Question

我需要创建两个严格交替的线程。以下是我使用的示例代码：

#include <Windows.h>
#include <iostream>
using std::cout;
using std::endl;

HANDLE g_hMutex1;
HANDLE g_hMutex2;

DWORD WINAPI ThreadFunc1(LPVOID lpParam);
DWORD WINAPI ThreadFunc2(LPVOID lpParam);

int main(void)
{
    int nCalcNumber = 10;
    DWORD dwThreadId;
    HANDLE pThreadHandles[2];

    g_hMutex1 = CreateMutex(NULL, FALSE, NULL);
    g_hMutex1 = CreateMutex(NULL, FALSE, NULL);

    pThreadHandles[0] = CreateThread(
        NULL,
        0,
        ThreadFunc1,
        static_cast<void*>(&nCalcNumber),
        0,
        &dwThreadId);

    pThreadHandles[1] = CreateThread(
        NULL,
        0,
        ThreadFunc2,
        static_cast<void*>(&nCalcNumber),
        0,
        &dwThreadId);

    WaitForMultipleObjects(2, pThreadHandles, TRUE, INFINITE);

    CloseHandle(pThreadHandles[0]);
    CloseHandle(pThreadHandles[1]);
    CloseHandle(g_hMutex1);
    CloseHandle(g_hMutex2);

    return 0;
}

DWORD WINAPI ThreadFunc1(LPVOID lpParam)
{
    int* nCalcNumber = static_cast<int*>(lpParam);

    for (int i = 0; i < *nCalcNumber; i++)
    {
        WaitForSingleObject(g_hMutex1, INFINITE);

        cout << "Func 1" << endl;

        ReleaseMutex(g_hMutex1);
    }

    return 0;
}

DWORD WINAPI ThreadFunc2(LPVOID lpParam)
{
    int* nCalcNumber = static_cast<int*>(lpParam);

    for (int i = 0; i < *nCalcNumber; i++)
    {
        WaitForSingleObject(g_hMutex1, INFINITE);

        cout << "Func 2" << endl;

        ReleaseMutex(g_hMutex1);
    }

    return 0;
}

结果，我希望收到：

 Func 1
 Func 2
 Func 1
 Func 2
 Func 1
 Func 2
 ...and so one

应该添加什么才能获得所需的结果。我可以使用第二个互斥锁吗？

Answer 1

正如其他答案所述，信号量是一种比互斥锁更好的选择。但作为纯粹的学术练习（家庭作业？），你也可以使用互斥量。（重点：这是一个纯粹的学术练习。一个真正的程序不应该使用这种技术。）

DWORD WINAPI ThreadFunc1(LPVOID lpParam)
{
    int* nCalcNumber = static_cast<int*>(lpParam);

    WaitForSingleObject(g_hMutex2, INFINITE);
    for (int i = 0; i < *nCalcNumber; i++)
    {
        WaitForSingleObject(g_hMutex1, INFINITE);
        ReleaseMutex(g_hMutex2);

        cout << "Func 1" << endl;

        ReleaseMutex(g_hMutex1);
        WaitForSingleObject(g_hMutex2, INFINITE);
    }

    return 0;
}

DWORD WINAPI ThreadFunc2(LPVOID lpParam)
{
    int* nCalcNumber = static_cast<int*>(lpParam);

    WaitForSingleObject(g_hMutex2, INFINITE);
    for (int i = 0; i < *nCalcNumber; i++)
    {
        WaitForSingleObject(g_hMutex1, INFINITE);
        ReleaseMutex(g_hMutex2);

        cout << "Func 2" << endl;

        ReleaseMutex(g_hMutex1);
        WaitForSingleObject(g_hMutex2, INFINITE);
    }

    return 0;
}

互斥锁1是“我拥有它”互斥锁，互斥锁2是“我想要它下一个”互斥锁。

Answer 2

如果您可以使用semaphore：您可以使用semaphore代替mutex，则可以轻松使用mutex。

此代码可以正常工作：

#include <windows.h>
#include <iostream>
using std::cout;
using std::endl;

PHANDLE sem1;
PHANDLE sem2;

DWORD WINAPI ThreadFunc1(LPVOID lpParam);
DWORD WINAPI ThreadFunc2(LPVOID lpParam);

int main(void)
{
    int nCalcNumber = 10;
    DWORD dwThreadId;
    HANDLE pThreadHandles[2];

    sem1 = (PHANDLE) CreateSemaphore(NULL, 1, 1, NULL);
    sem2 = (PHANDLE) CreateSemaphore(NULL, 0, 1, NULL);


    pThreadHandles[0] = CreateThread(
                                     NULL,
                                     0,
                                     ThreadFunc1,
                                     static_cast<void*> (&nCalcNumber),
                                     0,
                                     &dwThreadId);

    pThreadHandles[1] = CreateThread(
                                     NULL,
                                     0,
                                     ThreadFunc2,
                                     static_cast<void*> (&nCalcNumber),
                                     0,
                                     &dwThreadId);

    WaitForMultipleObjects(2, pThreadHandles, TRUE, INFINITE);

    CloseHandle(pThreadHandles[0]);
    CloseHandle(pThreadHandles[1]);
    CloseHandle(sem1);
    CloseHandle(sem2);

    return 0;
}

DWORD WINAPI ThreadFunc1(LPVOID lpParam)
{
    int* nCalcNumber = static_cast<int*> (lpParam);

    for (int i = 0; i < *nCalcNumber; i++)
    {
        WaitForSingleObject(sem1, INFINITE);

        cout << "Func 1" << endl;

        ReleaseSemaphore(sem2, 1 ,NULL);
    }

    return 0;
}

DWORD WINAPI ThreadFunc2(LPVOID lpParam)
{
    int* nCalcNumber = static_cast<int*> (lpParam);

    for (int i = 0; i < *nCalcNumber; i++)
    {

        WaitForSingleObject(sem2, INFINITE);

        cout << "Func 2" << endl;

        ReleaseSemaphore(sem1, 1 ,NULL);
    }

    return 0;
}

Answer 3

你假设操作系统实际上支持这个。 Windows没有。它无法保证调度，除此之外没有饥饿。

所以你需要做的是设置一个标志变量，这样每个线程都会改变它以允许另一个线程运行。例如，如果它是真的 - 运行，如果它是假的 - 释放互斥锁并睡眠一段时间，而另一个线程 - 完全相反。睡眠在这里很重要，以避免饥饿和僵局。我认为它可以是Sleep（0）（检查它是否意味着Windows中的“产量”，我不确定）。

当然，检查应该在执行互斥锁时完成，并且在运行结束时，每个线程都会将变量更改为相反的 - 允许另一个线程运行并阻塞自己直到另一个线程运行然后改回来。

通过使变量成为计数器模数的线程数，并且每个线程在运行结束时增加值，并将模数值检查为线程的数字顺序，可以很容易地将其更改为2个以上的线程开头的执行。

修改

volatile bool flag = false; DWORD WINAPI ThreadFunc1(LPVOID lpParam) { int* nCalcNumber = static_cast<int*>(lpParam); for (int i = 0; i < *nCalcNumber; /*no-op*/;) { WaitForSingleObject(g_hMutex1, INFINITE); if (flag) {Sleep(0); continue;} cout << "Func 1" << endl; flag = true; i++; ReleaseMutex(g_hMutex1); } return 0; } DWORD WINAPI ThreadFunc2(LPVOID lpParam) { int* nCalcNumber = static_cast<int*>(lpParam); for (int i = 0; i < *nCalcNumber; /*no-op*/;) { WaitForSingleObject(g_hMutex1, INFINITE); if (!flag) {Sleep(0); continue;} cout << "Func 2" << endl; flag = false; i++; ReleaseMutex(g_hMutex1); } return 0; }

如何使用互斥锁严格交替使用两个线程？

3 个答案: