c++ - 如何在互斥锁中进行循环类型排序？

Question

#include "stdafx.h"
#include <Windows.h>
#include <conio.h>
#include <fstream>
#include <iostream>
using namespace std;

int main ( int, char ** )
{
    HANDLE mutex = CreateMutex(NULL, FALSE, L"PRV");

    for (int j=0; j < 100; ++j)
    {
        WaitForSingleObject(mutex, INFINITE);

        ofstream file("c:\\write.txt", ios::app);
        for (int i=0; i < 10; ++i) {
            file << 1;
        }
        ReleaseMutex(mutex);
        Sleep(100);
    }

    CloseHandle(mutex);
}

我用 file << 1 创建了 4 个 pograms ... file << 4它们有效，但我需要循环排序。或者，至少，无需连续两次编写一个过程。

Answer 1

我不认为你可以用一个互斥量来实现你的目标，但你可以很容易地使用两个互斥量来确保没有一个进程在一个序列中写入两次。你需要做的就是始终有一个进程在等待队列中，一个进程处于写入状态。为此，您创建了两个互斥体，我们称它们为 queueMutex 和 writeMutex。伪代码中的迭代逻辑应该是这样的:

acquire(queueMutex) // The process is next to write
acquire(writeMutex) // The process can now write
release(queueMutex) // Some other process can enter the queue

// now we can write 
write_to_file()

// Let's hold on here until some other process 
// enters the queue
// we do it by trying to acquire the queueMutex
// until the acquisition fails
while try_acquire(queueMutex)
    release(queueMutex)
    sleep

// try_acquire(queueMutex) finally failed
// this means some other process has entered the queue
// we can release the writeMutex and finish this iteration
release(writeMutex)

我会将实现细节留给您。当您实现该算法时，请确保您正确处理了最后一个进程的最后一次迭代，否则它将挂起。祝你好运!