Re: [musl] pthread_cond_wait may has a error

[Patrick Oppenlander <patrick.oppenlander@gmail.com>]

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On Thu, 10 Aug 2023, 23:05 祁金全, <qijinquan@kaihong.com> wrote:

>
> Hi,
> In my demo, I create 3 threads, the first and the second thread are wait
> for the global pthread_cond with different pthread_mutex, the last thread
> will call pthread_cond_broadcast.
>
> When I build without musl, the output like this:
> task 3 pthread_cond_broadcast end time 0.000054
> task 2 pthread_cond_wait end time 0.000566
> task 1 pthread_cond_wait end time 0.000612
> task 2 pthread_cond_broadcast end time 1.012233
> task 1 pthread_cond_broadcast end time 1.012247
>
> But when I build with static lib of musl(libc.a) and run , the output like
> this:
> task 3 pthread_cond_broadcast end time 0.000026
> task 1 pthread_cond_wait end time 0.000074
> task 1 pthread_cond_broadcast end time 1.013168
> task 2 pthread_cond_wait end time 1.013245
> task 2 pthread_cond_broadcast end time 2.022992
>
> it seems that the second wakeup thead is waiting until the first wakeup
> thead unlock the mutex(different mute in the two threads).
>
>
>
>
>
> musl version : 1.2.4
>
> pthread_cond_t g_cond = PTHREAD_COND_INITIALIZER;
> pthread_mutex_t g_mutex1 = PTHREAD_MUTEX_INITIALIZER;
> pthread_mutex_t g_mutex2 = PTHREAD_MUTEX_INITIALIZER;
> auto tStart = std::chrono::steady_clock::now();
> void* ThreadTaskOne(void* arg)
> {
>     pthread_mutex_lock(&g_mutex1);
>     pthread_cond_wait(&g_cond, &g_mutex1);
>     auto tEnd = std::chrono::steady_clock::now();
>     auto diff = std::chrono::duration<double>(tEnd - tStart);
>     printf("task 1 pthread_cond_wait end time %f\n", diff.count());
>
>     sleep(1);
>     pthread_mutex_unlock(&g_mutex1);
>     tEnd = std::chrono::steady_clock::now();
>     diff = std::chrono::duration<double>(tEnd - tStart);
>     printf("task 1 pthread_cond_broadcast end time %f\n", diff.count());
>     return nullptr;
> }
>
> void* ThreadTaskTwo(void* arg)
> {
>     pthread_mutex_lock(&g_mutex2);
>     pthread_cond_wait(&g_cond, &g_mutex2);
>     auto tEnd = std::chrono::steady_clock::now();
>     auto diff = std::chrono::duration<double>(tEnd - tStart);
>     printf("task 2 pthread_cond_wait end time %f\n", diff.count());
>
>     sleep(1);
>     pthread_mutex_unlock(&g_mutex2);
>     tEnd = std::chrono::steady_clock::now();
>     diff = std::chrono::duration<double>(tEnd - tStart);
>     printf("task 2 pthread_cond_broadcast end time %f\n", diff.count());
>     return nullptr;
> }
>
> void* BroadcastNotifyMutex(void* arg)
> {
>     tStart = std::chrono::steady_clock::now();
>     pthread_cond_broadcast(&g_cond);
>     auto tEnd = std::chrono::steady_clock::now();
>     auto diff = std::chrono::duration<double>(tEnd - tStart);
>     printf("task 3 pthread_cond_broadcast end time %lf\n", diff.count());
>     return nullptr;
> }
> int main()
> {
>     pthread_t threadOne, threadTwo, threadThree;
>     pthread_create(&threadOne, nullptr, ThreadTaskOne, nullptr);
>     pthread_create(&threadTwo, nullptr, ThreadTaskTwo, nullptr);
>     sleep(3);
>     pthread_create(&threadThree, nullptr, BroadcastNotifyMutex, nullptr);
>     pthread_join(threadOne, nullptr);
>     pthread_join(threadTwo, nullptr);
>     pthread_join(threadThree, nullptr);
>     return 0;
> }
>


I don't think that's a valid test:

https://pubs.opengroup.org/onlinepubs/7908799/xsh/pthread_cond_wait.html

"The effect of using more than one mutex for concurrent
*pthread_cond_wait()* or *pthread_cond_timedwait()* operations on the same
condition variable is undefined; that is, a condition variable becomes
bound to a unique mutex when a thread waits on the condition variable, and
this (dynamic) binding ends when the wait returns."

Patrick

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