blob: db70b00a5fecbed1e3896ceb591d2fd6c15e3f49 [file] [log] [blame]
// 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 two or more extensions apply to the same member access, or if a
/// member of the receiver type takes precedence over an extension method, or if
/// the extension is imported with a prefix, then it is possible to force an
/// extension member invocation:
/// MyList(object).quickSort();
/// or if you don't want the type argument to the extension to be inferred:
/// MyList<String>(object).quickSort();
/// or if you imported the extension with a prefix to avoid name collision:
/// prefix.MyList<String>(object).quickSort();
/// The syntax looks like a constructor invocation, but it does not create a new
/// object.
/// If [object.quickSort()] would invoke an extension method of [MyList] if
/// [MyList] was the only extension in scope, then [MyList(object).quickSort()]
/// will invoke the exact same method in the same way.
/// @description Check that if several extensions can be applied to the same
/// member access, it's possible to force an extension member invocation
/// @author
import "../../Utils/expect.dart";
extension Ext1<T extends int> on List<T> {
bool get isExt1 => true;
extension Ext2<T extends num> on List<T> {
bool get isExt1 => false;
main() {
List<int> list = <int>[1, 2, 3];