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一个“count”变量。这个
第三个线程等待count变量达到指定值。
我的问题是-下面的代码如何确保在观察线程锁定互斥锁之前,两个工作线程中的一个不会锁定互斥锁?如果发生这种情况,观察线程将被锁定,并且pthread_cond_wait(&count_threshold_cv, &count_mutex)将永远不会被调用?
我假设pthread_create()实际上也开始线程。这是因为观察线程的pthread_create()在两个工作线程的pthread_create()之前开始的唯一原因吗?!这肯定不是铸铁的,调度可能导致工作线程在观察线程之前开始?甚至编译器也可能重新排序这些代码行?

#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);

}

最佳答案

我的问题是-下面的代码如何确保在观察线程锁定>互斥锁之前,两个工作线程中的一个不会锁定它?
代码不需要确保这一点。它不依赖于调用pthread cond_wait()的观察线程。
观察线程检查count<COUNT_LIMIT,这是线程关心的实际情况,或者相反,当count >= COUNT_LIMIT时,观察线程知道其他线程已经完成。
pthread_cond_wait()中使用的pthread条件变量只是在线程未完成时才需要的,因此可以将观察线程置于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watcher线程希望通过执行count += 125;

关于c - 这个pthread_cond_wait()示例如何工作?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23953849/

10-11 22:53
查看更多