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Bug 1240985 - Always enqueue OnMaybeDequeueOne task when receiving a message (r=dvander)

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This commit is contained in:
Bill McCloskey 2016-01-22 18:04:08 -08:00
parent 4015cde06b
commit bd056f2482
1 changed files with 11 additions and 10 deletions
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21
ipc/glue/MessageChannel.cpp
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@ -773,22 +773,23 @@ MessageChannel::OnMessageReceivedFromLink(const Message& aMsg)
if (shouldWakeUp) {
NotifyWorkerThread();
} else {
// Worker thread is either not blocked on a reply, or this is an
// incoming Interrupt that raced with outgoing sync, and needs to be
// deferred to a later event-loop iteration.
if (!compress) {
// If we compressed away the previous message, we'll re-use
// its pending task.
mWorkerLoop->PostTask(FROM_HERE, new DequeueTask(mDequeueOneTask));
}
}
// It's possible that a Send or Call call on the stack will process the
// message. However, it's difficult to ensure that Send always processes all
// messages in mPending before exiting, so it's safer to enqueue a task for
// each one. If the queue is empty, the task does nothing.
if (!compress) {
// If we compressed away the previous message, we'll re-use
// its pending task.
mWorkerLoop->PostTask(FROM_HERE, new DequeueTask(mDequeueOneTask));
}
}
void
MessageChannel::ProcessPendingRequests(int transaction, int prio)
{
IPC_LOG("ProcessPendingRequests");
IPC_LOG("ProcessPendingRequests for seqno=%d, xid=%d", seqno, transaction);
// Loop until there aren't any more priority messages to process.
for (;;) {