Question

这个问题是对this question的跟进。我希望线程执行一些工作并按顺序将句柄传递给下一个线程。在尝试执行以下代码时，我得到了

ConsoleApplication9.exe中0x0F7C1F5F（msvcp120d.dll）的未处理异常：0xC0000005：访问冲突读取位置0x00000004。

    #include "stdafx.h"
    #include <iostream>
    #include <thread>
    #include <mutex>
    #include <chrono>
    #include <condition_variable> 

    std::mutex* m_pMutexs;
    std::condition_variable* m_pCVs;
    int m_pCurrentWorker;

    void func(int i)
    {
        int cvCurrentInd = i;
        std::mutex* pCurMutex = &m_pMutexs[cvCurrentInd];

        std::condition_variable* pCuCV = (std::condition_variable*)(m_pCurrentWorker + i*sizeof(std::condition_variable));

        std::unique_lock<std::mutex> lk(m_pMutexs[i]);

        while (i != m_pCurrentWorker)
        {
            pCuCV->wait(lk);
        }

        std::cout << "entered thread " << std::this_thread::get_id() << std::endl;
        std::this_thread::sleep_for(std::chrono::seconds(rand() % 10));
        std::cout << "leaving thread " << std::this_thread::get_id() << std::endl;

        m_pCurrentWorker++;
        lk.unlock();
        pCuCV->notify_one();

    }

    int _tmain(int argc, _TCHAR* argv[])
    {
        m_pMutexs = new std::mutex[3];

        m_pCVs = new std::condition_variable[3];

        m_pCurrentWorker = 0;

        srand((unsigned int)time(0));

        std::thread t1(func,0);
        std::thread t2(func,1);
        std::thread t3(func,2);

        t1.join();
        t2.join();
        t3.join();

        return 0;
    }

Answer 1

不知道你想做什么但是

你是否将整数转换为指针？

std::condition_variable* pCuCV = (std::condition_variable*)(m_pCurrentWorker + i*sizeof(std::condition_variable));

我认为你应该写一下：

std::condition_variable* pCuCV = &m_pCVs[i];

整个功能可能是这样的：

void func(int i)
{
    std::mutex* pCurMutex = &m_pMutexs[i];

    std::condition_variable* pCuCV = &m_pCVs[i];

    std::unique_lock<std::mutex> lk(m_pMutexs[i]);

    while (i != m_pCurrentWorker) {
        pCuCV->wait(lk);
    }

    std::cout << "entered thread " << std::this_thread::get_id() << std::endl;
    std::this_thread::sleep_for(std::chrono::seconds(rand() % 2));
    std::cout << "leaving thread " << std::this_thread::get_id() << std::endl;

    m_pCurrentWorker++;
    lk.unlock();
    if (m_pCurrentWorker > 2) {
        return;
    }
    pCuCV = &m_pCVs[m_pCurrentWorker];
    pCuCV->notify_one();

}

Answer 2

我做了一些进一步的研究，似乎原始问题的代码不是线程安全的

C ++：使用互斥锁和条件变量控制线程执行的顺序

2 个答案: