c++ - 如何使 Win32/MFC 线程同步循环？

Question

我是 Windows 中多线程的新手，所以这可能是一个微不足道的问题:确保线程同步执行循环的最简单方法是什么？

我尝试将 Event 的共享数组传递给所有线程，并在循环结束时使用 WaitForMultipleObjects 来同步它们，但这让我在一个线程后出现死锁，有时是两个，周期。这是我当前代码的简化版本(只有两个线程，但我想让它具有可扩展性):

typedef struct
{
    int rank;
    HANDLE* step_events;
} IterationParams;

int main(int argc, char **argv)
{
    // ...

    IterationParams p[2];
    HANDLE step_events[2];
    for (int j=0; j<2; ++j)
    {
        step_events[j] = CreateEvent(NULL, FALSE, FALSE, NULL);
    }

    for (int j=0; j<2; ++j)
    {
        p[j].rank = j;
        p[j].step_events = step_events;
        AfxBeginThread(Iteration, p+j);
    }

    // ...
}

UINT Iteration(LPVOID pParam)
{
    IterationParams* p = (IterationParams*)pParam;
    int rank = p->rank;

    for (int i=0; i<100; i++)
    {
        if (rank == 0)
        {
            printf("%dth iteration\n",i);
            // do something
            SetEvent(p->step_events[0]);
            WaitForMultipleObjects(2, p->step_events, TRUE, INFINITE);
        }
        else if (rank == 1)
        {
            // do something else
            SetEvent(p->step_events[1]);
            WaitForMultipleObjects(2, p->step_events, TRUE, INFINITE);
        }
    }
    return 0;
}

(我知道我正在混合使用 C 和 C++，它实际上是我试图并行化的遗留 C 代码。)

阅读 MSDN 上的文档，我认为这应该可行。但是0号线程只打印一次，偶尔打印两次，然后程序就挂了。这是同步线程的正确方法吗？如果没有，您会推荐什么(MFC 中真的没有对屏障的内置支持吗？)。

---

编辑:这个解决方案错误，甚至包括Alessandro's fix .例如，考虑这种情况:

1. 线程 0 设置它的事件并调用 Wait，阻塞
2. 线程 1 设置它的事件并调用 Wait，阻塞
3. 线程 0 从 Wait 返回，重置其事件，并在线程 1 未获得控制权的情况下完成一个循环
4. 线程 0 设置自己的事件并调用 Wait。由于线程 1 还没有机会重置其事件，线程 0 的 Wait 立即返回并且线程不同步。

所以问题仍然存在:如何安全地确保线程保持同步？

Answer 1

介绍

我实现了一个简单的 C++ 程序供您考虑(在 Visual Studio 2010 中测试)。它仅使用 Win32 API(以及用于控制台输出和一些随机化的标准库)。您应该能够将它放入一个新的 Win32 控制台项目(没有预编译头文件)，编译并运行。

---

解决方案

#include <tchar.h>
#include <windows.h>


//---------------------------------------------------------
// Defines synchronization info structure. All threads will
// use the same instance of this struct to implement randezvous/
// barrier synchronization pattern.
struct SyncInfo
{
    SyncInfo(int threadsCount) : Awaiting(threadsCount), ThreadsCount(threadsCount), Semaphore(::CreateSemaphore(0, 0, 1024, 0)) {};
    ~SyncInfo() { ::CloseHandle(this->Semaphore); }
    volatile unsigned int Awaiting; // how many threads still have to complete their iteration
    const int ThreadsCount;
    const HANDLE Semaphore;
};


//---------------------------------------------------------
// Thread-specific parameters. Note that Sync is a reference
// (i.e. all threads share the same SyncInfo instance).
struct ThreadParams
{
    ThreadParams(SyncInfo &sync, int ordinal, int delay) : Sync(sync), Ordinal(ordinal), Delay(delay) {};
    SyncInfo &Sync;
    const int Ordinal;
    const int Delay;
};


//---------------------------------------------------------
// Called at the end of each itaration, it will "randezvous"
// (meet) all the threads before returning (so that next
// iteration can begin). In practical terms this function
// will block until all the other threads finish their iteration.
static void RandezvousOthers(SyncInfo &sync, int ordinal)
{
    if (0 == ::InterlockedDecrement(&(sync.Awaiting))) { // are we the last ones to arrive?
        // at this point, all the other threads are blocking on the semaphore
        // so we can manipulate shared structures without having to worry
        // about conflicts
        sync.Awaiting = sync.ThreadsCount;
        wprintf(L"Thread %d is the last to arrive, releasing synchronization barrier\n", ordinal);
        wprintf(L"---~~~---\n");

        // let's release the other threads from their slumber
        // by using the semaphore
        ::ReleaseSemaphore(sync.Semaphore, sync.ThreadsCount - 1, 0); // "ThreadsCount - 1" because this last thread will not block on semaphore
    }
    else { // nope, there are other threads still working on the iteration so let's wait
        wprintf(L"Thread %d is waiting on synchronization barrier\n", ordinal);
        ::WaitForSingleObject(sync.Semaphore, INFINITE); // note that return value should be validated at this point ;)
    }
}


//---------------------------------------------------------
// Define worker thread lifetime. It starts with retrieving
// thread-specific parameters, then loops through 5 iterations
// (randezvous-ing with other threads at the end of each),
// and then finishes (the thread can then be joined).
static DWORD WINAPI ThreadProc(void *p)
{
    ThreadParams *params = static_cast<ThreadParams *>(p);
    wprintf(L"Starting thread %d\n", params->Ordinal);

    for (int i = 1; i <= 5; ++i) {
        wprintf(L"Thread %d is executing iteration #%d (%d delay)\n", params->Ordinal, i, params->Delay);
        ::Sleep(params->Delay); 
        wprintf(L"Thread %d is synchronizing end of iteration #%d\n", params->Ordinal, i);
        RandezvousOthers(params->Sync, params->Ordinal);
    }

    wprintf(L"Finishing thread %d\n", params->Ordinal);
    return 0;
}


//---------------------------------------------------------
// Program to illustrate iteration-lockstep C++ solution.
int _tmain(int argc, _TCHAR* argv[])
{
    // prepare to run
    ::srand(::GetTickCount()); // pseudo-randomize random values :-)
    SyncInfo sync(4);
    ThreadParams p[] = {
        ThreadParams(sync, 1, ::rand() * 900 / RAND_MAX + 100), // a delay between 200 and 1000 milliseconds will simulate work that an iteration would do
        ThreadParams(sync, 2, ::rand() * 900 / RAND_MAX + 100),
        ThreadParams(sync, 3, ::rand() * 900 / RAND_MAX + 100),
        ThreadParams(sync, 4, ::rand() * 900 / RAND_MAX + 100),
    };

    // let the threads rip
    HANDLE t[] = {
        ::CreateThread(0, 0, ThreadProc, p + 0, 0, 0),
        ::CreateThread(0, 0, ThreadProc, p + 1, 0, 0),
        ::CreateThread(0, 0, ThreadProc, p + 2, 0, 0),
        ::CreateThread(0, 0, ThreadProc, p + 3, 0, 0),
    };

    // wait for the threads to finish (join)
    ::WaitForMultipleObjects(4, t, true, INFINITE);

    return 0;
}

---

示例输出

在我的机器(双核)上运行这个程序会产生以下输出:

Starting thread 1
Starting thread 2
Starting thread 4
Thread 1 is executing iteration #1 (712 delay)
Starting thread 3
Thread 2 is executing iteration #1 (798 delay)
Thread 4 is executing iteration #1 (477 delay)
Thread 3 is executing iteration #1 (104 delay)
Thread 3 is synchronizing end of iteration #1
Thread 3 is waiting on synchronization barrier
Thread 4 is synchronizing end of iteration #1
Thread 4 is waiting on synchronization barrier
Thread 1 is synchronizing end of iteration #1
Thread 1 is waiting on synchronization barrier
Thread 2 is synchronizing end of iteration #1
Thread 2 is the last to arrive, releasing synchronization barrier
---~~~---
Thread 2 is executing iteration #2 (798 delay)
Thread 3 is executing iteration #2 (104 delay)
Thread 1 is executing iteration #2 (712 delay)
Thread 4 is executing iteration #2 (477 delay)
Thread 3 is synchronizing end of iteration #2
Thread 3 is waiting on synchronization barrier
Thread 4 is synchronizing end of iteration #2
Thread 4 is waiting on synchronization barrier
Thread 1 is synchronizing end of iteration #2
Thread 1 is waiting on synchronization barrier
Thread 2 is synchronizing end of iteration #2
Thread 2 is the last to arrive, releasing synchronization barrier
---~~~---
Thread 4 is executing iteration #3 (477 delay)
Thread 3 is executing iteration #3 (104 delay)
Thread 1 is executing iteration #3 (712 delay)
Thread 2 is executing iteration #3 (798 delay)
Thread 3 is synchronizing end of iteration #3
Thread 3 is waiting on synchronization barrier
Thread 4 is synchronizing end of iteration #3
Thread 4 is waiting on synchronization barrier
Thread 1 is synchronizing end of iteration #3
Thread 1 is waiting on synchronization barrier
Thread 2 is synchronizing end of iteration #3
Thread 2 is the last to arrive, releasing synchronization barrier
---~~~---
Thread 2 is executing iteration #4 (798 delay)
Thread 3 is executing iteration #4 (104 delay)
Thread 1 is executing iteration #4 (712 delay)
Thread 4 is executing iteration #4 (477 delay)
Thread 3 is synchronizing end of iteration #4
Thread 3 is waiting on synchronization barrier
Thread 4 is synchronizing end of iteration #4
Thread 4 is waiting on synchronization barrier
Thread 1 is synchronizing end of iteration #4
Thread 1 is waiting on synchronization barrier
Thread 2 is synchronizing end of iteration #4
Thread 2 is the last to arrive, releasing synchronization barrier
---~~~---
Thread 3 is executing iteration #5 (104 delay)
Thread 4 is executing iteration #5 (477 delay)
Thread 1 is executing iteration #5 (712 delay)
Thread 2 is executing iteration #5 (798 delay)
Thread 3 is synchronizing end of iteration #5
Thread 3 is waiting on synchronization barrier
Thread 4 is synchronizing end of iteration #5
Thread 4 is waiting on synchronization barrier
Thread 1 is synchronizing end of iteration #5
Thread 1 is waiting on synchronization barrier
Thread 2 is synchronizing end of iteration #5
Thread 2 is the last to arrive, releasing synchronization barrier
---~~~---
Finishing thread 4
Finishing thread 3
Finishing thread 2
Finishing thread 1

请注意，为简单起见，每个线程都有随机的迭代持续时间，但该线程的所有迭代都将使用相同的随机持续时间(即它在迭代之间不会改变)。

---

它是如何工作的？

解决方案的“核心”在“RandezvousOthers”函数中。此函数将阻塞在共享信号量上(如果调用此函数的线程不是最后一个调用该函数的线程)或重置 Sync 结构并解除阻塞在共享信号量上的所有线程(如果此线程所在的线程)被调用的函数是最后一个调用该函数的函数)。