c - 尝试理解 pthread_cond_signal()

Question

嗨，我是 C 新手，正在尝试了解互斥体、条件和线程。我了解线程如何工作的基础知识。如果我错了，请纠正我，据我所知，一个线程正在等待另外两个线程发送信号以唤醒他。由于互斥体的原因，代码一次只能由一个线程运行，线程 2 获取锁并运行代码，sleep(1) 有助于交替线程，因此线程 2 不会运行再次执行，因此线程 3 获取锁并运行代码，依此类推。仅锁定和释放互斥锁的线程。一旦达到 COUNT_LIMIT 12，它就会向正在等待的线程(即线程 1)发送一个信号。如果我没有误解任何内容，我就理解到这里了

所以我的问题是，这里的几段代码位于 void *watch_count 中:

pthread_mutex_lock(&count_mutex);

pthread_cond_wait(&count_threshold_cv, &count_mutex);

pthread_mutex_unlock(&count_mutex);

它是否获取与void *inc_count中的锁相同的锁？如果是这样，它会锁定互斥锁，并且 pthread_cond_wait 正在等待 void *watch_count 中的信号，然后继续执行代码？ pthread_mutex_lock 和 pthread_mutex_unlock 在 void *watch_count 中做什么，这是否在等待线程 2 和 3 时阻止执行线程 1 完成它的工作吗？

我希望我说得有道理或解释了我正确的理解，提前感谢您提供的任何建议!

#define NUM_THREADS  3
#define TCOUNT 10
#define COUNT_LIMIT 12

int     count = 0;
pthread_mutex_t count_mutex;
pthread_cond_t count_threshold_cv;

void *inc_count(void *t) 
{
  int i;
  long my_id = (long)t;

  for (i=0; i < TCOUNT; i++) {
    pthread_mutex_lock(&count_mutex);
    count++;

    /* 
    Check the value of count and signal waiting thread when condition is
    reached.  Note that this occurs while mutex is locked. 
    */
    if (count == COUNT_LIMIT) {
      printf("inc_count(): thread %ld, count = %d  Threshold reached. ",
             my_id, count);
      pthread_cond_signal(&count_threshold_cv);
      printf("Just sent signal.\n");
      }
    printf("inc_count(): thread %ld, count = %d, unlocking mutex, i: %d\n", 
       my_id, count, i);
    pthread_mutex_unlock(&count_mutex);

    /* Do some work so threads can alternate on mutex lock */
    sleep(1);
    }
  pthread_exit(NULL);
}

void *watch_count(void *t) 
{
  long my_id = (long)t;

  printf("Starting watch_count(): thread %ld\n", my_id);

  pthread_mutex_lock(&count_mutex);
  while (count < COUNT_LIMIT) {
    printf("watch_count(): thread %ld Count= %d. Going into wait...\n", my_id,count);
    pthread_cond_wait(&count_threshold_cv, &count_mutex);
    printf("watch_count(): thread %ld Condition signal received. Count= %d\n", my_id,count);
    printf("watch_count(): thread %ld Updating the value of count...\n", my_id,count);
    count += 125;
    printf("watch_count(): thread %ld count now = %d.\n", my_id, count);
    }
  printf("watch_count(): thread %ld Unlocking mutex.\n", my_id);
  pthread_mutex_unlock(&count_mutex);
  pthread_exit(NULL);
}

int main(int argc, char *argv[])
{
  int i, rc; 
  long t1=1, t2=2, t3=3;
  pthread_t threads[3];
  pthread_attr_t attr;

  /* Initialize mutex and condition variable objects */
  pthread_mutex_init(&count_mutex, NULL);
  pthread_cond_init (&count_threshold_cv, NULL);

  /* For portability, explicitly create threads in a joinable state */
  pthread_attr_init(&attr);
  pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
  pthread_create(&threads[0], &attr, watch_count, (void *)t1);
  pthread_create(&threads[1], &attr, inc_count, (void *)t2);
  pthread_create(&threads[2], &attr, inc_count, (void *)t3);

  /* Wait for all threads to complete */
  for (i = 0; i < NUM_THREADS; i++) {
    pthread_join(threads[i], NULL);
  }
  printf ("Main(): Waited and joined with %d threads. Final value of count = %d. Done.\n", 
          NUM_THREADS, count);

  /* Clean up and exit */
  pthread_attr_destroy(&attr);
  pthread_mutex_destroy(&count_mutex);
  pthread_cond_destroy(&count_threshold_cv);
  pthread_exit (NULL);

}

Answer 1

是的，您的所有线程都获取相同的锁(该程序中只有一个互斥锁 - count_mutex )。

何时 watch_count()来电 pthread_cond_wait() ，它会解锁互斥体并在一个原子操作中暂停。这允许其他线程获取锁。当调用 pthread_cond_wait() 的线程时唤醒后，会重新获取pthread_cond_wait()之前的锁返回(这意味着如果另一个线程锁定了互斥锁，则 pthread_cond_wait() 无法返回，直到另一个线程首先释放锁)。

当您调用pthread_cond_wait()时，您必须锁定互斥锁。 - 这是 API 的要求。此外，watch_count()线程锁定 count_mutex这样它就可以访问count变量而不引入竞争条件(来自另一个线程的 count 的并发修改)。请注意，没有线程访问 count变量 - 用于读取或写入 - 无需 count_mutex当时锁定。