v8/test/debugger/debug/es6/debug-stepin-generators.js
yangguo 416e423fdb [debugger] step-next across yield should not leave the generator.
Stepping in a generator now behaves similar to stepping inside an
async function. Stepping in or next at a yield expression will result in
a break inside the same generator when we return to the generator.
Behavior of step-out does not change.

R=jgruber@chromium.org, neis@chromium.org
BUG=chromium:496865

Review-Url: https://codereview.chromium.org/2519853002
Cr-Commit-Position: refs/heads/master@{#41132}
2016-11-21 11:05:08 +00:00

49 lines
992 B
JavaScript

// Copyright 2014 the V8 project authors. All rights reserved.
// Use of this source code is governed by a BSD-style license that can be
// found in the LICENSE file.
Debug = debug.Debug
var exception = null;
var yields = 0;
function listener(event, exec_state, event_data, data) {
if (event != Debug.DebugEvent.Break) return;
try {
var source = exec_state.frame(0).sourceLineText();
print(source);
if (/stop stepping/.test(source)) return;
if (/yield/.test(source)) yields++;
if (yields == 4) {
exec_state.prepareStep(Debug.StepAction.StepOut);
} else {
exec_state.prepareStep(Debug.StepAction.StepIn);
}
} catch (e) {
print(e, e.stack);
exception = e;
}
};
Debug.setListener(listener);
function* g() {
for (var i = 0; i < 3; ++i) {
yield i;
}
}
var i = g();
debugger;
for (var num of g()) {}
i.next();
print(); // stop stepping
// Not stepped into.
i.next();
i.next();
assertNull(exception);
assertEquals(4, yields);