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如何解决C++:多线程同步经典案例之生产者消费者问题-创新互联

这篇文章主要介绍如何解决C++:多线程同步经典案例之生产者消费者问题,文中介绍的非常详细,具有一定的参考价值,感兴趣的小伙伴们一定要看完!

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生产者消费者问题(英语:Producer-consumer problem),也称有限缓冲问题(英语:Bounded-buffer problem),是一个多线程同步问题的经典案例。该问题描述了共享固定大小缓冲区的两个线程——即所谓的“生产者”和“消费者”——在实际运行时会发生的问题。生产者的主要作用是生成一定量的数据放到缓冲区中,然后重复此过程。与此同时,消费者也在缓冲区消耗这些数据。该问题的关键就是要保证生产者不会在缓冲区满时加入数据,消费者也不会在缓冲区中空时消耗数据。

要解决该问题,就必须让生产者在缓冲区满时休眠(要么干脆就放弃数据),等到下次消费者消耗缓冲区中的数据的时候,生产者才能被唤醒,开始往缓冲区添加数据。同样,也可以让消费者在缓冲区空时进入休眠,等到生产者往缓冲区添加数据之后,再唤醒消费者。

本文用一个ItemRepository类表示产品仓库,其中包含一个数组和两个坐标表示的环形队列、一个std::mutex成员、用来保证每次只被一个线程读写操作 (为了保证打印出来的消息是一行一行的,在它空闲的时候也借用的这个互斥量╮(╯▽╰)╭)、两个std::condition_variable表示队列不满和不空的状态,进而保证生产的时候不满,消耗的时候不空。

#pragma once
#include //std::chrono
#include //std::mutex,std::unique_lock,std::lock_guard
#include //std::thread
#include //std::condition_variable
#include //std::cout,std::endl
#include //std::map
namespace MyProducerToConsumer {
    static const int gRepositorySize = 10;//total size of the repository
    static const int gItemNum = 97;//number of products to produce
    std::mutex produce_mtx, consume_mtx;//mutex for all the producer thread or consumer thread
    std::map threadPerformance;//records of every thread's producing/consuming number
    struct ItemRepository {//repository class
        int m_ItemBuffer[gRepositorySize];//Repository itself (as a circular queue)
        int m_ProducePos;//rear position of circular queue
        int m_ConsumePos;//head position of circular queue
        std::mutex m_mtx;//mutex for operating the repository
        std::condition_variable m_RepoUnfull;//indicating that this repository is unfull(then producers can produce items)
        std::condition_variable m_RepoUnempty;//indicating that this repository is unempty(then consumers can produce items)
    }gItemRepo;

    void ProduceItem(ItemRepository *ir, int item) {
        std::unique_lock ulk(ir->m_mtx);
        while ((ir->m_ProducePos + 1) % gRepositorySize == ir->m_ConsumePos) {//full(spare one slot for indicating)
            std::cout << "Reposity is full. Waiting for consumers..." << std::endl;
            ir->m_RepoUnfull.wait(ulk);//unlocking ulk and waiting for unfull condition
        }
        //when unfull
        ir->m_ItemBuffer[ir->m_ProducePos++] = item;//procude and shift
        std::cout << "Item No." << item << " produced successfully by "
            <m_ProducePos == gRepositorySize)//loop
            ir->m_ProducePos = 0;
        ir->m_RepoUnempty.notify_all();//item produced, so it's unempty; notify all consumers
    }

    int ConsumeItem(ItemRepository *ir) {
        std::unique_lockulk(ir->m_mtx);
        while (ir->m_ConsumePos == ir->m_ProducePos) {//empty
            std::cout << "Repository is empty.Waiting for producing..." << std::endl;
            ir->m_RepoUnempty.wait(ulk);
        }
        int item = ir->m_ItemBuffer[ir->m_ConsumePos++];
        std::cout << "Item No." << item << " consumed successfully by "
            <m_ConsumePos == gRepositorySize)
            ir->m_ConsumePos = 0;
        ir->m_RepoUnfull.notify_all();//item consumed, so it's unempty; notify all consumers
        return item;
    }

    void ProducerThread() {
        static int produced = 0;//static variable to indicate the number of produced items
        while (1) {
            std::this_thread::sleep_for(std::chrono::milliseconds(10));//sleep long enough in case it runs too fast for other threads to procude
            std::lock_guardlck(produce_mtx);//auto unlock when break
            produced++;
            if (produced > gItemNum)break;
            gItemRepo.m_mtx.lock();
            std::cout << "Producing item No." << produced << "..." << std::endl;
            gItemRepo.m_mtx.unlock();
            ProduceItem(&gItemRepo, produced);
        }
        gItemRepo.m_mtx.lock();
        std::cout << "Producer thread " << std::this_thread::get_id()
            << " exited." << std::endl;
        gItemRepo.m_mtx.unlock();
    }

    void ConsumerThread() {
        static int consumed = 0;
        while (1) {
            std::this_thread::sleep_for(std::chrono::milliseconds(10));
            std::lock_guardlck(consume_mtx);
            consumed++;
            if (consumed > gItemNum)break;
            gItemRepo.m_mtx.lock();
            std::cout << "Consuming item available..." << std::endl;
            gItemRepo.m_mtx.unlock();
            ConsumeItem(&gItemRepo);
        }
        gItemRepo.m_mtx.lock();
        std::cout << "Consumer thread " << std::this_thread::get_id()
            << " exited." << std::endl;
        gItemRepo.m_mtx.unlock();
    }

    void InitItemRepository(ItemRepository* ir) {
        ir->m_ConsumePos = 0;
        ir->m_ProducePos = 0;
    }

    void Run() {
        InitItemRepository(&gItemRepo);
        std::thread thdConsume[11];
        std::thread thdProduce[11];
        for (auto& t : thdConsume)t = std::thread(ConsumerThread);
        for (auto& t : thdProduce)t = std::thread(ProducerThread);
        for (auto& t : thdConsume)t.join();
        for (auto& t : thdProduce)t.join();
        for (auto& iter : threadPerformance)cout << iter.first << ":" << iter.second << endl;
    }
}

以上是如何解决C++:多线程同步经典案例之生产者消费者问题的所有内容,感谢各位的阅读!希望分享的内容对大家有帮助,更多相关知识,欢迎关注创新互联行业资讯频道!


本文名称:如何解决C++:多线程同步经典案例之生产者消费者问题-创新互联
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