正确实现condition_variable timed_wait

Question

我正在阅读我的STL实现(标准问题g++ 4.6.2),并在以下内容中遇到了这种竞争条件condition_variable:

template<typename _Rep, typename _Period>
cv_status
wait_for(unique_lock<mutex>& __lock,
         const chrono::duration<_Rep, _Period>& __rtime)
{
    return wait_until(__lock, __clock_t::now() + __rtime);
}

因为__clock_t是一个std::chrono::system_clock,我们与NTP之类的东西有关(如果时钟在一天之后被移回__clock_t::now() + __rtime,那么我们将等待一天).

C++标准(30.5.1)看起来是正确的:

26

效果:好像

return wait_until(lock, chrono::steady_clock::now() + rel_time);

Boost的condition_variable实现有同样的问题:

template<typename duration_type>
bool timed_wait(unique_lock<mutex>& m,duration_type const& wait_duration)
{
    return timed_wait(m,get_system_time()+wait_duration);
}

实际上,底层的pthreads实现似乎是问题所在:

int pthread_cond_timedwait(pthread_cond_t *restrict cond,
   pthread_mutex_t *restrict mutex,
   const struct timespec *restrict abstime);

因为abstime被指定为"系统时间",而不是单调时钟.

所以我的问题是:如何才能std::condition_variable::wait_for正确实现？是否有现成的实现可以做到这一点？或者我错过了什么？

Answer 1

诀窍是使用a pthread_condattr_setclock告诉pthread_condattr_t使用CLOCK_MONOTONIC.这样做的C代码非常简单:

#include <time.h>
#include <pthread.h>

#include <errno.h>
#include <stdio.h>

int main()
{
    // Set the clock to be CLOCK_MONOTONIC
    pthread_condattr_t attr;
    pthread_condattr_init(&attr);
    if (int err = pthread_condattr_setclock(&attr, CLOCK_MONOTONIC))
    {
        printf("Error setting clock: %d\n", err);
    }

    // Now we can initialize the pthreads objects with that condattr
    pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
    pthread_cond_t  cond;
    pthread_cond_init(&cond, &attr);

    // when getting the time, we must poll from CLOCK_MONOTONIC
    struct timespec timeout;
    struct timespec now;
    clock_gettime(CLOCK_MONOTONIC, &now);
    timeout.tv_sec = now.tv_sec + 5;
    timeout.tv_nsec = now.tv_nsec;

    // business as usual...
    pthread_mutex_lock(&mutex);
    int rc = pthread_cond_timedwait(&cond, &mutex, &timeout);
    if (rc == ETIMEDOUT)
        printf("Success!\n");
    else
        printf("Got return that wasn't timeout: %d\n", rc);
    pthread_mutex_unlock(&mutex);

    return 0;
}

我打算暂时离开,因为有人可能会有一个更简单的答案.我对此不满意的是,它意味着wait_until使用实时时钟实现起来相当困难(我最好的解决方案是将提供Clock的time_point转换为steady_clock时间并从那里开始...它仍然受到时间变化竞争条件的影响,但如果你实时指定超时,那你就已经犯了一个可怕的错误.