哲学家用餐问题 - 只有 2 个线程工作

Question

以我为例，每个哲学家都应该吃一百万次。问题是好像只有“1”，“3”吃完了。我正在使用带临界区锁的线程，这是我的代码：

CRITICAL_SECTION ghCARITICALSection1;
CRITICAL_SECTION ghCARITICALSection2;
CRITICAL_SECTION ghCARITICALSection3;
CRITICAL_SECTION ghCARITICALSection4;
CRITICAL_SECTION ghCARITICALSection5;
DWORD WINAPI func(int* phiphilosopher)
{
    if (1 == *phiphilosopher && TryEnterCriticalSection(&ghCARITICALSection1) && TryEnterCriticalSection(&ghCARITICALSection2))
    {
        std::cout << "1 is eating...\n";
        for (int i = 0; i < 1000000; i++)
        {
            i = i;
        }
        LeaveCriticalSection(&ghCARITICALSection1);
        LeaveCriticalSection(&ghCARITICALSection2);
    }
    if (2 == *phiphilosopher && TryEnterCriticalSection(&ghCARITICALSection2) && TryEnterCriticalSection(&ghCARITICALSection3))
    {
        std::cout << "2 is eating...\n";
        for (int i = 0; i < 1000000; i++)
        {
        }
        LeaveCriticalSection(&ghCARITICALSection2);
        LeaveCriticalSection(&ghCARITICALSection3);
    }
    if (3 == *phiphilosopher && TryEnterCriticalSection(&ghCARITICALSection3) && TryEnterCriticalSection(&ghCARITICALSection4))
    {
        std::cout << "3 is eating...\n";
        for (int i = 0; i < 1000000; i++)
        {
        }
        LeaveCriticalSection(&ghCARITICALSection3);
        LeaveCriticalSection(&ghCARITICALSection4);
    }
    //...also for 4,5
    return 0;
}

    int philosopher1 = 1;
    int* philosopher1ptr = &philosopher1;
    int philosopher2 = 2;
    int* philosopher2ptr = &philosopher2;
    //...Also for philosopher 3,4,5

    InitializeCriticalSection(&ghCARITICALSection1);
    InitializeCriticalSection(&ghCARITICALSection2);
//...aslo for ghCARITICALSection 3,4,5

    HANDLE WINAPI th1 = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)func, philosopher1ptr, 0, NULL);
    HANDLE WINAPI th2 = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)func, philosopher2ptr, 0, NULL);
////...aslo for th3,4,5
    WaitForSingleObject(th1, INFINITE);
    WaitForSingleObject(th2, INFINITE);
    //...also for th3,4,5

每个哲学家都必须交替思考和吃饭。然而，哲学家只有在同时拥有左右叉子的情况下才能吃到意大利面。每把叉子只能由一位哲学家持有，因此只有当另一位哲学家没有使用时，一位哲学家才能使用叉子。

Answer 1

想想这里的逻辑

    if (TryEnterCriticalSection(&a) && TryEnterCriticalSection(&b)) {
        // . . .
        LeaveCriticalSection(&a);
        LeaveCriticalSection(&b);
    }

如果 TryEnterCriticalSection(&a) 成功而 TryEnterCriticalSection(&b) 失败会发生什么； CS a 永远保持进入状态。

它应该看起来像

    if (TryEnterCriticalSection(&a)) {
        if (TryEnterCriticalSection(&b)) {
            // . . .
            LeaveCriticalSection(&b);
        }
        LeaveCriticalSection(&a);
    }

哲学家用餐问题 - 只有 2 个线程工作

Dining philosophers problem - only 2 thread worked

c++

multithreading

dining-philosopher