如何阻止线程在 pthread_join 上停止?

How do I stop threads stalling on pthread_join?

我有一个项目,我正在将作业添加到队列中,并且我有多个线程接受作业,并计算它们自己的独立结果。

我的程序处理 SIGINT 信号,我正在尝试加入线程以将结果相加,打印到屏幕,然后退出。我的问题是当我发送信号时线程似乎停止运行,或者它们在 mutex_lock 上被阻塞。为了简明扼要,下面是我程序的重要部分。

main.c

//the thread pool has a queue of jobs inside
//called jobs (which is a struct)
struct thread_pool * pool;

void signal_handler(int signo) {
    pool->jobs->running = 0; //stop the thread pool
    pthread_cond_broadcast(pool->jobs->cond);

    for (i = 0; i < tpool->thread_count; i++) {
        pthread_join(tpool->threads[i], retval);
        //do stuff with retval
    }

    //print results then exit
    exit(EXIT_SUCCESS);
}

int main() {
    signal(SIGINT, signal_handler);
    //set up threadpool and jobpool
    //start threads (they all run the workerThread function)
    while (1) {
        //send jobs to the job pool
    }
    return 0;
}

thread_stuff.c

void add_job(struct jobs * j) {
    if (j->running) {
        pthread_mutex_lock(j->mutex);
        //add job to queue and update count and empty
        pthread_cond_signal(j->cond);
        pthread_mutex_unlock(j->mutex);
    }
}

struct job * get_job(struct jobs * j) {

    pthread_mutex_lock(j->mutex);

    while (j->running && j->empty)
        pthread_cond_wait(j->cond, j->mutex);

    if (!j->running || j->empty) return NULL;

    //get the next job from the queue
    //unlock mutex and send a signal to other threads
    //waiting on the condition
    pthread_cond_signal(j->cond);
    pthread_mutex_unlock(j->mutex);
    //return new job
}

void * workerThread(void * arg) {
    struct jobs * j = (struct jobs *) arg;
    int results = 0;
    while (j->running) {
        //get next job and process results
    }
    return results;
}

谢谢你的帮助,这让我很头疼!

您不应从处理异步生成的信号(例如 SIGINT)的信号处理程序调用 pthread_cond_waitpthread_join。相反,您应该为所有线程阻塞 SIGINT,生成一个专用线程,然后在那里调用 sigwait。这意味着您在信号处理程序上下文之外检测到 SIGINT 信号的到达,因此您不限于 async-signal-safe functions。如果信号被传送到其中一个工作线程,您还可以避免自死锁的风险。

此时,您只需有序地关闭工作queue/thread 池即可。根据详细信息,您现有的带有 running 标志的方法甚至可能不会改变。