In pthread, how to reliably pass signal to another thread?

Question

I'm trying to write a simple thread pool program in pthread. However, it seems that pthread_cond_signal doesn't block, which creates a problem. For example, let's say I have a "producer-consumer" program:

pthread_cond_t my_cond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t my_cond_m = PTHREAD_MUTEX_INITIALIZER;

void * liberator(void * arg)
{
    // XXX make sure he is ready to be freed
    sleep(1);

    pthread_mutex_lock(&my_cond_m);
    pthread_cond_signal(&my_cond);
    pthread_mutex_unlock(&my_cond_m);

    return NULL;
}

int main()
{
    pthread_t t1;
    pthread_create(&t1, NULL, liberator, NULL);

    // XXX Don't take too long to get ready. Otherwise I'll miss 
    // the wake up call forever
    //sleep(3);

    pthread_mutex_lock(&my_cond_m);
    pthread_cond_wait(&my_cond, &my_cond_m);
    pthread_mutex_unlock(&my_cond_m);

    pthread_join(t1, NULL);

    return 0;
}

As described in the two XXX marks, if I take away the sleep calls, then main() may stall because it has missed the wake up call from liberator(). Of course, sleep isn't a very robust way to ensure that either.

In real life situation, this would be a worker thread telling the manager thread that it is ready for work, or the manager thread announcing that new work is available.

How would you do this reliably in pthread?

---

精

@Borealid的答案有点作品,但他对这个问题的解释可能会更好.我建议任何看这个问题的人阅读评论中的讨论,以了解正在发生的事情.

特别是,我自己会修改他的答案和这样的代码示例,以使其更清楚.(因为Borealid的原始答案,虽然编译和工作,但我很困惑)

// In main
pthread_mutex_lock(&my_cond_m);

// If the flag is not set, it means liberator has not 
// been run yet. I'll wait for him through pthread's signaling 
// mechanism

// If it _is_ set, it means liberator has been run. I'll simply 
// skip waiting since I've already synchronized. I don't need to 
// use pthread's signaling mechanism
if(!flag) pthread_cond_wait(&my_cond, &my_cond_m);

pthread_mutex_unlock(&my_cond_m);

// In liberator thread
pthread_mutex_lock(&my_cond_m);

// Signal anyone who's sleeping. If no one is sleeping yet, 
// they should check this flag which indicates I have already 
// sent the signal. This is needed because pthread's signals 
// is not like a message queue -- a sent signal is lost if 
// nobody's waiting for a condition when it's sent.
// You can think of this flag as a "persistent" signal
flag = 1;
pthread_cond_signal(&my_cond);
pthread_mutex_unlock(&my_cond_m);

Answer 1

使用同步变量.

在main:

pthread_mutex_lock(&my_cond_m);
while (!flag) {
    pthread_cond_wait(&my_cond, &my_cond_m);
}
pthread_mutex_unlock(&my_cond_m);

在线程中:

pthread_mutex_lock(&my_cond_m);
flag = 1;
pthread_cond_broadcast(&my_cond);
pthread_mutex_unlock(&my_cond_m);

对于生产者 - 消费者问题,这将是消费者在缓冲区为空时休眠,并且生产者在其满时时休眠.记得在访问全局变量之前获取锁.