Re: Condition Mutexes

[Jérôme M. Berger]

dsimcha wrote:

== Quote from Bartosz Milewski (bartosz-nos...@relisoft.com)'s article

dsimcha Wrote:

void main() {
    condition = new Condition( new Mutex() );
    auto T = new Thread(&waitThenPrint);
    T.start();
    condition.notify();  // Never wakes up and prints FOO.
}

Your program terminates immediately after sending the notification. You need to

stall the exit until the other thread has a chance to wake up.

Thanks.  I've implemented this, along w/ one other suggestion from another 
poster.
 Here's the new program.  It still doesn't work.  Has anyone successfully used
core.sync.condition from druntime *on D2, not the Tango version on D1*?  If it
works, then it should be better documented so people who aren't already 
threading
gurus can figure out how to use it.  If it doesn't work, then as soon as I can
confirm that I'm not the problem, I'll go file a bug report.

Bartosz, since you're a threading guru, could you please write a simple test
program using core.sync.condition and see if it works, and if not either file a
bug report or let me know?

import core.sync.mutex, core.sync.condition, core.thread, std.stdio;

__gshared Condition condition;
__gshared Mutex mutex;

void waitThenPrint() {
    mutex.lock;
    condition.wait();
    mutex.unlock;
    writeln("FOO");
}

void main() {
    mutex = new Mutex;
    condition = new Condition(mutex);
    auto T = new Thread(&waitThenPrint);
    T.start();
    condition.notify();  // Never wakes up and prints FOO.
    T.join;
}

Well, you should lock the mutex before calling condition.notify and release it afterwards.

                Jerome
--
mailto:jeber...@free.fr
http://jeberger.free.fr
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Note that there is always a chance that "notify" will be called before the thread starts waiting. In that case your program will deadlock (randomly).

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