type_test_with_undefined_name

The name '{0}' isn't defined, so it can't be used in an 'is' expression.

The analyzer produces this diagnostic when the name following the is in a type test expression isn't defined.

The following code produces this diagnostic because the name Srting isn't defined:

void f(Object o) {
  if (o is Srting) {
    // ...
  }
}

Replace the name with the name of a type:

void f(Object o) {
  if (o is String) {
    // ...
  }
}

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