blob: 291e6cd420211e7a81e46f6e30a9f86b5f69080d [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 if a
/// member of the receiver type tis imported with a prefix.
/// @author
import "../../Utils/expect.dart";
import "extension_conflict_resolution_lib.dart" as testlib;
extension MySimpleExt on String {
bool get isLibraryVersion => false;
main() {
dynamic aString = "testme";