type_test_with_non_type

The name '{0}' isn't a type and can't be used in an 'is' expression.

The analyzer produces this diagnostic when the right-hand side of an is or is! test isn't a type.

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;
  }
}

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;
  }
}

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