Question

这是在C ++中创建线程安全队列的正确方法，它可以处理二进制数据的 unsigned char * arrays 吗？

请注意，数据是从主线程生成的，而不是创建的pthread，这让我怀疑pthread_mutex_t是否会在push和pop上正常工作。

线程安全队列

#include <queue>
#include <pthread.h>

class ts_queue
{

private:

    std::queue<unsigned char*> _queue_;
    pthread_mutex_t mutex;
    pthread_cond_t cond;

public:

    ts_queue()
    {
        pthread_mutex_init(&mutex, NULL);
        pthread_cond_init(&cond, NULL);
    }

    void push(unsigned char* data)
    {
        pthread_mutex_lock(&mutex);

        _queue_.push(data);

        pthread_cond_signal(&cond);
        pthread_mutex_unlock(&mutex);
    }

    void pop(unsigned char** popped_data)
    {
        pthread_mutex_lock(&mutex);

        while (_queue_.empty() == true)
        {
            pthread_cond_wait(&cond, &mutex);
        }

        *popped_data = _queue_.front();
        _queue_.pop();

        pthread_mutex_unlock(&mutex);
    }
};

消费者测试：

void *consumer_thread(void *arguments)
{
    ts_queue *tsq = static_cast<ts_queue*>(arguments);

    while (true)
    {
        unsigned char* data = NULL;

        tsq->pop(&data);

        if (data != NULL)
        {
            // Eureka! Received from the other thread!!!
            // Delete it so memory keeps free.
            // NOTE: In the real scenario for which I need
            // this class, the data received are bitmap pixels
            // and at this point it would be processed
            delete[] data;
        }
    }

    return 0;
}

生产者测试：

void main()
{
    ts_queue tsq;

    // Create the consumer
    pthread_t consumer;
    pthread_create(&consumer, NULL, consumer_thread, &tsq));

    // Start producing
    while(true)
    {
        // Push data. 
        // Expected behaviour: memory should never run out, as the
        // consumer should receive the data and delete it.
        // NOTE: test_data in the real purpose scenario for which I 
        // need this class would hold bitmap pixels, so it's meant to 
        // hold binary data and not a string

        unsigned char* test_data = new unsigned char [8192];
        tsq.push(test_data);
    }

    return 0;
}

Answer 1

您如何知道消费者永远不会获得数据？当我尝试你的程序时，我得到一个分段错误，GDB告诉我消费者确实得到了一个指针，但它是一个无效的。

我相信您的问题是您在_queue_成员上进行了数据竞争。在push()和_queue_.push(data)来电_queue_时，push_mutex拨打pop()（写在_queue_.front()上）（阅读_queue_）并且_queue_.pop()（另一个写在_queue_上）同时持有pop_mutex，但push()和pop()可以同时发生，导致两个线程都在写（和阅读）_queue_同时是一个经典的数据竞赛。

C ++中的线程安全队列

1 个答案: