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