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.

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.

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>();

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>();

Unless stated otherwise, the documentation on this site reflects Dart 3.8.1. Page last updated on 2025-05-08. View source or report an issue.