Question

我从这个网站获得了以下代码：

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＆＃34;计数＆＃34;变量。该第三个线程等待，直到count变量达到指定值。

我的问题是 - 下面的代码如何确保两个工作线程中的一个在观察者线程锁定之前不会锁定互斥锁？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)永远不会被调用？

我假设pthread_create()实际上也开始了这个问题。这是唯一的原因，因为观察者线程的pthread_create()在两个工作线程的pthread_create()之前开始？！当然这不是铸铁，调度可能导致工作线程在观察者线程之前开始？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);

}

Answer 1

我的问题是 - 以下代码如何确保两个工作线程中的一个在观察者线程锁定之前不会锁定＆gt;互斥锁？

代码并不需要确保这一点。它并不依赖于调用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条件 - 或者更确切地说是反向，当count >= COUNT_LIMIT时 - 观察者线程知道其他线程已完成。

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

那就是说，这个例子看起来有些愚蠢，但通过count += 125;

进行观察者线程想要实现的目标并不十分清楚

Answer 2

代码中的注释说明您不必担心：

另请注意，如果在此例程运行之前达到COUNT_LIMIT 等待线程，将跳过循环以防止pthread_cond_wait 从不回来。

实际上，如果您注意到，只有在COUNT_LIMIT尚未到达count时才会运行while循环。如果是这种情况，则根本不会调用pthread_cond_signal。

这个pthread_cond_wait（）示例如何工作？

2 个答案: