试图理解 pthread_cond_signal()

Question

您好，我是 C 新手，正在尝试了解互斥锁、条件和线程。我有线程如何工作的基础知识。如果我错了，请纠正我，据我了解，一个线程正在等待另一个 2 个线程发送信号以唤醒他。由于互斥锁，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 不会再次运行它，因此线程 3 获取锁和运行代码等等。只有锁定和释放互斥锁的线程。一旦它到达COUNT_LIMIT 12，它就会向线程 1 的等待线程发送一个信号。如果我没有误解任何内容，我到此为止了解

所以我的问题是void *watch_count 中的这几段代码：

pthread_mutex_lock(&count_mutex);

pthread_cond_wait(&count_threshold_cv, &count_mutex);

pthread_mutex_unlock(&count_mutex);

它是否获得了与void *inc_count 中相同的锁？如果是这样，它会锁定互斥体并且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 中做了什么，这是否在等待线程 1 完成工作时阻止线程 2 和 3 执行？

我希望我能理解或解释我正确理解的内容，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，你所有的线程都在获取同一个锁（这个程序中只有一个互斥锁 - count_mutex）。

当watch_count() 调用pthread_cond_wait() 时，它会解锁互斥锁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()的线程唤醒时，它会在pthread_cond_wait()返回之前重新获取锁（这意味着如果另一个线程锁定了互斥锁，pthread_cond_wait()不能返回，直到另一个线程先释放锁)。

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() 线程锁定了count_mutex，以便它可以访问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，没有个线程访问count变量——无论是读还是写——都没有锁定count_mutex。