Strange error when using extension types as generic types

[jellynoone]

This example is analyzed without issues.

void main() {
  final sink = MySink<ExtType<String>>().spying(
    onAdd: (value) {
      print('Spied added value: $value');
    },
  );

  sink.add(ExtType<String>('Hello'));
  sink.add(ExtType<String>('World'));
}

extension type ExtType<T>(T value) {}

abstract interface class MySink<T extends ExtType> {
  void add(T value);

  factory MySink() = _MySink<T>;
}

class _MySink<T extends ExtType> implements MySink<T> {
  @override
  void add(T value) {
    print('Added: $value');
  }
}

extension MySinkExt<T extends ExtType> on MySink<T> {
  MySink<T> spying({required void Function(T value) onAdd}) {
    return _Spying(this, onAdd); // <- missing <T>, causes compile error
  }
}

class _Spying<T extends ExtType> implements MySink<T> {
  final MySink<T> _delegate;
  final void Function(T value) _onAdd;

  _Spying(this._delegate, this._onAdd);

  @override
  void add(T value) {
    _delegate.add(value);
    _onAdd(value);
  }
}

However, when running it, the program doesn't compile. It produces a cryptic compile time error message:

compileNewDDC
main.dart:29:20: Error: The argument type 'MySink<T>' can't be assigned to the parameter type 'MySink<T>'.
 - 'MySink' is from 'package:dartpad_sample/main.dart' ('/tmp/dartpadKTFFEK/lib/main.dart').
    return _Spying(this, onAdd);
                   ^

However, if we manually type out: _Spying<T>, everything works as expected.
This issue is present in all versions of Dart available on DartPad today.

[SeifShaheen]

Hi, I am new contributor, And I want to be assigned to this issue