Question

我有以下代码：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <queue>
using namespace std;

queue<int> myqueue;

pthread_mutex_t count_mutex     = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t  condition_var   = PTHREAD_COND_INITIALIZER;

void *consumer(void*);
void *producer(void*);

#define COUNT_DONE 10
int count = 0;

main()
{
   pthread_t thread1, thread2;

   pthread_create( &thread2, NULL, &consumer, NULL);
   pthread_create( &thread1, NULL, &producer, NULL);

   pthread_join( thread1, NULL);
   pthread_join( thread2, NULL);

   printf("Final count: %d\n",count);

   system("PAUSE");
   return EXIT_SUCCESS;
}

void *consumer(void*)
{
   for(;;)
   {
      // Lock mutex and then wait for signal to relase mutex
      printf("consumer mutex lock \n");
      pthread_mutex_lock( &count_mutex );
      printf("consumer mutex locked\n");

      // Wait while functionCount2() operates on count
      // mutex unlocked if condition varialbe in functionCount2() signaled.
      printf("consumer wait\n");
      pthread_cond_wait( &condition_var, &count_mutex );
      printf("consumer condition woke up\n");
      myqueue.pop();count--;
      printf("Counter value consumer: %d\n",count);

      printf("consumer mutex unlock\n");
      pthread_mutex_unlock( &count_mutex );

      if(count >= COUNT_DONE) return(NULL);
    }
}

void * producer(void*)
{
    for(;;)
    {
       printf("producer mutex lock\n");
       pthread_mutex_lock( &count_mutex );
       printf("producer mutex locked\n");

       if( count < COUNT_DONE)
       {
           myqueue.push(1);
           count++;
           printf("Counter value producer: %d\n",count);
           printf("producer signal\n");
           pthread_cond_signal( &condition_var );
       }

       printf("producer mutex unlock\n");
       pthread_mutex_unlock( &count_mutex );

       if(count >= COUNT_DONE) return(NULL);

       Sleep(5000);
    }

}

当消费者线程首先使用互斥锁时，这个示例工作正常。但是当生产者线程最初首先获取互斥锁时，我将总是在队列中有一个整数，消费者无法弹出。

如何让消费者线程最初在生产者之前获取互斥锁。

注意：我正在寻找比在另一个之前启动一个线程更好的方法。

谢谢，

Answer 1

我看到的一个问题是你的消费者实际上没有检查要做的工作，只是盲目地从队列中弹出。

我看到的第二个问题是你在一个中递增计数而在另一个中减少它，那么你如何达到终止条件呢？

将忍者“count--”从消费者手中夺走，它应该有效。不过，您可能希望在消费者中做以下事情：

// Wait for producer to do its thing and tell us there is work to do.
while ( myqueue.empty() ) {
    pthread_cond_wait(&condition_var, &count_mutex);
}
// we've been told there's work to do with the queue,
// and we know there's something ON the queue.
// consume the entire queue.
while ( !myqueue.empty() ) {
  myqueue.pop();
}

// treat count as protected by the mutex, so hoist this test into the lock.
bool workDone = (count >= COUNT_DONE);
pthread_mutex_unlock(&count_mutex);

if(workDone)
    return break;

编辑：消费者的首选版本：

bool workDone = false;
while(workDone == false)
{
    // Lock mutex and then wait for signal to relase mutex
    pthread_mutex_lock( &count_mutex );

    // Wait for producer to do its thing and tell us there is work to do.
    while ( myqueue.empty() )
    pthread_cond_wait( &condition_var, &count_mutex );

    // we've been told there's work to do with the queue,
    // and we know there's something ON the queue.
    // consume the entire queue.
    while ( myqueue.empty() == false ) {
        myqueue.pop();
    }

    // count is protected by the lock so check if we're done before we unlock.
    workDone = (count >= COUNT_DONE);
    pthread_mutex_unlock( &count_mutex );

}
return NULL;

如何同步线程（消费者/生产者）

1 个答案: