C ++多线程生产者－消费者问题

Question

我写了一个用条件变量乘以生产者和消费者的代码。即使当我只有一个生产者和一个消费者时，它也不起作用。生产者和消费者都应在while（true）中运行。当我运行代码时，它阻塞了大约50％的运行。我想它会因等待过多而陷入僵局。我无法成功调试卡住的位置以及如何解锁conds。根据要求，我必须创建带有等待，信号和广播的代码。

如果队列已满，则生产者正在等待。如果队列为空，则消费者正在等待。

void WaitableQueue::enqueue(size_t a_item)
{
    (m_cond.getMutex()).lock();

    while(m_itemsCounter==m_capacity && !m_isBeingDestroyed)
    {
        ++m_numberOfWaiting;
        m_cond.wait();
        --m_numberOfWaiting;
    }

    std::cout<<"enqueue "<<a_item<<"\n";

    m_queue.push(a_item);
    ++m_itemsCounter;
    ++m_numbOfProduced;
    if(m_isBeingDestroyed)
    {
        m_cond.broadcast(); 
    }

    (m_cond.getMutex()).unlock();
    m_cond.broadcast();
}

void WaitableQueue::dequeue()
{
    (m_cond.getMutex()).lock();

    while(m_itemsCounter==0 && !m_isBeingDestroyed)
    {
        ++m_numberOfWaiting;
        std::cout<<"Waiting\n";
        m_cond.wait();
        std::cout<<"Done waiting\n";
        --m_numberOfWaiting;
    }

    if (m_isBeingDestroyed)
    {
        (m_cond.getMutex()).unlock();
        m_cond.broadcast();
        return;
    }
    std::cout<<"dequeue "<<m_queue.front()<<"\n";
    m_queue.pop();
    --m_itemsCounter;
    ++m_numbOfConsumed;
    (m_cond.getMutex()).unlock();
    m_cond.broadcast();
}

void WaitableQueue::destroy()
{
    (m_cond.getMutex()).lock();
    m_isBeingDestroyed=true;
    (m_cond.getMutex()).unlock();
}



void Producer::run()
{
    for(size_t i=0;i<m_numOfItemsToProduce;++i)
    {
        usleep(m_delay);
        size_t item=produce();
        m_wq.enqueue(item);
    }
}


Producer::produce() const
{
    return rand()%m_numOfItemsToProduce;
}

void Consumer::run()
{
    m_numOfProducersMutex.lock();
    while(m_numOfProducers>0)
    {
        m_numOfProducersMutex.unlock();
        usleep(m_delay);
        m_wq.dequeue();
        m_numOfProducersMutex.lock();
    }
    m_numOfProducersMutex.unlock();
}


int main()
{
    size_t numProducers=1;
    size_t numConsumers=3;
    Mutex mutex;
    ConditionalVariable cond(mutex);

    WaitableQueue<size_t> wq(NUM_OF_ITEMS,cond);
    std::vector<Producer<size_t>*> producerArray;
    std::vector<Consumer<size_t>*> consumerArray;
    Mutex numOfProducersMutex;

    for(size_t i=0;i<numProducers;++i)
    {
        Producer<size_t>* tempP=new Producer<size_t>(wq,NUM_OF_ITEMS,DELAY);
        producerArray.push_back(tempP);
    }

    for(size_t i=0;i<numConsumers;++i)
    {
        Consumer<size_t>* tempC=new Consumer<size_t>(wq,numProducers,numOfProducersMutex,DELAY);
        consumerArray.push_back(tempC);
    }

    for(size_t i=0;i<numProducers;++i)
    {
        producerArray[i]->start();
    }

    for(size_t i=0;i<numConsumers;++i)
    {
        consumerArray[i]->start();
    }

    for(size_t i=0;i<numProducers;++i)
    {
        producerArray[i]->join();
        numOfProducersMutex.lock();
        --numProducers;
        numOfProducersMutex.unlock();
    }
    usleep(100);

    //tell the consumers stop waiting
    wq.destroy();
   for(size_t i=0;i<numConsumers;++i)
    {
        consumerArray[i]->join();
    }

   for(size_t i=0;i<numProducers;++i)
   {
        delete producerArray[i];
   }

    for(size_t i=0;i<numConsumers;++i)
   {
        delete consumerArray[i];
   }
}

它运行约50％。另外50％的用户则被卡住。

Answer 1

要使用条件变量解决生产者-消费者问题，首先需要了解有界缓冲区问题。

Answer 2

您已经找到了C ++如何从概念上简单的问题中解决难题的另一个示例。