这个 pthread_cond_wait() 示例如何工作？

Question

我从这个网站得到了以下代码：

https://computing.llnl.gov/tutorials/pthreads/#Abstract

这个简单的示例代码演示了几个 Pthread 的使用 条件变量例程。主例程创建三个线程。 其中两个线程执行工作并更新“计数”变量。这 第三个线程一直等到计数变量达到指定值。

我的问题是 - 下面的代码如何确保在观察者线程锁定互斥体之前，两个工作线程之一不会锁定互斥体？如果发生这种情况，观察者线程将被锁定，pthread_cond_wait(&count_threshold_cv, &count_mutex) 将永远不会被调用？

我假设pthread_create() 实际上也开始了线程。这是因为观察者线程的pthread_create() 在两个工作线程的pthread_create() 之前开始的唯一原因吗？！当然这不是铸铁，调度可能会导致工作线程在观察者线程之前开始？甚至编译器也可能对这些代码行重新排序？

#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>

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

int     count = 0;
int     thread_ids[3] = {0,1,2};
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) {
      pthread_cond_signal(&count_threshold_cv);
      printf("inc_count(): thread %ld, count = %d  Threshold reached.\n", 
             my_id, count);
      }
    printf("inc_count(): thread %ld, count = %d, unlocking mutex\n", 
       my_id, count);
    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);

  /*
  Lock mutex and wait for signal.  Note that the pthread_cond_wait 
  routine will automatically and atomically unlock mutex while it waits. 
  Also, note that if COUNT_LIMIT is reached before this routine is run by
  the waiting thread, the loop will be skipped to prevent pthread_cond_wait
  from never returning. 
  */
  pthread_mutex_lock(&count_mutex);
  while (count<COUNT_LIMIT) {
    pthread_cond_wait(&count_threshold_cv, &count_mutex);
    printf("watch_count(): thread %ld Condition signal received.\n", my_id);
    count += 125;
    printf("watch_count(): thread %ld count now = %d.\n", my_id, count);
    }
  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 on %d  threads. Done.\n", NUM_THREADS);

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

}

Answer 1

我的问题是 - 下面的代码如何确保两个工作线程之一在观察者线程锁定之前不会锁定 >mutex？

代码不需要确保这一点。它不依赖于调用 pthread_cond_wait() 的观察者线程。

观察者线程检查count<COUNT_LIMIT，这是线程关心的实际情况 - 或者相反，当count >= COUNT_LIMIT - 观察者线程知道其他线程已完成。

pthread_cond_wait() 中使用的 pthread 条件变量只是在线程未完成的情况下需要，因此观察者线程可以进入睡眠状态并唤醒以检查它关心的条件。

也就是说，这个例子看起来有点傻，不太清楚观察者线程想要通过count += 125;实现什么

Answer 2

您的代码中的注释说明您不必担心：

另外，请注意，如果在运行此例程之前达到 COUNT_LIMIT 等待线程，将跳过循环以防止 pthread_cond_wait 从永远不会回来。

事实上，如果您注意到，while 循环仅在COUNT_LIMIT 尚未到达count 时运行。如果是这种情况，则根本不会调用 pthread_cond_signal。