5

我想实现一个生产者/消费者场景，该场景遵循大致如下的接口：

class Consumer {
private:
    vector<char> read(size_t n) {
        // If the internal buffer has `n` elements, then dequeue them
        // Otherwise wait for more data and try again
    }
public:
    void run() {
        read(10);
        read(4839);
        // etc
    }
    void feed(const vector<char> &more) {
        // Safely queue the data
        // Notify `read` that there is now more data
    }
};

在这种情况下，feedandrun将在单独的线程上运行，并且read应该是阻塞读取（如recvand fread）。显然，我需要在我的双端队列上进行某种互斥，并且我需要某种通知系统来通知read重试。

我听说条件变量是要走的路，但我所有的多线程体验都在 Windows 上，我很难理解它们。

谢谢你的帮助！

（是的，我知道返回向量是低效的。我们不要讨论这个。）

4

6 回答 6

8

此代码尚未准备好生产。不对任何库调用的结果进行错误检查。

我已经在 LockThread 中包装了互斥锁的锁定/解锁，所以它是异常安全的。但仅此而已。

此外，如果我认真地这样做，我会将互斥锁和条件变量包装在对象中，这样它们就可以在 Consumer 的其他方法中被滥用。但是只要您注意必须在使用条件变量（以任何方式）之前获取锁，那么这种简单的情况就可以保持原样。

出于兴趣，您检查过 boost 线程库吗？

#include <iostream>
#include <vector>
#include <pthread.h>

class LockThread
{
    public:
    LockThread(pthread_mutex_t& m)
        :mutex(m)
    {
        pthread_mutex_lock(&mutex);
    }
    ~LockThread()
    {
        pthread_mutex_unlock(&mutex);
    }
    private:
        pthread_mutex_t& mutex;
};
class Consumer
{
    pthread_mutex_t     lock;
    pthread_cond_t      cond;
    std::vector<char>   unreadData;
    public:
    Consumer()
    {
        pthread_mutex_init(&lock,NULL);
        pthread_cond_init(&cond,NULL);
    }
    ~Consumer()
    {
        pthread_cond_destroy(&cond);
        pthread_mutex_destroy(&lock);
    }

    private:
        std::vector<char> read(size_t n)
        {
            LockThread  locker(lock);
            while (unreadData.size() < n)
            {
                // Must wait until we have n char.
                // This is a while loop because feed may not put enough in.

                // pthread_cond() releases the lock.
                // Thread will not be allowed to continue until
                // signal is called and this thread reacquires the lock.

                pthread_cond_wait(&cond,&lock);

                // Once released from the condition you will have re-aquired the lock.
                // Thus feed() must have exited and released the lock first.
            }

            /*
             * Not sure if this is exactly what you wanted.
             * But the data is copied out of the thread safe buffer
             * into something that can be returned.
             */
            std::vector<char>   result(n); // init result with size n
            std::copy(&unreadData[0],
                      &unreadData[n],
                      &result[0]);

            unreadData.erase(unreadData.begin(),
                             unreadData.begin() + n);
            return (result);
        }
public:
    void run()
    {
        read(10);
        read(4839);
        // etc
    }
    void feed(const std::vector<char> &more)
    {
        LockThread  locker(lock);

        // Once we acquire the lock we can safely modify the buffer.
        std::copy(more.begin(),more.end(),std::back_inserter(unreadData));

        // Only signal the thread if you have the lock
        // Otherwise race conditions happen.
        pthread_cond_signal(&cond);

        // destructor releases the lock and thus allows read thread to continue.
    }
};


int main()
{
    Consumer    c;
}
于 2008-10-15T23:42:48.093 回答
2

我倾向于使用我所说的“同步队列”。我包装普通队列并使用 Semaphore 类来锁定和读取块，就像你想要的那样：

#ifndef SYNCQUEUE_20061005_H_
#define SYNCQUEUE_20061005_H_

#include <queue>
#include "Semaphore.h"

// similar, but slightly simpler interface to std::queue
// this queue implementation will serialize pushes and pops
// and block on a pop while empty (as apposed to throwing an exception)
// it also locks as neccessary on insertion and removal to avoid race 
// conditions

template <class T, class C = std::deque<T> > class SyncQueue {
protected:
    std::queue<T, C>    m_Queue;
    Semaphore           m_Semaphore;
    Mutex               m_Mutex;

public:
    typedef typename std::queue<T, C>::value_type value_type;
    typedef typename std::queue<T, C>::size_type size_type;

    explicit SyncQueue(const C& a = C()) : m_Queue(a), m_Semaphore(0) {}

    bool empty() const              { return m_Queue.empty(); }
    size_type size() const          { return m_Queue.size(); }

    void push(const value_type& x);
    value_type pop();
};

template <class T, class C>
void SyncQueue<T, C>::push(const SyncQueue<T, C>::value_type &x) {
    // atomically push item
    m_Mutex.lock(); 
    m_Queue.push(x); 
    m_Mutex.unlock(); 

    // let blocking semaphore know another item has arrived
    m_Semaphore.v();
}

template <class T, class C>
typename SyncQueue<T, C>::value_type SyncQueue<T, C>::pop() {
    // block until we have at least one item
    m_Semaphore.p();

    // atomically read and pop front item
    m_Mutex.lock();
    value_type ret = m_Queue.front();
    m_Queue.pop();
    m_Mutex.unlock();

    return ret;
}

#endif

您可以在线程实现中使用适当的原语来实现信号量和互斥锁。

注意：此实现是队列中单个元素的示例，但您可以轻松地将其包装为缓冲结果直到提供 N 的函数。如果它是一个字符队列，则如下所示：

std::vector<char> func(int size) {
    std::vector<char> result;
    while(result.size() != size) {
        result.push_back(my_sync_queue.pop());
    }
    return result;
}
于 2008-10-15T23:57:04.117 回答
1

我会扔掉一些半伪代码。以下是我的评论：

1）这里有非常大的锁定颗粒。如果您需要更快的访问速度，您将需要重新考虑您的数据结构。STL 不是线程安全的。

2）锁定将阻塞，直到互斥锁允许它通过。互斥体结构是它使用锁定/解锁机制一次允许 1 个线程通过它。不需要轮询或某种异常式结构。

3）这是一个在句法上相当老套的问题。我对 API 和 C++ 语法并不精确，但我相信它提供了语义上正确的解决方案。

4）根据评论进行编辑。

class piper
{
pthread_mutex queuemutex;
pthread_mutex readymutex;
bool isReady; //init to false by constructor

//whatever else
};

piper::read()
{//whatever
pthread_mutex_lock(&queuemutex)
if(myqueue.size() >= n)
{ 
   return_queue_vector.push_back(/* you know what to do here */)

    pthread_mutex_lock(&readymutex)
    isReady = false;
    pthread_mutex_unlock(&readymutex)
}
pthread_mutex_unlock(&queuemutex)
}

piper::push_em_in()
{
//more whatever
pthread_mutex_lock(&queuemutex)
//push push push
if(myqueue.size() >= n)
{
    pthread_mutex_lock(&readymutex)
    isReady = true;
    pthread_mutex_unlock(&readymutex)
}
pthread_mutex_unlock(&queuemutex)
}
于 2008-10-15T23:13:29.743 回答
1

只是为了好玩，这里是一个使用 Boost 的快速而肮脏的实现。它在支持它的平台上使用 pthreads，在 windows 上使用 windows 操作。

boost::mutex access;
boost::condition cond;

// consumer
data read()
{
  boost::mutex::scoped_lock lock(access);
  // this blocks until the data is ready
  cond.wait(lock);

  // queue is ready
  return data_from_queue();
}

// producer
void push(data)
{
  boost::mutex::scoped_lock lock(access);
  // add data to queue

  if (queue_has_enough_data())
    cond.notify_one();  
}
于 2008-10-16T00:04:31.873 回答
1

为了更有趣，这是我的最终版本。没有充分理由的 STL 化。:-)

#include <algorithm>
#include <deque>
#include <pthread.h>

template<typename T>
class MultithreadedReader {
    std::deque<T>   buffer;
    pthread_mutex_t moreDataMutex;
    pthread_cond_t  moreDataCond;

protected:
    template<typename OutputIterator>
    void read(size_t count, OutputIterator result) {
        pthread_mutex_lock(&moreDataMutex);

        while (buffer.size() < count) {
            pthread_cond_wait(&moreDataCond, &moreDataMutex);
        }
        std::copy(buffer.begin(), buffer.begin() + count, result);
        buffer.erase(buffer.begin(), buffer.begin() + count);

        pthread_mutex_unlock(&moreDataMutex);
    }

public:
    MultithreadedReader() {
        pthread_mutex_init(&moreDataMutex, 0);
        pthread_cond_init(&moreDataCond, 0);
    }

    ~MultithreadedReader() {
        pthread_cond_destroy(&moreDataCond);
        pthread_mutex_destroy(&moreDataMutex);
    }

    template<typename InputIterator>
    void feed(InputIterator first, InputIterator last) {
        pthread_mutex_lock(&moreDataMutex);

        buffer.insert(buffer.end(), first, last);
        pthread_cond_signal(&moreDataCond);

        pthread_mutex_unlock(&moreDataMutex);
    }
};
于 2008-10-16T00:36:49.483 回答
0

Glib 异步队列在读取您正在寻找的空队列时提供锁定和睡眠。请参阅http://library.gnome.org/devel/glib/2.20/glib-Asynchronous-Queues.html您可以将它们与 gthread 或 gthread 池结合使用。

于 2009-04-30T16:44:36.993 回答