Question

这是this question的后续行动。我正在尝试确保多个线程按特定顺序更改公共资源（m_pCurrentWorker）。我在每个线程使用互斥和条件变量，并且在线程执行结束时，我通知下一个线程的条件变量以便继续。工作代码如下，我运行了几十次，总是得到正确的执行顺序，似乎没有死锁：

entered thread 0
leaving thread 0
entered thread 1
leaving thread 1
entered thread 2
leaving thread 2

然而，在原始问题下面讨论了这种实现是错误的。很高兴听到其他人对这个设计的看法以及改进它的建议。

#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)
{
    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 " << i << std::endl;
    std::this_thread::sleep_for(std::chrono::seconds(rand() % 2));
    std::cout << "leaving thread " << i << std::endl;

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

C ++确保线程以特定顺序更改公共资源

0 个答案: