如何解決C++：多線程同步經(jīng)典案例之生產(chǎn)者消費者問題-創(chuàng)新互聯(lián)

這篇文章主要介紹如何解決C++：多線程同步經(jīng)典案例之生產(chǎn)者消費者問題，文中介紹的非常詳細，具有一定的參考價值，感興趣的小伙伴們一定要看完！

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生產(chǎn)者消費者問題（英語：Producer-consumer problem），也稱有限緩沖問題（英語：Bounded-buffer problem），是一個多線程同步問題的經(jīng)典案例。該問題描述了共享固定大小緩沖區(qū)的兩個線程——即所謂的“生產(chǎn)者”和“消費者”——在實際運行時會發(fā)生的問題。生產(chǎn)者的主要作用是生成一定量的數(shù)據(jù)放到緩沖區(qū)中，然后重復此過程。與此同時，消費者也在緩沖區(qū)消耗這些數(shù)據(jù)。該問題的關鍵就是要保證生產(chǎn)者不會在緩沖區(qū)滿時加入數(shù)據(jù)，消費者也不會在緩沖區(qū)中空時消耗數(shù)據(jù)。

要解決該問題，就必須讓生產(chǎn)者在緩沖區(qū)滿時休眠（要么干脆就放棄數(shù)據(jù)），等到下次消費者消耗緩沖區(qū)中的數(shù)據(jù)的時候，生產(chǎn)者才能被喚醒，開始往緩沖區(qū)添加數(shù)據(jù)。同樣，也可以讓消費者在緩沖區(qū)空時進入休眠，等到生產(chǎn)者往緩沖區(qū)添加數(shù)據(jù)之后，再喚醒消費者。

本文用一個ItemRepository類表示產(chǎn)品倉庫，其中包含一個數(shù)組和兩個坐標表示的環(huán)形隊列、一個std::mutex成員、用來保證每次只被一個線程讀寫操作 （為了保證打印出來的消息是一行一行的，在它空閑的時候也借用的這個互斥量╮(╯▽╰)╭）、兩個std::condition_variable表示隊列不滿和不空的狀態(tài)，進而保證生產(chǎn)的時候不滿，消耗的時候不空。

#pragma once
#include <chrono>//std::chrono
#include <mutex>//std::mutex,std::unique_lock,std::lock_guard
#include <thread>//std::thread
#include <condition_variable>//std::condition_variable
#include <iostream>//std::cout,std::endl
#include <map>//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<std::thread::id, int> 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 <std::mutex>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 "
            <<std::this_thread::get_id()<<"!" << std::endl;
        threadPerformance[std::this_thread::get_id()]++;
        if (ir->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_lock<std::mutex>ulk(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 "
            <<std::this_thread::get_id()<<"!" << std::endl;
        threadPerformance[std::this_thread::get_id()]++;
        if (ir->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_guard<std::mutex>lck(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_guard<std::mutex>lck(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;
    }
}

以上是如何解決C++：多線程同步經(jīng)典案例之生產(chǎn)者消費者問題的所有內容，感謝各位的閱讀！希望分享的內容對大家有幫助，更多相關知識，歡迎關注創(chuàng)新互聯(lián)行業(yè)資訊頻道！

分享名稱：如何解決C++：多線程同步經(jīng)典案例之生產(chǎn)者消費者問題-創(chuàng)新互聯(lián)
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