Question

这是一个pthread代码，即使我已经设置了一个条件变量来等待信号，它也不会跳出while循环。

感谢任何帮助。

  #include <pthread.h>
  #include <iostream>
  using namespace std; 

  pthread_mutex_t myMutex = PTHREAD_MUTEX_INITIALIZER;

  int counter = 0; int WATCHCOUNT = 12 ;int totalCount = 10 ;
  int threadsID[3] = {0,1,2};
  pthread_cond_t cout_theashold_cv = PTHREAD_COND_INITIALIZER ; 

  void* watchCount(void *a) ;

  void* increaseCount(void *a)
 {
for (int i =0; i < totalCount ; ++i)
{
    pthread_mutex_lock(&myMutex);
    ++counter ;         

    cout << "I am thread " << *((int *)a) << " I increase count to " << counter << endl ;
    if (counter == WATCHCOUNT)
    {
        cout << "I am thread " << *((int *)a) << " I am before send signal with counter as " << counter << endl ;
        pthread_cond_signal(&cout_theashold_cv);
    }
    pthread_mutex_unlock(&myMutex);
  }
 } 

   int main()
   {
  pthread_t threads[3];
  pthread_create(&threads[0], NULL, increaseCount, (void *) &(threadsID[0]));
  pthread_create(&threads[1], NULL, increaseCount, (void *) &(threadsID[1]));   
  pthread_create(&threads[2], NULL, watchCount, (void *) &(threadsID[2]));
  for (int i = 0 ; i < 3 ; ++i)
  {
      pthread_join(threads[i], NULL);
  }
   return 0 ; 

 }
  // wait for cond var 
  void* watchCount(void *a)
  {
        pthread_mutex_lock(&myMutex);
        cout << "I am thread " << *((int *)a) << " I am watching count " << counter << endl ;

   //while (counter <= WATCHCOUNT)
   while (counter <= INT_MAX)
   {
        cout << "I am thread " << *((int *)a) << " before watch count current count is " << counter << endl ;
    pthread_cond_wait(&cout_theashold_cv, &myMutex);

    break ;

    cout << "I am thread " << *((int *)a) << " after watch count current count is " << counter << endl ;

}

cout << "I am thread " << *((int *)a) << " after loop and watch count current count is " << counter << endl ;

pthread_mutex_unlock(&myMutex);
}

Answer 1

线程3和2都可能在线程3开始等待条件变量之前完成。

这意味着它永远不会收到信号。

首先启动线程3。在尝试启动1和2之前，确保它已到达并且正在等待条件。

你真正想要的是counted semaphore（就像一个条件变量，但它保留了signal（）和wait（）调用次数的计数）。如果它有比等待更多的信号，则线程不会停止。

Answer 2

您的计划是竞争条件的受害者。有几个问题，但主要问题似乎是：

很多时候，当你调用“watchCount”的第三个线程进入条件wait（）时，前两个线程已经传递了调用信号的计数器的值。因此，即使线程开始等待，也会调用一个信号。因此，你的第三个线程永远等待。如果您将条件更改为信号计数器＆gt; = WATCHCOUNT，您可能会有更好的运气

一个pthread程序，在收到条件信号后不能跳出循环

2 个答案: