Common fixes

If the switch is using the correct value, then change the case to compare the value to the right type of object:

class C {
  const C();

  bool operator >(C other) => true;
}

void f(C c) {
  switch (c) {
    case > const C():
      print('positive');
  }
}

If the switch is using the wrong value, then change the expression used to compute the value being matched:

class C {
  const C();

  bool operator >(C other) => true;

  int get toInt => 0;
}

void f(C c) {
  switch (c.toInt) {
    case > 0:
      print('positive');
  }
}

relational_pattern_operator_return_type_not_assignable_to_bool

The return type of operators used in relational patterns must be assignable to ‘bool’.

Description

The analyzer produces this diagnostic when a relational pattern references an operator that doesn’t produce a value of type bool.

Example

The following code produces this diagnostic because the operator >, used in the relational pattern > c2, returns a value of type int rather than a bool:

class C {
  const C();

  int operator >(C c) => 3;

  bool operator <(C c) => false;
}

const C c2 = C();

void f(C c1) {
  if (c1 case > c2) {}
}

Common fixes

If there’s a different operator that should be used, then change the operator:

class C {
  const C();

  int operator >(C c) => 3;

  bool operator <(C c) => false;
}

const C c2 = C();

void f(C c1) {
  if (c1 case < c2) {}
}

If the operator is expected to return bool, then update the declaration of the operator:

class C {
  const C();

  bool operator >(C c) => true;

  bool operator <(C c) => false;
}

const C c2 = C();

void f(C c1) {
  if (c1 case > c2) {}
}

rest_element_in_map_pattern

A map pattern can’t contain a rest pattern.

Description

The analyzer produces this diagnostic when a map pattern contains a rest pattern. Map patterns match a map with more keys than those explicitly given in the pattern (as long as the given keys match), so a rest pattern is unnecessary.

Example

The following code produces this diagnostic because the map pattern contains a rest pattern:

void f(Map<int, String> x) {
  if (x case {0: _, ...}) {}
}

Common fixes

Remove the rest pattern:

void f(Map<int, String> x) {
  if (x case {0: _}) {}
}

rethrow_outside_catch

A rethrow must be inside of a catch clause.

Description

The analyzer produces this diagnostic when a rethrow statement is outside a catch clause. The rethrow statement is used to throw a caught exception again, but there’s no caught exception outside of a catch clause.

Example

The following code produces this diagnostic because therethrow statement is outside of a catch clause:

void f() {
  rethrow;
}

Common fixes

If you’re trying to rethrow an exception, then wrap the rethrow statement in a catch clause:

void f() {
  try {
    // ...
  } catch (exception) {
    rethrow;
  }
}

If you’re trying to throw a new exception, then replace the rethrow statement with a throw expression:

void f() {
  throw UnsupportedError('Not yet implemented');
}

return_in_generative_constructor

Constructors can’t return values.

Description

The analyzer produces this diagnostic when a generative constructor contains a return statement that specifies a value to be returned. Generative constructors always return the object that was created, and therefore can’t return a different object.

Example

The following code produces this diagnostic because the return statement has an expression:

class C {
  C() {
    return this;
  }
}

Common fixes

If the constructor should create a new instance, then remove either the return statement or the expression:

class C {
  C();
}

If the constructor shouldn’t create a new instance, then convert it to be a factory constructor:

class C {
  factory C() {
    return _instance;
  }

  static C _instance = C._();

  C._();
}

return_in_generator

Can’t return a value from a generator function that uses the ‘async*’ or ‘sync*’ modifier.

Description

The analyzer produces this diagnostic when a generator function (one whose body is marked with either async* or sync*) uses either a return statement to return a value or implicitly returns a value because of using =>. In any of these cases, they should use yield instead of return.

Examples

The following code produces this diagnostic because the method f is a generator and is using return to return a value:

Iterable<int> f() sync* {
  return 3;
}

The following code produces this diagnostic because the function f is a generator and is implicitly returning a value:

Stream<int> f() async* => 3;

Common fixes

If the function is using => for the body of the function, then convert it to a block function body, and use yield to return a value:

Stream<int> f() async* {
  yield 3;
}

If the method is intended to be a generator, then use yield to return a value:

Iterable<int> f() sync* {
  yield 3;
}

If the method isn’t intended to be a generator, then remove the modifier from the body (or use async if you’re returning a future):

int f() {
  return 3;
}

return_of_do_not_store

‘{0}’ is annotated with ‘doNotStore’ and shouldn’t be returned unless ‘{1}’ is also annotated.

Description

The analyzer produces this diagnostic when a value that is annotated with the doNotStore annotation is returned from a method, getter, or function that doesn’t have the same annotation.

Example

The following code produces this diagnostic because the result of invoking f shouldn’t be stored, but the function g isn’t annotated to preserve that semantic:

import 'package:meta/meta.dart';

@doNotStore
int f() => 0;

int g() => f();

Common fixes

If the value that shouldn’t be stored is the correct value to return, then mark the function with the doNotStore annotation:

import 'package:meta/meta.dart';

@doNotStore
int f() => 0;

@doNotStore
int g() => f();

Otherwise, return a different value from the function:

import 'package:meta/meta.dart';

@doNotStore
int f() => 0;

int g() => 0;

return_of_invalid_type

A value of type ‘{0}’ can’t be returned from the constructor ‘{2}’ because it has a return type of ‘{1}’.

A value of type ‘{0}’ can’t be returned from the function ‘{2}’ because it has a return type of ‘{1}’.

A value of type ‘{0}’ can’t be returned from the method ‘{2}’ because it has a return type of ‘{1}’.

Description

The analyzer produces this diagnostic when a method or function returns a value whose type isn’t assignable to the declared return type.

Example

The following code produces this diagnostic because f has a return type of String but is returning an int:

String f() => 3;

Common fixes

If the return type is correct, then replace the value being returned with a value of the correct type, possibly by converting the existing value:

String f() => 3.toString();

If the value is correct, then change the return type to match:

int f() => 3;

return_of_invalid_type_from_closure

The return type ‘{0}’ isn’t a ‘{1}’, as required by the closure’s context.

Description

The analyzer produces this diagnostic when the static type of a returned expression isn’t assignable to the return type that the closure is required to have.

Example

The following code produces this diagnostic because f is defined to be a function that returns a String, but the closure assigned to it returns an int:

String Function(String) f = (s) => 3;

Common fixes

If the return type is correct, then replace the returned value with a value of the correct type, possibly by converting the existing value:

String Function(String) f = (s) => 3.toString();

return_without_value

The return value is missing after ‘return’.

Description

The analyzer produces this diagnostic when it finds a return statement without an expression in a function that declares a return type.

Example

The following code produces this diagnostic because the function f is expected to return an int, but no value is being returned:

int f() {
  return;
}

Common fixes

Add an expression that computes the value to be returned:

int f() {
  return 0;
}

sdk_version_async_exported_from_core

The class ‘{0}’ wasn’t exported from ‘dart:core’ until version 2.1, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when either the class Future or Stream is referenced in a library that doesn’t import dart:async in code that has an SDK constraint whose lower bound is less than 2.1.0. In earlier versions, these classes weren’t defined in dart:core, so the import was necessary.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.1.0:

environment:
  sdk: '>=2.0.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

void f(Future f) {}

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the classes to be referenced:

environment:
  sdk: '>=2.1.0 <2.4.0'

If you need to support older versions of the SDK, then import the dart:async library.

import 'dart:async';

void f(Future f) {}

sdk_version_as_expression_in_const_context

The use of an as expression in a constant expression wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an as expression inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using an as expression in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const num n = 3;
const int i = n as int;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the expression to be used:

environment:
  sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use an as expression, or change the code so that the as expression isn’t in a constant context:

num x = 3;
int y = x as int;

sdk_version_bool_operator_in_const_context

The use of the operator ‘{0}’ for ‘bool’ operands in a constant context wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when any use of the &, |, or ^ operators on the class bool inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using these operators in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const bool a = true;
const bool b = false;
const bool c = a & b;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the operators to be used:

environment:
 sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use these operators, or change the code so that the expression isn’t in a constant context:

const bool a = true;
const bool b = false;
bool c = a & b;

sdk_version_constructor_tearoffs

Tearing off a constructor requires the ‘constructor-tearoffs’ language feature.

Description

The analyzer produces this diagnostic when a constructor tear-off is found in code that has an SDK constraint whose lower bound is less than 2.15. Constructor tear-offs weren’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.15:

environment:
  sdk: '>=2.9.0 <2.15.0'

In the package that has that pubspec, code like the following produces this diagnostic:

var setConstructor = Set.identity;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the operator to be used:

environment:
  sdk: '>=2.15.0 <2.16.0'

If you need to support older versions of the SDK, then rewrite the code to not use constructor tear-offs:

var setConstructor = () => Set.identity();

sdk_version_eq_eq_operator_in_const_context

Using the operator ‘==’ for non-primitive types wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when the operator == is used on a non-primitive type inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using this operator in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

class C {}
const C a = null;
const C b = null;
const bool same = a == b;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the operator to be used:

environment:
  sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use the == operator, or change the code so that the expression isn’t in a constant context:

class C {}
const C a = null;
const C b = null;
bool same = a == b;

sdk_version_extension_methods

Extension methods weren’t supported until version 2.6.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an extension declaration or an extension override is found in code that has an SDK constraint whose lower bound is less than 2.6.0. Using extensions wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.6.0:

environment:
 sdk: '>=2.4.0 <2.7.0'

In the package that has that pubspec, code like the following produces this diagnostic:

extension E on String {
  void sayHello() {
    print('Hello $this');
  }
}

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.6.0 <2.7.0'

If you need to support older versions of the SDK, then rewrite the code to not make use of extensions. The most common way to do this is to rewrite the members of the extension as top-level functions (or methods) that take the value that would have been bound to this as a parameter:

void sayHello(String s) {
  print('Hello $s');
}

sdk_version_gt_gt_gt_operator

The operator ‘»>’ wasn’t supported until version 2.14.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when the operator >>> is used in code that has an SDK constraint whose lower bound is less than 2.14.0. This operator wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.14.0:

environment:
 sdk: '>=2.0.0 <2.15.0'

In the package that has that pubspec, code like the following produces this diagnostic:

int x = 3 >>> 4;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the operator to be used:

environment:
  sdk: '>=2.14.0 <2.15.0'

If you need to support older versions of the SDK, then rewrite the code to not use the >>> operator:

int x = logicalShiftRight(3, 4);

int logicalShiftRight(int leftOperand, int rightOperand) {
  int divisor = 1 << rightOperand;
  if (divisor == 0) {
    return 0;
  }
  return leftOperand ~/ divisor;
}

sdk_version_is_expression_in_const_context

The use of an is expression in a constant context wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an is expression inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using an is expression in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const Object x = 4;
const y = x is int ? 0 : 1;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the expression to be used:

environment:
  sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use the is operator, or, if that isn’t possible, change the code so that the is expression isn’t in a constant context:

const Object x = 4;
var y = x is int ? 0 : 1;

sdk_version_never

The type ‘Never’ wasn’t supported until version 2.12.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when a reference to the class Never is found in code that has an SDK constraint whose lower bound is less than 2.12.0. This class wasn’t defined in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.12.0:

environment:
  sdk: '>=2.5.0 <2.6.0'

In the package that has that pubspec, code like the following produces this diagnostic:

Never n;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the type to be used:

environment:
  sdk: '>=2.12.0 <2.13.0'

If you need to support older versions of the SDK, then rewrite the code to not reference this class:

dynamic x;

sdk_version_set_literal

Set literals weren’t supported until version 2.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when a set literal is found in code that has an SDK constraint whose lower bound is less than 2.2.0. Set literals weren’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.2.0:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

var s = <int>{};

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.2.0 <2.4.0'

If you do need to support older versions of the SDK, then replace the set literal with code that creates the set without the use of a literal:

var s = new Set<int>();

sdk_version_ui_as_code

The for, if, and spread elements weren’t supported until version 2.3.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when a for, if, or spread element is found in code that has an SDK constraint whose lower bound is less than 2.3.0. Using a for, if, or spread element wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.0:

environment:
  sdk: '>=2.2.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

var digits = [for (int i = 0; i < 10; i++) i];

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.3.0 <2.4.0'

If you need to support older versions of the SDK, then rewrite the code to not make use of those elements:

var digits = _initializeDigits();

List<int> _initializeDigits() {
  var digits = <int>[];
  for (int i = 0; i < 10; i++) {
    digits.add(i);
  }
  return digits;
}

sdk_version_ui_as_code_in_const_context

The if and spread elements weren’t supported in constant expressions until version 2.5.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an if or spread element inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.5.0. Using an if or spread element inside a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Example

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.5.0:

environment:
  sdk: '>=2.4.0 <2.6.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const a = [1, 2];
const b = [...a];

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.5.0 <2.6.0'

If you need to support older versions of the SDK, then rewrite the code to not make use of those elements:

const a = [1, 2];
const b = [1, 2];

If that isn’t possible, change the code so that the element isn’t in a constant context:

const a = [1, 2];
var b = [...a];

set_element_type_not_assignable

The element type ‘{0}’ can’t be assigned to the set type ‘{1}’.

Description

The analyzer produces this diagnostic when an element in a set literal has a type that isn’t assignable to the element type of the set.

Example

The following code produces this diagnostic because the type of the string literal '0' is String, which isn’t assignable to int, the element type of the set:

var s = <int>{'0'};

Common fixes

If the element type of the set literal is wrong, then change the element type of the set:

var s = <String>{'0'};

If the type of the element is wrong, then change the element:

var s = <int>{'0'.length};

shared_deferred_prefix

The prefix of a deferred import can’t be used in other import directives.

Description

The analyzer produces this diagnostic when a prefix in a deferred import is also used as a prefix in other imports (whether deferred or not). The prefix in a deferred import can’t be shared with other imports because the prefix is used to load the imported library.

Example

The following code produces this diagnostic because the prefix x is used as the prefix for a deferred import and is also used for one other import:

import 'dart:math' deferred as x;
import 'dart:convert' as x;

var y = x.json.encode(x.min(0, 1));

Common fixes

If you can use a different name for the deferred import, then do so:

import 'dart:math' deferred as math;
import 'dart:convert' as x;

var y = x.json.encode(math.min(0, 1));

If you can use a different name for the other imports, then do so:

import 'dart:math' deferred as x;
import 'dart:convert' as convert;

var y = convert.json.encode(x.min(0, 1));

size_annotation_dimensions

‘Array’s must have an ‘Array’ annotation that matches the dimensions.

Description

The analyzer produces this diagnostic when the number of dimensions specified in an Array annotation doesn’t match the number of nested arrays specified by the type of a field.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the field a0 has a type with three nested arrays, but only two dimensions are given in the Array annotation:

import 'dart:ffi';

final class C extends Struct {
  @Array(8, 8)
  external Array<Array<Array<Uint8>>> a0;
}

Common fixes

If the type of the field is correct, then fix the annotation to have the required number of dimensions:

import 'dart:ffi';

final class C extends Struct {
  @Array(8, 8, 4)
  external Array<Array<Array<Uint8>>> a0;
}

If the type of the field is wrong, then fix the type of the field:

import 'dart:ffi';

final class C extends Struct {
  @Array(8, 8)
  external Array<Array<Uint8>> a0;
}

static_access_to_instance_member

Instance member ‘{0}’ can’t be accessed using static access.

Description

The analyzer produces this diagnostic when a class name is used to access an instance field. Instance fields don’t exist on a class; they exist only on an instance of the class.

Example

The following code produces this diagnostic because x is an instance field:

class C {
  static int a;

  int b;
}

int f() => C.b;

Common fixes

If you intend to access a static field, then change the name of the field to an existing static field:

class C {
  static int a;

  int b;
}

int f() => C.a;

If you intend to access the instance field, then use an instance of the class to access the field:

class C {
  static int a;

  int b;
}

int f(C c) => c.b;

subtype_of_base_or_final_is_not_base_final_or_sealed

The mixin ‘{0}’ must be ‘base’ because the supertype ‘{1}’ is ‘base’.

The mixin ‘{0}’ must be ‘base’ because the supertype ‘{1}’ is ‘final’.

The type ‘{0}’ must be ‘base’, ‘final’ or ‘sealed’ because the supertype ‘{1}’ is ‘base’.

The type ‘{0}’ must be ‘base’, ‘final’ or ‘sealed’ because the supertype ‘{1}’ is ‘final’.

Description

The analyzer produces this diagnostic when a class or mixin has a direct or indirect supertype that is either base or final, but the class or mixin itself isn’t marked either base, final, or sealed.

Example

The following code produces this diagnostic because the class B is a subtype of A, and A is a base class, but B is neither base, final or sealed:

base class A {}
class B extends A {}

Common fixes

Add either base, final or sealed to the class or mixin declaration:

base class A {}
final class B extends A {}

subtype_of_deferred_class

Classes and mixins can’t implement deferred classes.

Classes can’t extend deferred classes.

Classes can’t mixin deferred classes.

Description

The analyzer produces this diagnostic when a type (class or mixin) is a subtype of a class from a library being imported using a deferred import. The supertypes of a type must be compiled at the same time as the type, and classes from deferred libraries aren’t compiled until the library is loaded.

For more information, check out Lazily loading a library.

Example

Given a file a.dart that defines the class A:

class A {}

The following code produces this diagnostic because the superclass of B is declared in a deferred library:

import 'a.dart' deferred as a;

class B extends a.A {}

Common fixes

If you need to create a subtype of a type from the deferred library, then remove the deferred keyword:

import 'a.dart' as a;

class B extends a.A {}

subtype_of_disallowed_type

‘{0}’ can’t be used as a superclass constraint.

Classes and mixins can’t implement ‘{0}’.

Classes can’t extend ‘{0}’.

Classes can’t mixin ‘{0}’.

Description

The analyzer produces this diagnostic when one of the restricted classes is used in either an extends, implements, with, or on clause. The classes bool, double, FutureOr, int, Null, num, and String are all restricted in this way, to allow for more efficient implementations.

Examples

The following code produces this diagnostic because String is used in an extends clause:

class A extends String {}

The following code produces this diagnostic because String is used in an implements clause:

class B implements String {}

The following code produces this diagnostic because String is used in a with clause:

class C with String {}

The following code produces this diagnostic because String is used in an on clause:

mixin M on String {}

Common fixes

If a different type should be specified, then replace the type:

class A extends Object {}

If there isn’t a different type that would be appropriate, then remove the type, and possibly the whole clause:

class B {}

subtype_of_ffi_class

The class ‘{0}’ can’t extend ‘{1}’.

The class ‘{0}’ can’t implement ‘{1}’.

The class ‘{0}’ can’t mix in ‘{1}’.

Description

The analyzer produces this diagnostic when a class extends any FFI class other than Struct or Union, or implements or mixes in any FFI class. Struct and Union are the only FFI classes that can be subtyped, and then only by extending them.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class C extends Double:

import 'dart:ffi';

final class C extends Double {}

Common fixes

If the class should extend either Struct or Union, then change the declaration of the class:

import 'dart:ffi';

final class C extends Struct {
  @Int32()
  external int i;
}

If the class shouldn’t extend either Struct or Union, then remove any references to FFI classes:

final class C {}

subtype_of_sealed_class

The class ‘{0}’ shouldn’t be extended, mixed in, or implemented because it’s sealed.

Description

The analyzer produces this diagnostic when a sealed class (one that either has the sealed annotation or inherits or mixes in a sealed class) is referenced in either the extends, implements, or with clause of a class or mixin declaration if the declaration isn’t in the same package as the sealed class.

Example

Given a library in a package other than the package being analyzed that contains the following:

import 'package:meta/meta.dart';

class A {}

@sealed
class B {}

The following code produces this diagnostic because C, which isn’t in the same package as B, is extending the sealed class B:

import 'package:a/a.dart';

class C extends B {}

Common fixes

If the class doesn’t need to be a subtype of the sealed class, then change the declaration so that it isn’t:

import 'package:a/a.dart';

class B extends A {}

If the class needs to be a subtype of the sealed class, then either change the sealed class so that it’s no longer sealed or move the subclass into the same package as the sealed class.

subtype_of_struct_class

The class ‘{0}’ can’t extend ‘{1}’ because ‘{1}’ is a subtype of ‘Struct’, ‘Union’, or ‘AbiSpecificInteger’.

The class ‘{0}’ can’t implement ‘{1}’ because ‘{1}’ is a subtype of ‘Struct’, ‘Union’, or ‘AbiSpecificInteger’.

The class ‘{0}’ can’t mix in ‘{1}’ because ‘{1}’ is a subtype of ‘Struct’, ‘Union’, or ‘AbiSpecificInteger’.

Description

The analyzer produces this diagnostic when a class extends, implements, or mixes in a class that extends either Struct or Union. Classes can only extend either Struct or Union directly.

For more information about FFI, see C interop using dart:ffi.

Example

The following code produces this diagnostic because the class C extends S, and S extends Struct:

import 'dart:ffi';

final class S extends Struct {
  external Pointer f;
}

final class C extends S {
  external Pointer g;
}

Common fixes

If you’re trying to define a struct or union that shares some fields declared by a different struct or union, then extend Struct or Union directly and copy the shared fields:

import 'dart:ffi';

final class S extends Struct {
  external Pointer f;
}

final class C extends Struct {
  external Pointer f;

  external Pointer g;
}

supertype_expands_to_type_parameter

A type alias that expands to a type parameter can’t be implemented.

A type alias that expands to a type parameter can’t be mixed in.

A type alias that expands to a type parameter can’t be used as a superclass constraint.

A type alias that expands to a type parameter can’t be used as a superclass.

Description

The analyzer produces this diagnostic when a type alias that expands to a type parameter is used in an extends, implements, with, or on clause.

Example

The following code produces this diagnostic because the type alias T, which expands to the type parameter S, is used in the extends clause of the class C:

typedef T<S> = S;

class C extends T<Object> {}

Common fixes

Use the value of the type argument directly:

typedef T<S> = S;

class C extends Object {}

super_formal_parameter_type_is_not_subtype_of_associated

The type ‘{0}’ of this parameter isn’t a subtype of the type ‘{1}’ of the associated super constructor parameter.

Description

The analyzer produces this diagnostic when the type of a super parameter isn’t a subtype of the corresponding parameter from the super constructor.

Example

The following code produces this diagnostic because the type of the super parameter x in the constructor for B isn’t a subtype of the parameter x in the constructor for A:

class A {
  A(num x);
}

class B extends A {
  B(String super.x);
}

Common fixes

If the type of the super parameter can be the same as the parameter from the super constructor, then remove the type annotation from the super parameter (if the type is implicit, it is inferred from the type in the super constructor):

class A {
  A(num x);
}

class B extends A {
  B(super.x);
}

If the type of the super parameter can be a subtype of the corresponding parameter’s type, then change the type of the super parameter:

class A {
  A(num x);
}

class B extends A {
  B(int super.x);
}

If the type of the super parameter can’t be changed, then use a normal parameter instead of a super parameter:

class A {
  A(num x);
}

class B extends A {
  B(String x) : super(x.length);
}

super_formal_parameter_without_associated_named

No associated named super constructor parameter.

Description

The analyzer produces this diagnostic when there’s a named super parameter in a constructor and the implicitly or explicitly invoked super constructor doesn’t have a named parameter with the same name.

Named super parameters are associated by name with named parameters in the super constructor.

Example

The following code produces this diagnostic because the constructor in A doesn’t have a parameter named y:

class A {
  A({int? x});
}

class B extends A {
  B({super.y});
}

Common fixes

If the super parameter should be associated with an existing parameter from the super constructor, then change the name to match the name of the corresponding parameter:

class A {
  A({int? x});
}

class B extends A {
  B({super.x});
}

If the super parameter should be associated with a parameter that hasn’t yet been added to the super constructor, then add it:

class A {
  A({int? x, int? y});
}

class B extends A {
  B({super.y});
}

If the super parameter doesn’t correspond to a named parameter from the super constructor, then change it to be a normal parameter:

class A {
  A({int? x});
}

class B extends A {
  B({int? y});
}

super_formal_parameter_without_associated_positional

No associated positional super constructor parameter.

Description

The analyzer produces this diagnostic when there’s a positional super parameter in a constructor and the implicitly or explicitly invoked super constructor doesn’t have a positional parameter at the corresponding index.

Positional super parameters are associated with positional parameters in the super constructor by their index. That is, the first super parameter is associated with the first positional parameter in the super constructor, the second with the second, and so on.

Examples

The following code produces this diagnostic because the constructor in B has a positional super parameter, but there’s no positional parameter in the super constructor in A:

class A {
  A({int? x});
}

class B extends A {
  B(super.x);
}

The following code produces this diagnostic because the constructor in B has two positional super parameters, but there’s only one positional parameter in the super constructor in A, which means that there’s no corresponding parameter for y:

class A {
  A(int x);
}

class B extends A {
  B(super.x, super.y);
}

Common fixes

If the super constructor should have a positional parameter corresponding to the super parameter, then update the super constructor appropriately:

class A {
  A(int x, int y);
}

class B extends A {
  B(super.x, super.y);
}

If the super constructor is correct, or can’t be changed, then convert the super parameter into a normal parameter:

class A {
  A(int x);
}

class B extends A {
  B(super.x, int y);
}

super_invocation_not_last

(Previously known as invalid_super_invocation)

The superconstructor call must be last in an initializer list: ‘{0}’.

Description

The analyzer produces this diagnostic when the initializer list of a constructor contains an invocation of a constructor in the superclass, but the invocation isn’t the last item in the initializer list.

Example

The following code produces this diagnostic because the invocation of the superclass’ constructor isn’t the last item in the initializer list:

class A {
  A(int x);
}

class B extends A {
  B(int x) : super(x), assert(x >= 0);
}

Common fixes

Move the invocation of the superclass’ constructor to the end of the initializer list:

class A {
  A(int x);
}

class B extends A {
  B(int x) : assert(x >= 0), super(x);
}

super_in_enum_constructor

The enum constructor can’t have a ‘super’ initializer.

Description

The analyzer produces this diagnostic when the initializer list in a constructor in an enum contains an invocation of a super constructor.

Example

The following code produces this diagnostic because the constructor in the enum E has a super constructor invocation in the initializer list:

enum E {
  e;

  const E() : super();
}

Common fixes

Remove the super constructor invocation:

enum E {
  e;

  const E();
}

super_in_extension

The ‘super’ keyword can’t be used in an extension because an extension doesn’t have a superclass.

Description

The analyzer produces this diagnostic when a member declared inside an extension uses the super keyword . Extensions aren’t classes and don’t have superclasses, so the super keyword serves no purpose.

Example

The following code produces this diagnostic because super can’t be used in an extension:

extension E on Object {
  String get displayString => super.toString();
}

Common fixes

Remove the super keyword :

extension E on Object {
  String get displayString => toString();
}

super_in_invalid_context

Invalid context for ‘super’ invocation.

Description

The analyzer produces this diagnostic when the keyword super is used outside of an instance method.

Example

The following code produces this diagnostic because super is used in a top-level function:

void f() {
  super.f();
}

Common fixes

Rewrite the code to not use super.

super_in_redirecting_constructor

The redirecting constructor can’t have a ‘super’ initializer.

Description

The analyzer produces this diagnostic when a constructor that redirects to another constructor also attempts to invoke a constructor from the superclass. The superclass constructor will be invoked when the constructor that the redirecting constructor is redirected to is invoked.

Example

The following code produces this diagnostic because the constructor C.a both redirects to C.b and invokes a constructor from the superclass:

class C {
  C.a() : this.b(), super();
  C.b();
}

Common fixes

Remove the invocation of the super constructor:

class C {
  C.a() : this.b();
  C.b();
}

switch_case_completes_normally

The ‘case’ shouldn’t complete normally.

Description

The analyzer produces this diagnostic when the statements following a case label in a switch statement could fall through to the next case or default label.

Example

The following code produces this diagnostic because the case label with a value of zero (0) falls through to the default statements:

void f(int a) {
  switch (a) {
    case 0:
      print(0);
    default:
      return;
  }
}

Common fixes

Change the flow of control so that the case won’t fall through. There are several ways that this can be done, including adding one of the following at the end of the current list of statements:

• a return statement,
• a throw expression,
• a break statement,
• a continue, or
• an invocation of a function or method whose return type is Never.

switch_expression_not_assignable

Type ‘{0}’ of the switch expression isn’t assignable to the type ‘{1}’ of case expressions.

Description

The analyzer produces this diagnostic when the type of the expression in a switch statement isn’t assignable to the type of the expressions in the case clauses.

Example

The following code produces this diagnostic because the type of s (String) isn’t assignable to the type of 0 (int):

void f(String s) {
  switch (s) {
    case 0:
      break;
  }
}

Common fixes

If the type of the case expressions is correct, then change the expression in the switch statement to have the correct type:

void f(String s) {
  switch (int.parse(s)) {
    case 0:
      break;
  }
}

If the type of the switch expression is correct, then change the case expressions to have the correct type:

void f(String s) {
  switch (s) {
    case '0':
      break;
  }
}

tearoff_of_generative_constructor_of_abstract_class

A generative constructor of an abstract class can’t be torn off.

Description

The analyzer produces this diagnostic when a generative constructor from an abstract class is being torn off. This isn’t allowed because it isn’t valid to create an instance of an abstract class, which means that there isn’t any valid use for the torn off constructor.

Example

The following code produces this diagnostic because the constructor C.new is being torn off and the class C is an abstract class:

abstract class C {
  C();
}

void f() {
  C.new;
}

Common fixes

Tear off the constructor of a concrete class.

text_direction_code_point_in_comment

The Unicode code point ‘U+{0}’ changes the appearance of text from how it’s interpreted by the compiler.

Description

The analyzer produces this diagnostic when it encounters source that contains text direction Unicode code points. These code points cause source code in either a string literal or a comment to be interpreted and compiled differently than how it appears in editors, leading to possible security vulnerabilities.

Example

The following code produces this diagnostic twice because there are hidden characters at the start and end of the label string:

var label = 'Interactive text';

Common fixes

If the code points are intended to be included in the string literal, then escape them:

var label = '\u202AInteractive text\u202C';

If the code points aren’t intended to be included in the string literal, then remove them:

var label = 'Interactive text';

text_direction_code_point_in_literal

The Unicode code point ‘U+{0}’ changes the appearance of text from how it’s interpreted by the compiler.

Description

The analyzer produces this diagnostic when it encounters source that contains text direction Unicode code points. These code points cause source code in either a string literal or a comment to be interpreted and compiled differently than how it appears in editors, leading to possible security vulnerabilities.

Example

The following code produces this diagnostic twice because there are hidden characters at the start and end of the label string:

var label = 'Interactive text';

Common fixes

If the code points are intended to be included in the string literal, then escape them:

var label = '\u202AInteractive text\u202C';

If the code points aren’t intended to be included in the string literal, then remove them:

var label = 'Interactive text';

throw_of_invalid_type

The type ‘{0}’ of the thrown expression must be assignable to ‘Object’.

Description

The analyzer produces this diagnostic when the type of the expression in a throw expression isn’t assignable to Object. It isn’t valid to throw null, so it isn’t valid to use an expression that might evaluate to null.

Example

The following code produces this diagnostic because s might be null:

void f(String? s) {
  throw s;
}

Common fixes

Add an explicit null-check to the expression:

void f(String? s) {
  throw s!;
}

top_level_cycle

The type of ‘{0}’ can’t be inferred because it depends on itself through the cycle: {1}.

Description

The analyzer produces this diagnostic when a top-level variable has no type annotation and the variable’s initializer refers to the variable, either directly or indirectly.

Example

The following code produces this diagnostic because the variables x and y are defined in terms of each other, and neither has an explicit type, so the type of the other can’t be inferred:

var x = y;
var y = x;

Common fixes

If the two variables don’t need to refer to each other, then break the cycle:

var x = 0;
var y = x;

If the two variables need to refer to each other, then give at least one of them an explicit type:

int x = y;
var y = x;

Note, however, that while this code doesn’t produce any diagnostics, it will produce a stack overflow at runtime unless at least one of the variables is assigned a value that doesn’t depend on the other variables before any of the variables in the cycle are referenced.

type_alias_cannot_reference_itself

Typedefs can’t reference themselves directly or recursively via another typedef.

Description

The analyzer produces this diagnostic when a typedef refers to itself, either directly or indirectly.

Example

The following code produces this diagnostic because F depends on itself indirectly through G:

typedef F = void Function(G);
typedef G = void Function(F);

Common fixes

Change one or more of the typedefs in the cycle so that none of them refer to themselves:

typedef F = void Function(G);
typedef G = void Function(int);

type_annotation_deferred_class

The deferred type ‘{0}’ can’t be used in a declaration, cast, or type test.

Description

The analyzer produces this diagnostic when the type annotation is in a variable declaration, or the type used in a cast (as) or type test (is) is a type declared in a library that is imported using a deferred import. These types are required to be available at compile time, but aren’t.

For more information, check out Lazily loading a library.

Example

The following code produces this diagnostic because the type of the parameter f is imported from a deferred library:

import 'dart:io' deferred as io;

void f(io.File f) {}

Common fixes

If you need to reference the imported type, then remove the deferred keyword:

import 'dart:io' as io;

void f(io.File f) {}

If the import is required to be deferred and there’s another type that is appropriate, then use that type in place of the type from the deferred library.

type_argument_not_matching_bounds

‘{0}’ doesn’t conform to the bound ‘{2}’ of the type parameter ‘{1}’.

Description

The analyzer produces this diagnostic when a type argument isn’t the same as or a subclass of the bounds of the corresponding type parameter.

Example

The following code produces this diagnostic because String isn’t a subclass of num:

class A<E extends num> {}

var a = A<String>();

Common fixes

Change the type argument to be a subclass of the bounds:

class A<E extends num> {}

var a = A<int>();

type_check_with_null

Tests for non-null should be done with ‘!= null’.

Tests for null should be done with ‘== null’.

Description

The analyzer produces this diagnostic when there’s a type check (using the as operator) where the type is Null. There’s only one value whose type is Null, so the code is both more readable and more performant when it tests for null explicitly.

Examples

The following code produces this diagnostic because the code is testing to see whether the value of s is null by using a type check:

void f(String? s) {
  if (s is Null) {
    return;
  }
  print(s);
}

The following code produces this diagnostic because the code is testing to see whether the value of s is something other than null by using a type check:

void f(String? s) {
  if (s is! Null) {
    print(s);
  }
}

Common fixes

Replace the type check with the equivalent comparison with null:

void f(String? s) {
  if (s == null) {
    return;
  }
  print(s);
}

type_parameter_referenced_by_static

Static members can’t reference type parameters of the class.

Description

The analyzer produces this diagnostic when a static member references a type parameter that is declared for the class. Type parameters only have meaning for instances of the class.

Example

The following code produces this diagnostic because the static method hasType has a reference to the type parameter T:

class C<T> {
  static bool hasType(Object o) => o is T;
}

Common fixes

If the member can be an instance member, then remove the keyword static:

class C<T> {
  bool hasType(Object o) => o is T;
}

If the member must be a static member, then make the member be generic:

class C<T> {
  static bool hasType<S>(Object o) => o is S;
}

Note, however, that there isn’t a relationship between T and S, so this second option changes the semantics from what was likely to be intended.

type_parameter_supertype_of_its_bound

‘{0}’ can’t be a supertype of its upper bound.

Description

The analyzer produces this diagnostic when the bound of a type parameter (the type following the extends keyword) is either directly or indirectly the type parameter itself. Stating that the type parameter must be the same as itself or a subtype of itself or a subtype of itself isn’t helpful because it will always be the same as itself.

Examples

The following code produces this diagnostic because the bound of T is T:

class C<T extends T> {}

The following code produces this diagnostic because the bound of T1 is T2, and the bound of T2 is T1, effectively making the bound of T1 be T1:

class C<T1 extends T2, T2 extends T1> {}

Common fixes

If the type parameter needs to be a subclass of some type, then replace the bound with the required type:

class C<T extends num> {}

If the type parameter can be any type, then remove the extends clause:

class C<T> {}

type_test_with_non_type

The name ‘{0}’ isn’t a type and can’t be used in an ‘is’ expression.

Description

The analyzer produces this diagnostic when the right-hand side of an is or is! test isn’t a type.

Example

The following code produces this diagnostic because the right-hand side is a parameter, not a type:

typedef B = int Function(int);

void f(Object a, B b) {
  if (a is b) {
    return;
  }
}

Common fixes

If you intended to use a type test, then replace the right-hand side with a type:

typedef B = int Function(int);

void f(Object a, B b) {
  if (a is B) {
    return;
  }
}

If you intended to use a different kind of test, then change the test:

typedef B = int Function(int);

void f(Object a, B b) {
  if (a == b) {
    return;
  }
}

type_test_with_undefined_name

The name ‘{0}’ isn’t defined, so it can’t be used in an ‘is’ expression.

Description

The analyzer produces this diagnostic when the name following the is in a type test expression isn’t defined.

Example

The following code produces this diagnostic because the name Srting isn’t defined:

void f(Object o) {
  if (o is Srting) {
    // ...
  }
}

Common fixes

Replace the name with the name of a type:

void f(Object o) {
  if (o is String) {
    // ...
  }
}

unchecked_use_of_nullable_value

A nullable expression can’t be used as a condition.

A nullable expression can’t be used as an iterator in a for-in loop.

A nullable expression can’t be used in a spread.

A nullable expression can’t be used in a yield-each statement.

The function can’t be unconditionally invoked because it can be ‘null’.

The method ‘{0}’ can’t be unconditionally invoked because the receiver can be ‘null’.

The operator ‘{0}’ can’t be unconditionally invoked because the receiver can be ‘null’.

The property ‘{0}’ can’t be unconditionally accessed because the receiver can be ‘null’.

Description

The analyzer produces this diagnostic when an expression whose type is potentially non-nullable is dereferenced without first verifying that the value isn’t null.

Example

The following code produces this diagnostic because s can be null at the point where it’s referenced:

void f(String? s) {
  if (s.length > 3) {
    // ...
  }
}

Common fixes

If the value really can be null, then add a test to ensure that members are only accessed when the value isn’t null:

void f(String? s) {
  if (s != null && s.length > 3) {
    // ...
  }
}

If the expression is a variable and the value should never be null, then change the type of the variable to be non-nullable:

void f(String s) {
  if (s.length > 3) {
    // ...
  }
}

If you believe that the value of the expression should never be null, but you can’t change the type of the variable, and you’re willing to risk having an exception thrown at runtime if you’re wrong, then you can assert that the value isn’t null:

void f(String? s) {
  if (s!.length > 3) {
    // ...
  }
}

undefined_annotation

Undefined name ‘{0}’ used as an annotation.

Description

The analyzer produces this diagnostic when a name that isn’t defined is used as an annotation.

Example

The following code produces this diagnostic because the name undefined isn’t defined:

@undefined
void f() {}

Common fixes

If the name is correct, but it isn’t declared yet, then declare the name as a constant value:

const undefined = 'undefined';

@undefined
void f() {}

If the name is wrong, replace the name with the name of a valid constant:

@deprecated
void f() {}

Otherwise, remove the annotation.

undefined_class

Undefined class ‘{0}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a class but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Example

The following code produces this diagnostic because Piont isn’t defined:

class Point {}

void f(Piont p) {}

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a class that is defined. The example above can be corrected by fixing the spelling of the class:

class Point {}

void f(Point p) {}

If the class is defined but isn’t visible, then you probably need to add an import.

undefined_constructor_in_initializer

The class ‘{0}’ doesn’t have a constructor named ‘{1}’.

The class ‘{0}’ doesn’t have an unnamed constructor.

Description

The analyzer produces this diagnostic when a superclass constructor is invoked in the initializer list of a constructor, but the superclass doesn’t define the constructor being invoked.

Examples

The following code produces this diagnostic because A doesn’t have an unnamed constructor:

class A {
  A.n();
}
class B extends A {
  B() : super();
}

The following code produces this diagnostic because A doesn’t have a constructor named m:

class A {
  A.n();
}
class B extends A {
  B() : super.m();
}

Common fixes

If the superclass defines a constructor that should be invoked, then change the constructor being invoked:

class A {
  A.n();
}
class B extends A {
  B() : super.n();
}

If the superclass doesn’t define an appropriate constructor, then define the constructor being invoked:

class A {
  A.m();
  A.n();
}
class B extends A {
  B() : super.m();
}

undefined_enum_constant

There’s no constant named ‘{0}’ in ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of an enum constant, and the name either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Example

The following code produces this diagnostic because E doesn’t define a constant named c:

enum E {a, b}

var e = E.c;

Common fixes

If the constant should be defined, then add it to the declaration of the enum:

enum E {a, b, c}

var e = E.c;

If the constant shouldn’t be defined, then change the name to the name of an existing constant:

enum E {a, b}

var e = E.b;

undefined_enum_constructor

The enum doesn’t have a constructor named ‘{0}’.

The enum doesn’t have an unnamed constructor.

Description

The analyzer produces this diagnostic when the constructor invoked to initialize an enum constant doesn’t exist.

Examples

The following code produces this diagnostic because the enum constant c is being initialized by the unnamed constructor, but there’s no unnamed constructor defined in E:

enum E {
  c();

  const E.x();
}

The following code produces this diagnostic because the enum constant c is being initialized by the constructor named x, but there’s no constructor named x defined in E:

enum E {
  c.x();

  const E.y();
}

Common fixes

If the enum constant is being initialized by the unnamed constructor and one of the named constructors should have been used, then add the name of the constructor:

enum E {
  c.x();

  const E.x();
}

If the enum constant is being initialized by the unnamed constructor and none of the named constructors are appropriate, then define the unnamed constructor:

enum E {
  c();

  const E();
}

If the enum constant is being initialized by a named constructor and one of the existing constructors should have been used, then change the name of the constructor being invoked (or remove it if the unnamed constructor should be used):

enum E {
  c.y();

  const E();
  const E.y();
}

If the enum constant is being initialized by a named constructor and none of the existing constructors should have been used, then define a constructor with the name that was used:

enum E {
  c.x();

  const E.x();
}

undefined_extension_getter

The getter ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a getter, but the getter isn’t defined by the specified extension. The analyzer also produces this diagnostic when a static getter is referenced but isn’t defined by the specified extension.

Examples

The following code produces this diagnostic because the extension E doesn’t declare an instance getter named b:

extension E on String {
  String get a => 'a';
}

extension F on String {
  String get b => 'b';
}

void f() {
  E('c').b;
}

The following code produces this diagnostic because the extension E doesn’t declare a static getter named a:

extension E on String {}

var x = E.a;

Common fixes

If the name of the getter is incorrect, then change it to the name of an existing getter:

extension E on String {
  String get a => 'a';
}

extension F on String {
  String get b => 'b';
}

void f() {
  E('c').a;
}

If the name of the getter is correct but the name of the extension is wrong, then change the name of the extension to the correct name:

extension E on String {
  String get a => 'a';
}

extension F on String {
  String get b => 'b';
}

void f() {
  F('c').b;
}

If the name of the getter and extension are both correct, but the getter isn’t defined, then define the getter:

extension E on String {
  String get a => 'a';
  String get b => 'z';
}

extension F on String {
  String get b => 'b';
}

void f() {
  E('c').b;
}

undefined_extension_method

The method ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a method, but the method isn’t defined by the specified extension. The analyzer also produces this diagnostic when a static method is referenced but isn’t defined by the specified extension.

Examples

The following code produces this diagnostic because the extension E doesn’t declare an instance method named b:

extension E on String {
  String a() => 'a';
}

extension F on String {
  String b() => 'b';
}

void f() {
  E('c').b();
}

The following code produces this diagnostic because the extension E doesn’t declare a static method named a:

extension E on String {}

var x = E.a();

Common fixes

If the name of the method is incorrect, then change it to the name of an existing method:

extension E on String {
  String a() => 'a';
}

extension F on String {
  String b() => 'b';
}

void f() {
  E('c').a();
}

If the name of the method is correct, but the name of the extension is wrong, then change the name of the extension to the correct name:

extension E on String {
  String a() => 'a';
}

extension F on String {
  String b() => 'b';
}

void f() {
  F('c').b();
}

If the name of the method and extension are both correct, but the method isn’t defined, then define the method:

extension E on String {
  String a() => 'a';
  String b() => 'z';
}

extension F on String {
  String b() => 'b';
}

void f() {
  E('c').b();
}

undefined_extension_operator

The operator ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an operator is invoked on a specific extension when that extension doesn’t implement the operator.

Example

The following code produces this diagnostic because the extension E doesn’t define the operator *:

var x = E('') * 4;

extension E on String {}

Common fixes

If the extension is expected to implement the operator, then add an implementation of the operator to the extension:

var x = E('') * 4;

extension E on String {
  int operator *(int multiplier) => length * multiplier;
}

If the operator is defined by a different extension, then change the name of the extension to the name of the one that defines the operator.

If the operator is defined on the argument of the extension override, then remove the extension override:

var x = '' * 4;

extension E on String {}

undefined_extension_setter

The setter ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a setter, but the setter isn’t defined by the specified extension. The analyzer also produces this diagnostic when a static setter is referenced but isn’t defined by the specified extension.

Examples

The following code produces this diagnostic because the extension E doesn’t declare an instance setter named b:

extension E on String {
  set a(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  E('c').b = 'd';
}

The following code produces this diagnostic because the extension E doesn’t declare a static setter named a:

extension E on String {}

void f() {
  E.a = 3;
}

Common fixes

If the name of the setter is incorrect, then change it to the name of an existing setter:

extension E on String {
  set a(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  E('c').a = 'd';
}

If the name of the setter is correct, but the name of the extension is wrong, then change the name of the extension to the correct name:

extension E on String {
  set a(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  F('c').b = 'd';
}

If the name of the setter and extension are both correct, but the setter isn’t defined, then define the setter:

extension E on String {
  set a(String v) {}
  set b(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  E('c').b = 'd';
}

undefined_function

The function ‘{0}’ isn’t defined.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a function but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Example

The following code produces this diagnostic because the name emty isn’t defined:

List<int> empty() => [];

void main() {
  print(emty());
}

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a function that is defined. The example above can be corrected by fixing the spelling of the function:

List<int> empty() => [];

void main() {
  print(empty());
}

If the function is defined but isn’t visible, then you probably need to add an import or re-arrange your code to make the function visible.

undefined_getter

The getter ‘{0}’ isn’t defined for the ‘{1}’ function type.

The getter ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a getter but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Example

The following code produces this diagnostic because String has no member named len:

int f(String s) => s.len;

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a getter that is defined. The example above can be corrected by fixing the spelling of the getter:

int f(String s) => s.length;

undefined_hidden_name

The library ‘{0}’ doesn’t export a member with the hidden name ‘{1}’.

Description

The analyzer produces this diagnostic when a hide combinator includes a name that isn’t defined by the library being imported.

Example

The following code produces this diagnostic because dart:math doesn’t define the name String:

import 'dart:math' hide String, max;

var x = min(0, 1);

Common fixes

If a different name should be hidden, then correct the name. Otherwise, remove the name from the list:

import 'dart:math' hide max;

var x = min(0, 1);

undefined_identifier

Undefined name ‘{0}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Example

The following code produces this diagnostic because the name rihgt isn’t defined:

int min(int left, int right) => left <= rihgt ? left : right;

Common fixes

If the identifier isn’t defined, then either define it or replace it with an identifier that is defined. The example above can be corrected by fixing the spelling of the variable:

int min(int left, int right) => left <= right ? left : right;

If the identifier is defined but isn’t visible, then you probably need to add an import or re-arrange your code to make the identifier visible.

undefined_identifier_await

Undefined name ‘await’ in function body not marked with ‘async’.

Description

The analyzer produces this diagnostic when the name await is used in a method or function body without being declared, and the body isn’t marked with the async keyword. The name await only introduces an await expression in an asynchronous function.

Example

The following code produces this diagnostic because the name await is used in the body of f even though the body of f isn’t marked with the async keyword:

void f(p) { await p; }

Common fixes

Add the keyword async to the function body:

void f(p) async { await p; }

undefined_method

The method ‘{0}’ isn’t defined for the ‘{1}’ function type.

The method ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a method but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Example

The following code produces this diagnostic because the identifier removeMiddle isn’t defined:

int f(List<int> l) => l.removeMiddle();

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a method that is defined. The example above can be corrected by fixing the spelling of the method:

int f(List<int> l) => l.removeLast();

undefined_named_parameter

The named parameter ‘{0}’ isn’t defined.

Description

The analyzer produces this diagnostic when a method or function invocation has a named argument, but the method or function being invoked doesn’t define a parameter with the same name.

Example

The following code produces this diagnostic because m doesn’t declare a named parameter named a:

class C {
  m({int b}) {}
}

void f(C c) {
  c.m(a: 1);
}

Common fixes

If the argument name is mistyped, then replace it with the correct name. The example above can be fixed by changing a to b:

class C {
  m({int b}) {}
}

void f(C c) {
  c.m(b: 1);
}

If a subclass adds a parameter with the name in question, then cast the receiver to the subclass:

class C {
  m({int b}) {}
}

class D extends C {
  m({int a, int b}) {}
}

void f(C c) {
  (c as D).m(a: 1);
}

If the parameter should be added to the function, then add it:

class C {
  m({int a, int b}) {}
}

void f(C c) {
  c.m(a: 1);
}

undefined_operator

The operator ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when a user-definable operator is invoked on an object for which the operator isn’t defined.

Example

The following code produces this diagnostic because the class C doesn’t define the operator +:

class C {}

C f(C c) => c + 2;

Common fixes

If the operator should be defined for the class, then define it:

class C {
  C operator +(int i) => this;
}

C f(C c) => c + 2;

undefined_prefixed_name

The name ‘{0}’ is being referenced through the prefix ‘{1}’, but it isn’t defined in any of the libraries imported using that prefix.

Description

The analyzer produces this diagnostic when a prefixed identifier is found where the prefix is valid, but the identifier isn’t declared in any of the libraries imported using that prefix.

Example

The following code produces this diagnostic because dart:core doesn’t define anything named a:

import 'dart:core' as p;

void f() {
  p.a;
}

Common fixes

If the library in which the name is declared isn’t imported yet, add an import for the library.

If the name is wrong, then change it to one of the names that’s declared in the imported libraries.

undefined_referenced_parameter

The parameter ‘{0}’ isn’t defined by ‘{1}’.

Description

The analyzer produces this diagnostic when an annotation of the form UseResult.unless(parameterDefined: parameterName) specifies a parameter name that isn’t defined by the annotated function.

Example

The following code produces this diagnostic because the function f doesn’t have a parameter named b:

import 'package:meta/meta.dart';

@UseResult.unless(parameterDefined: 'b')
int f([int? a]) => a ?? 0;

Common fixes

Change the argument named parameterDefined to match the name of one of the parameters to the function:

import 'package:meta/meta.dart';

@UseResult.unless(parameterDefined: 'a')
int f([int? a]) => a ?? 0;

undefined_setter

The setter ‘{0}’ isn’t defined for the ‘{1}’ function type.

The setter ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a setter but either isn’t defined or isn’t visible in the scope in which the identifier is being referenced.

Example

The following code produces this diagnostic because there isn’t a setter named z:

class C {
  int x = 0;
  void m(int y) {
    this.z = y;
  }
}

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a setter that is defined. The example above can be corrected by fixing the spelling of the setter:

class C {
  int x = 0;
  void m(int y) {
    this.x = y;
  }
}

undefined_shown_name

The library ‘{0}’ doesn’t export a member with the shown name ‘{1}’.

Description

The analyzer produces this diagnostic when a show combinator includes a name that isn’t defined by the library being imported.

Example

The following code produces this diagnostic because dart:math doesn’t define the name String:

import 'dart:math' show min, String;

var x = min(0, 1);

Common fixes

If a different name should be shown, then correct the name. Otherwise, remove the name from the list:

import 'dart:math' show min;

var x = min(0, 1);

undefined_super_member

(Previously known as undefined_super_method)

The getter ‘{0}’ isn’t defined in a superclass of ‘{1}’.

The method ‘{0}’ isn’t defined in a superclass of ‘{1}’.

The operator ‘{0}’ isn’t defined in a superclass of ‘{1}’.

The setter ‘{0}’ isn’t defined in a superclass of ‘{1}’.

Description

The analyzer produces this diagnostic when an inherited member (method, getter, setter, or operator) is referenced using super, but there’s no member with that name in the superclass chain.

Examples

The following code produces this diagnostic because Object doesn’t define a method named n:

class C {
  void m() {
    super.n();
  }
}

The following code produces this diagnostic because Object doesn’t define a getter named g:

class C {
  void m() {
    super.g;
  }
}

Common fixes

If the inherited member you intend to invoke has a different name, then make the name of the invoked member match the inherited member.

If the member you intend to invoke is defined in the same class, then remove the super..

If the member isn’t defined, then either add the member to one of the superclasses or remove the invocation.

unknown_platform

The platform ‘{0}’ is not a recognized platform.

Description

The analyzer produces this diagnostic when an unknown platform name is used as a key in the platforms map. To learn more about specifying your package’s supported platforms, check out the documentation on platform declarations.

Example

The following pubspec.yaml produces this diagnostic because the platform browser is unknown.

name: example
platforms:
  browser:

Common fixes

If you can rely on automatic platform detection, then omit the top-level platforms key.

name: example

If you need to manually specify the list of supported platforms, then write the platforms field as a map with known platform names as keys.

name: example
platforms:
  # These are the known platforms
  android:
  ios:
  linux:
  macos:
  web:
  windows:

unnecessary_cast

Unnecessary cast.

Description

The analyzer produces this diagnostic when the value being cast is already known to be of the type that it’s being cast to.

Example

The following code produces this diagnostic because n is already known to be an int as a result of the is test:

void f(num n) {
  if (n is int) {
    (n as int).isEven;
  }
}

Common fixes

Remove the unnecessary cast:

void f(num n) {
  if (n is int) {
    n.isEven;
  }
}

unnecessary_dev_dependency

The dev dependency on {0} is unnecessary because there is also a normal dependency on that package.

Description

The analyzer produces this diagnostic when there’s an entry under dev_dependencies for a package that is also listed under dependencies. The packages under dependencies are available to all of the code in the package, so there’s no need to also list them under dev_dependencies.

Example

The following code produces this diagnostic because the package meta is listed under both dependencies and dev_dependencies:

name: example
dependencies:
  meta: ^1.0.2
dev_dependencies:
  meta: ^1.0.2

Common fixes

Remove the entry under dev_dependencies (and the dev_dependencies key if that’s the only package listed there):

name: example
dependencies:
  meta: ^1.0.2

unnecessary_final

The keyword ‘final’ isn’t necessary because the parameter is implicitly ‘final’.

Description

The analyzer produces this diagnostic when either a field initializing parameter or a super parameter in a constructor has the keyword final. In both cases the keyword is unnecessary because the parameter is implicitly final.

Examples

The following code produces this diagnostic because the field initializing parameter has the keyword final:

class A {
  int value;

  A(final this.value);
}

The following code produces this diagnostic because the super parameter in B has the keyword final:

class A {
  A(int value);
}

class B extends A {
  B(final super.value);
}

Common fixes

Remove the unnecessary final keyword:

class A {
  A(int value);
}

class B extends A {
  B(super.value);
}

unnecessary_import

The import of ‘{0}’ is unnecessary because all of the used elements are also provided by the import of ‘{1}’.

Description

The analyzer produces this diagnostic when an import isn’t needed because all of the names that are imported and referenced within the importing library are also visible through another import.

Example

Given a file a.dart that contains the following:

class A {}

And, given a file b.dart that contains the following:

export 'a.dart';

class B {}

The following code produces this diagnostic because the class A, which is imported from a.dart, is also imported from b.dart. Removing the import of a.dart leaves the semantics unchanged:

import 'a.dart';
import 'b.dart';

void f(A a, B b) {}

Common fixes

If the import isn’t needed, then remove it.

If some of the names imported by this import are intended to be used but aren’t yet, and if those names aren’t imported by other imports, then add the missing references to those names.

unnecessary_nan_comparison

A double can’t equal ‘double.nan’, so the condition is always ‘false’.

A double can’t equal ‘double.nan’, so the condition is always ‘true’.

Description

The analyzer produces this diagnostic when a value is compared to double.nan using either == or !=.

Dart follows the IEEE 754 floating-point standard for the semantics of floating point operations, which states that, for any floating point value x (including NaN, positive infinity, and negative infinity),

• NaN == x is always false
• NaN != x is always true

As a result, comparing any value to NaN is pointless because the result is already known (based on the comparison operator being used).

Example

The following code produces this diagnostic because d is being compared to double.nan:

bool isNaN(double d) => d == double.nan;

Common fixes

Use the getter double.isNaN instead:

bool isNaN(double d) => d.isNaN;

unnecessary_non_null_assertion

The ‘!’ will have no effect because the receiver can’t be null.

Description

The analyzer produces this diagnostic when the operand of the ! operator can’t be null.

Example

The following code produces this diagnostic because x can’t be null:

int f(int x) {
  return x!;
}

Common fixes

Remove the null check operator (!):

int f(int x) {
  return x;
}

unnecessary_no_such_method

Unnecessary ‘noSuchMethod’ declaration.

Description

The analyzer produces this diagnostic when there’s a declaration of noSuchMethod, the only thing the declaration does is invoke the overridden declaration, and the overridden declaration isn’t the declaration in Object.

Overriding the implementation of Object’s noSuchMethod (no matter what the implementation does) signals to the analyzer that it shouldn’t flag any inherited abstract methods that aren’t implemented in that class. This works even if the overriding implementation is inherited from a superclass, so there’s no value to declare it again in a subclass.

Example

The following code produces this diagnostic because the declaration of noSuchMethod in A makes the declaration of noSuchMethod in B unnecessary:

class A {
  @override
  dynamic noSuchMethod(x) => super.noSuchMethod(x);
}
class B extends A {
  @override
  dynamic noSuchMethod(y) {
    return super.noSuchMethod(y);
  }
}

Common fixes

Remove the unnecessary declaration:

class A {
  @override
  dynamic noSuchMethod(x) => super.noSuchMethod(x);
}
class B extends A {}

unnecessary_null_assert_pattern

The null-assert pattern will have no effect because the matched type isn’t nullable.

Description

The analyzer produces this diagnostic when a null-assert pattern is used to match a value that isn’t nullable.

Example

The following code produces this diagnostic because the variable x isn’t nullable:

void f(int x) {
  if (x case var a! when a > 0) {}
}

Common fixes

Remove the null-assert pattern:

void f(int x) {
  if (x case var a when a > 0) {}
}

unnecessary_null_check_pattern

The null-check pattern will have no effect because the matched type isn’t nullable.

Description

The analyzer produces this diagnostic when a null-check pattern is used to match a value that isn’t nullable.

Example

The following code produces this diagnostic because the value x isn’t nullable:

void f(int x) {
  if (x case var a? when a > 0) {}
}

Common fixes

Remove the null-check pattern:

void f(int x) {
  if (x case var a when a > 0) {}
}

unnecessary_null_comparison

The operand can’t be null, so the condition is always ‘false’.

The operand can’t be null, so the condition is always ‘true’.

Description

The analyzer produces this diagnostic when it finds an equality comparison (either == or !=) with one operand of null and the other operand can’t be null. Such comparisons are always either true or false, so they serve no purpose.

Examples

The following code produces this diagnostic because x can never be null, so the comparison always evaluates to true:

void f(int x) {
  if (x != null) {
    print(x);
  }
}

The following code produces this diagnostic because x can never be null, so the comparison always evaluates to false:

void f(int x) {
  if (x == null) {
    throw ArgumentError("x can't be null");
  }
}

Common fixes

If the other operand should be able to be null, then change the type of the operand:

void f(int? x) {
  if (x != null) {
    print(x);
  }
}

If the other operand really can’t be null, then remove the condition:

void f(int x) {
  print(x);
}

unnecessary_question_mark

The ‘?’ is unnecessary because ‘{0}’ is nullable without it.

Description

The analyzer produces this diagnostic when either the type dynamic or the type Null is followed by a question mark. Both of these types are inherently nullable so the question mark doesn’t change the semantics.

Example

The following code produces this diagnostic because the question mark following dynamic isn’t necessary:

dynamic? x;

Common fixes

Remove the unneeded question mark:

dynamic x;

unnecessary_set_literal

Braces unnecessarily wrap this expression in a set literal.

Description

The analyzer produces this diagnostic when a function that has a return type of void, Future<void>, or FutureOr<void> uses an expression function body (=>) and the returned value is a literal set containing a single element.

Although the language allows it, returning a value from a void function isn’t useful because it can’t be used at the call site. In this particular case the return is often due to a misunderstanding about the syntax. The braces aren’t necessary and can be removed.

Example

The following code produces this diagnostic because the closure being passed to g has a return type of void, but is returning a set:

void f() {
  g(() => {1});
}

void g(void Function() p) {}

Common fixes

Remove the braces from around the value:

void f() {
  g(() => 1);
}

void g(void Function() p) {}

unnecessary_type_check

Unnecessary type check; the result is always ‘false’.

Unnecessary type check; the result is always ‘true’.

Description

The analyzer produces this diagnostic when the value of a type check (using either is or is!) is known at compile time.

Example

The following code produces this diagnostic because the test a is Object? is always true:

bool f<T>(T a) => a is Object?;

Common fixes

If the type check doesn’t check what you intended to check, then change the test:

bool f<T>(T a) => a is Object;

If the type check does check what you intended to check, then replace the type check with its known value or completely remove it:

bool f<T>(T a) => true;

unqualified_reference_to_non_local_static_member

Static members from supertypes must be qualified by the name of the defining type.

Description

The analyzer produces this diagnostic when code in one class references a static member in a superclass without prefixing the member’s name with the name of the superclass. Static members can only be referenced without a prefix in the class in which they’re declared.

Example

The following code produces this diagnostic because the static field x is referenced in the getter g without prefixing it with the name of the defining class:

class A {
  static int x = 3;
}

class B extends A {
  int get g => x;
}

Common fixes

Prefix the name of the static member with the name of the declaring class:

class A {
  static int x = 3;
}

class B extends A {
  int get g => A.x;
}

unqualified_reference_to_static_member_of_extended_type

Static members from the extended type or one of its superclasses must be qualified by the name of the defining type.

Description

The analyzer produces this diagnostic when an undefined name is found, and the name is the same as a static member of the extended type or one of its superclasses.

Example

The following code produces this diagnostic because m is a static member of the extended type C:

class C {
  static void m() {}
}

extension E on C {
  void f() {
    m();
  }
}

Common fixes

If you’re trying to reference a static member that’s declared outside the extension, then add the name of the class or extension before the reference to the member:

class C {
  static void m() {}
}

extension E on C {
  void f() {
    C.m();
  }
}

If you’re referencing a member that isn’t declared yet, add a declaration:

class C {
  static void m() {}
}

extension E on C {
  void f() {
    m();
  }

  void m() {}
}

unreachable_switch_case

This case is covered by the previous cases.

Description

The analyzer produces this diagnostic when a case clause in a switch statement doesn’t match anything because all of the matchable values are matched by an earlier case clause.

Example

The following code produces this diagnostic because the value 1 was matched in the preceding case:

void f(int x) {
  switch (x) {
    case 1:
      print('one');
    case 1:
      print('two');
  }
}

Common fixes

Change one or both of the conflicting cases to match different values:

void f(int x) {
  switch (x) {
    case 1:
      print('one');
    case 2:
      print('two');
  }
}

unused_catch_clause

The exception variable ‘{0}’ isn’t used, so the ‘catch’ clause can be removed.

Description

The analyzer produces this diagnostic when a catch clause is found, and neither the exception parameter nor the optional stack trace parameter are used in the catch block.

Example

The following code produces this diagnostic because e isn’t referenced:

void f() {
  try {
    int.parse(';');
  } on FormatException catch (e) {
    // ignored
  }
}

Common fixes

Remove the unused catch clause:

void f() {
  try {
    int.parse(';');
  } on FormatException {
    // ignored
  }
}

unused_catch_stack

The stack trace variable ‘{0}’ isn’t used and can be removed.

Description

The analyzer produces this diagnostic when the stack trace parameter in a catch clause isn’t referenced within the body of the catch block.

Example

The following code produces this diagnostic because stackTrace isn’t referenced:

void f() {
  try {
    // ...
  } catch (exception, stackTrace) {
    // ...
  }
}

Common fixes

If you need to reference the stack trace parameter, then add a reference to it. Otherwise, remove it:

void f() {
  try {
    // ...
  } catch (exception) {
    // ...
  }
}

unused_element

A value for optional parameter ‘{0}’ isn’t ever given.

The declaration ‘{0}’ isn’t referenced.

Description

The analyzer produces this diagnostic when a private declaration isn’t referenced in the library that contains the declaration. The following kinds of declarations are analyzed:

• Private top-level declarations and all of their members
• Private members of public declarations
• Optional parameters of private functions for which a value is never passed

Not all references to an element will mark it as “used”:

• Assigning a value to a top-level variable (with a standard = assignment, or a null-aware ??= assignment) does not count as using it.
• Referring to an element in a doc comment reference does not count as using it.
• Referring to a class, mixin, or enum on the right side of an is expression does not count as using it.

Example

Assuming that no code in the library references _C, the following code produces this diagnostic:

class _C {}

Assuming that no code in the library passes a value for y in any invocation of _m, the following code produces this diagnostic:

class C {
  void _m(int x, [int y]) {}

  void n() => _m(0);
}

Common fixes

If the declaration isn’t needed, then remove it:

class C {
  void _m(int x) {}

  void n() => _m(0);
}

If the declaration is intended to be used, then add the code to use it.

unused_field

The value of the field ‘{0}’ isn’t used.

Description

The analyzer produces this diagnostic when a private field is declared but never read, even if it’s written in one or more places.

Example

The following code produces this diagnostic because the field _originalValue isn’t read anywhere in the library:

class C {
  final String _originalValue;
  final String _currentValue;

  C(this._originalValue) : _currentValue = _originalValue;

  String get value => _currentValue;
}

It might appear that the field _originalValue is being read in the initializer (_currentValue = _originalValue), but that is actually a reference to the parameter of the same name, not a reference to the field.

Common fixes

If the field isn’t needed, then remove it.

If the field was intended to be used, then add the missing code.

unused_import

Unused import: ‘{0}’.

Description

The analyzer produces this diagnostic when an import isn’t needed because none of the names that are imported are referenced within the importing library.

Example

The following code produces this diagnostic because nothing defined in dart:async is referenced in the library:

import 'dart:async';

void main() {}

Common fixes

If the import isn’t needed, then remove it.

If some of the imported names are intended to be used, then add the missing code.

unused_label

The label ‘{0}’ isn’t used.

Description

The analyzer produces this diagnostic when a label that isn’t used is found.

Example

The following code produces this diagnostic because the label loop isn’t referenced anywhere in the method:

void f(int limit) {
  loop: for (int i = 0; i < limit; i++) {
    print(i);
  }
}

Common fixes

If the label isn’t needed, then remove it:

void f(int limit) {
  for (int i = 0; i < limit; i++) {
    print(i);
  }
}

If the label is needed, then use it:

void f(int limit) {
  loop: for (int i = 0; i < limit; i++) {
    print(i);
    if (i != 0) {
      break loop;
    }
  }
}

unused_local_variable

The value of the local variable ‘{0}’ isn’t used.

Description

The analyzer produces this diagnostic when a local variable is declared but never read, even if it’s written in one or more places.

Example

The following code produces this diagnostic because the value of count is never read:

void main() {
  int count = 0;
}

Common fixes

If the variable isn’t needed, then remove it.

If the variable was intended to be used, then add the missing code.

unused_result

‘{0}’ should be used. {1}.

The value of ‘{0}’ should be used.

Description

The analyzer produces this diagnostic when a function annotated with useResult is invoked, and the value returned by that function isn’t used. The value is considered to be used if a member of the value is invoked, if the value is passed to another function, or if the value is assigned to a variable or field.

Example

The following code produces this diagnostic because the invocation of c.a() isn’t used, even though the method a is annotated with useResult:

import 'package:meta/meta.dart';

class C {
  @useResult
  int a() => 0;

  int b() => 0;
}

void f(C c) {
  c.a();
}

Common fixes

If you intended to invoke the annotated function, then use the value that was returned:

import 'package:meta/meta.dart';

class C {
  @useResult
  int a() => 0;

  int b() => 0;
}

void f(C c) {
  print(c.a());
}

If you intended to invoke a different function, then correct the name of the function being invoked:

import 'package:meta/meta.dart';

class C {
  @useResult
  int a() => 0;

  int b() => 0;
}

void f(C c) {
  c.b();
}

unused_shown_name

The name {0} is shown, but isn’t used.

Description

The analyzer produces this diagnostic when a show combinator includes a name that isn’t used within the library. Because it isn’t referenced, the name can be removed.

Example

The following code produces this diagnostic because the function max isn’t used:

import 'dart:math' show min, max;

var x = min(0, 1);

Common fixes

Either use the name or remove it:

import 'dart:math' show min;

var x = min(0, 1);

uri_does_not_exist

Target of URI doesn’t exist: ‘{0}’.

Description

The analyzer produces this diagnostic when an import, export, or part directive is found where the URI refers to a file that doesn’t exist.

Example

If the file lib.dart doesn’t exist, the following code produces this diagnostic:

import 'lib.dart';

Common fixes

If the URI was mistyped or invalid, then correct the URI.

If the URI is correct, then create the file.

uri_has_not_been_generated

Target of URI hasn’t been generated: ‘{0}’.

Description

The analyzer produces this diagnostic when an import, export, or part directive is found where the URI refers to a file that doesn’t exist and the name of the file ends with a pattern that’s commonly produced by code generators, such as one of the following:

• .g.dart
• .pb.dart
• .pbenum.dart
• .pbserver.dart
• .pbjson.dart
• .template.dart

Example

If the file lib.g.dart doesn’t exist, the following code produces this diagnostic:

import 'lib.g.dart';

Common fixes

If the file is a generated file, then run the generator that generates the file.

If the file isn’t a generated file, then check the spelling of the URI or create the file.

uri_with_interpolation

URIs can’t use string interpolation.

Description

The analyzer produces this diagnostic when the string literal in an import, export, or part directive contains an interpolation. The resolution of the URIs in directives must happen before the declarations are compiled, so expressions can’t be evaluated while determining the values of the URIs.

Example

The following code produces this diagnostic because the string in the import directive contains an interpolation:

import 'dart:$m';

const m = 'math';

Common fixes

Remove the interpolation from the URI:

import 'dart:math';

var zero = min(0, 0);

use_of_native_extension

Dart native extensions are deprecated and aren’t available in Dart 2.15.

Description

The analyzer produces this diagnostic when a library is imported using the dart-ext scheme.

Example

The following code produces this diagnostic because the native library x is being imported using a scheme of dart-ext:

import 'dart-ext:x';

Common fixes

Rewrite the code to use dart:ffi as a way of invoking the contents of the native library.

use_of_void_result

This expression has a type of ‘void’ so its value can’t be used.

Description

The analyzer produces this diagnostic when it finds an expression whose type is void, and the expression is used in a place where a value is expected, such as before a member access or on the right-hand side of an assignment.

Example

The following code produces this diagnostic because f doesn’t produce an object on which toString can be invoked:

void f() {}

void g() {
  f().toString();
}

Common fixes

Either rewrite the code so that the expression has a value or rewrite the code so that it doesn’t depend on the value.

values_declaration_in_enum

A member named ‘values’ can’t be declared in an enum.

Description

The analyzer produces this diagnostic when an enum declaration defines a member named values, whether the member is an enum constant, an instance member, or a static member.

Any such member conflicts with the implicit declaration of the static getter named values that returns a list containing all the enum constants.

Example

The following code produces this diagnostic because the enum E defines an instance member named values:

enum E {
  v;
  void values() {}
}

Common fixes

Change the name of the conflicting member:

enum E {
  v;
  void getValues() {}
}

variable_pattern_keyword_in_declaration_context

Variable patterns in declaration context can’t specify ‘var’ or ‘final’ keyword.

Description

The analyzer produces this diagnostic when a variable pattern is used within a declaration context.

Example

The following code produces this diagnostic because the variable patterns in the record pattern are in a declaration context:

void f((int, int) r) {
  var (var x, y) = r;
  print(x + y);
}

Common fixes

Remove the var or final keyword(s) within the variable pattern:

void f((int, int) r) {
  var (x, y) = r;
  print(x + y);
}

variable_type_mismatch

A value of type ‘{0}’ can’t be assigned to a const variable of type ‘{1}’.

Description

The analyzer produces this diagnostic when the evaluation of a constant expression would result in a CastException.

Example

The following code produces this diagnostic because the value of x is an int, which can’t be assigned to y because an int isn’t a String:

const Object x = 0;
const String y = x;

Common fixes

If the declaration of the constant is correct, then change the value being assigned to be of the correct type:

const Object x = 0;
const String y = '$x';

If the assigned value is correct, then change the declaration to have the correct type:

const Object x = 0;
const int y = x;

wrong_number_of_parameters_for_operator

Operator ‘-‘ should declare 0 or 1 parameter, but {0} found.

Operator ‘{0}’ should declare exactly {1} parameters, but {2} found.

Description

The analyzer produces this diagnostic when a declaration of an operator has the wrong number of parameters.

Example

The following code produces this diagnostic because the operator + must have a single parameter corresponding to the right operand:

class C {
  int operator +(a, b) => 0;
}

Common fixes

Add or remove parameters to match the required number:

class C {
  int operator +(a) => 0;
}

wrong_number_of_parameters_for_setter

Setters must declare exactly one required positional parameter.

Description

The analyzer produces this diagnostic when a setter is found that doesn’t declare exactly one required positional parameter.

Examples

The following code produces this diagnostic because the setter s declares two required parameters:

class C {
  set s(int x, int y) {}
}

The following code produces this diagnostic because the setter s declares one optional parameter:

class C {
  set s([int x]) {}
}

Common fixes

Change the declaration so that there’s exactly one required positional parameter:

class C {
  set s(int x) {}
}

wrong_number_of_type_arguments

The type ‘{0}’ is declared with {1} type parameters, but {2} type arguments were given.

Description

The analyzer produces this diagnostic when a type that has type parameters is used and type arguments are provided, but the number of type arguments isn’t the same as the number of type parameters.

The analyzer also produces this diagnostic when a constructor is invoked and the number of type arguments doesn’t match the number of type parameters declared for the class.

Examples

The following code produces this diagnostic because C has one type parameter but two type arguments are provided when it is used as a type annotation:

class C<E> {}

void f(C<int, int> x) {}

The following code produces this diagnostic because C declares one type parameter, but two type arguments are provided when creating an instance:

class C<E> {}

var c = C<int, int>();

Common fixes

Add or remove type arguments, as necessary, to match the number of type parameters defined for the type:

class C<E> {}

void f(C<int> x) {}

wrong_number_of_type_arguments_constructor

The constructor ‘{0}.{1}’ doesn’t have type parameters.

Description

The analyzer produces this diagnostic when type arguments are provided after the name of a named constructor. Constructors can’t declare type parameters, so invocations can only provide the type arguments associated with the class, and those type arguments are required to follow the name of the class rather than the name of the constructor.

Example

The following code produces this diagnostic because the type parameters (<String>) follow the name of the constructor rather than the name of the class:

class C<T> {
  C.named();
}
C f() => C.named<String>();

Common fixes

If the type arguments are for the class’ type parameters, then move the type arguments to follow the class name:

class C<T> {
  C.named();
}
C f() => C<String>.named();

If the type arguments aren’t for the class’ type parameters, then remove them:

class C<T> {
  C.named();
}
C f() => C.named();

wrong_number_of_type_arguments_enum

The enum is declared with {0} type parameters, but {1} type arguments were given.

Description

The analyzer produces this diagnostic when an enum constant in an enum that has type parameters is instantiated and type arguments are provided, but the number of type arguments isn’t the same as the number of type parameters.

Example

The following code produces this diagnostic because the enum constant c provides one type argument even though the enum E is declared to have two type parameters:

enum E<T, U> {
  c<int>()
}

Common fixes

If the number of type parameters is correct, then change the number of type arguments to match the number of type parameters:

enum E<T, U> {
  c<int, String>()
}

If the number of type arguments is correct, then change the number of type parameters to match the number of type arguments:

enum E<T> {
  c<int>()
}

wrong_number_of_type_arguments_extension

The extension ‘{0}’ is declared with {1} type parameters, but {2} type arguments were given.

Description

The analyzer produces this diagnostic when an extension that has type parameters is used and type arguments are provided, but the number of type arguments isn’t the same as the number of type parameters.

Example

The following code produces this diagnostic because the extension E is declared to have a single type parameter (T), but the extension override has two type arguments:

extension E<T> on List<T> {
  int get len => length;
}

void f(List<int> p) {
  E<int, String>(p).len;
}

Common fixes

Change the type arguments so that there are the same number of type arguments as there are type parameters:

extension E<T> on List<T> {
  int get len => length;
}

void f(List<int> p) {
  E<int>(p).len;
}

wrong_number_of_type_arguments_method

The method ‘{0}’ is declared with {1} type parameters, but {2} type arguments are given.

Description

The analyzer produces this diagnostic when a method or function is invoked with a different number of type arguments than the number of type parameters specified in its declaration. There must either be no type arguments or the number of arguments must match the number of parameters.

Example

The following code produces this diagnostic because the invocation of the method m has two type arguments, but the declaration of m only has one type parameter:

class C {
  int m<A>(A a) => 0;
}

int f(C c) => c.m<int, int>(2);

Common fixes

If the type arguments are necessary, then make them match the number of type parameters by either adding or removing type arguments:

class C {
  int m<A>(A a) => 0;
}

int f(C c) => c.m<int>(2);

If the type arguments aren’t necessary, then remove them:

class C {
  int m<A>(A a) => 0;
}

int f(C c) => c.m(2);

yield_in_non_generator

Yield statements must be in a generator function (one marked with either ‘async*’ or ‘sync*’).

Yield-each statements must be in a generator function (one marked with either ‘async*’ or ‘sync*’).

Description

The analyzer produces this diagnostic when a yield or yield* statement appears in a function whose body isn’t marked with one of the async* or sync* modifiers.

Examples

The following code produces this diagnostic because yield is being used in a function whose body doesn’t have a modifier:

Iterable<int> get digits {
  yield* [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
}

The following code produces this diagnostic because yield* is being used in a function whose body has the async modifier rather than the async* modifier:

Stream<int> get digits async {
  yield* [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
}

Common fixes

Add a modifier, or change the existing modifier to be either async* or sync*:

Iterable<int> get digits sync* {
  yield* [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
}

yield_of_invalid_type

A yielded value of type ‘{0}’ must be assignable to ‘{1}’.

The type ‘{0}’ implied by the ‘yield*’ expression must be assignable to ‘{1}’.

Description

The analyzer produces this diagnostic when the type of object produced by a yield or yield* expression doesn’t match the type of objects that are to be returned from the Iterable or Stream types that are returned from a generator (a function or method marked with either sync* or async*).

Example

The following code produces this diagnostic because the getter zero is declared to return an Iterable that returns integers, but the yield is returning a string from the iterable:

Iterable<int> get zero sync* {
  yield '0';
}

Common fixes

If the return type of the function is correct, then fix the expression following the keyword yield to return the correct type:

Iterable<int> get zero sync* {
  yield 0;
}

If the expression following the yield is correct, then change the return type of the function to allow it:

Iterable<String> get zero sync* {
  yield '0';
}