New test, checking allowed usages of void typed expressions

Change-Id: I9a3096bb3fb3deca178688029657cd71f8aedf7d
Reviewed-on: https://dart-review.googlesource.com/30840
Reviewed-by: Lasse R.H. Nielsen <lrn@google.com>
Commit-Queue: Erik Ernst <eernst@google.com>
This commit is contained in:
Erik Ernst 2017-12-20 14:04:13 +00:00 committed by commit-bot@chromium.org
parent 4e162c5318
commit aa9765604e

View file

@ -0,0 +1,37 @@
// Copyright (c) 2017, 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.
// Dart test for static checks on situations where expressions of type void
// can be used. The point is simply that there are no compile-time errors.
// We need expressions of type `void`; `void x` would do, but using
// `void get x` will work with tools that do not yet handle `void x`.
void get x => null;
void use(dynamic x) {}
main() {
// In an expressionStatement `e;`, e may have type void.
x;
// In the initialization and increment expressions of a for-loop,
// `for (e1; e2; e3) {..}`, `e1` and `e3` may have type void.
for (x;; x) {
break;
}
// In a typeCast `e as T`, `e` may have type void.
var y = x as Object;
// In a parenthesized expression `(e)`, `e` may have type void.
(x);
// In a return statement `return e;`, when the return type of the
// innermost enclosing function is the type void, e may have type void.
void f() => x;
void g() {
return x;
}
}