如何同步线程（消费者/生产者）

Question

I have the following code:

#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);
    }

}

This is example works fine when consumer thread takes the mutex first. But then when the producer thread acquires the mutex initially at first, I will always have 1 integer in the queue that the consumer can't pop.

How can I let the consumer thread, initially acquire the mutex before the producer.

Note: I am looking for a better way than launching one thread before the other.

Thanks,

Answer 1

One problem I see is that your consumer doesn't actually check for work to do, it just blindly pops from the queue.

The second problem I see is that you increment count in one and decrement it in the other, so how do you ever reach the termination condition?

Take the ninja "count--" out of consumer and it should work. Still, you might want to do the following inside consumer:

// 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;

edit: preferred version of consumer:

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;