LanguageFeatures/Extension-methods/extension_conflict_resolution_t06.dart

/*
 * Copyright (c) 2019, the Dart project authors.  Please see the AUTHORS file
 * for details. All rights reserved. Use of this source code is governed by a
 * BSD-style license that can be found in the LICENSE file.
 */
/**
 * @assertion If more than one extension applies to a specific member
 * invocation, then we resort to a heuristic to choose one of the extensions to
 * apply. If exactly one of them is "more specific" than all the others, that
 * one is chosen. Otherwise it is a compile-time error.
 *
 * An extension with [on] type clause [T1] is more specific than another
 * extension with [on] type clause [T2] iff
 *
 *   1. The latter extension is declared in a platform library, and the former
 *      extension is not
 *   2. they are both declared in platform libraries or both declared in
 *      non-platform libraries, and
 *   3. the instantiated type (the type after applying type inference from the
 *      receiver) of [T1] is a subtype of the instantiated type of [T2] and
 *      either
 *   4. not vice versa, or
 *   5. the instantiate-to-bounds type of [T1] is a subtype of the
 *      instantiate-to-bounds type of [T2] and not vice versa.
 * @description Check that compile time error is thrown if two extension members
 * with [on] type clause are equitable and a member from the conflicting
 * extensions is referenced.
 * @compile-error
 * @author iarkh@unipro.ru
 */
// SharedOptions=--enable-experiment=extension-methods

extension MyFancyList1<T> on List<T> {
  bool get isTested => false;
}

extension MyIntFancyList2<T> on List<T> {
  bool get isTested => false;
}

main() {
  List aList = [1, 2, 3];
  bool res = aList.isTested;
}