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：确保线程以锁步方式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，但这会在一个周期（有时是两个）周期后出现死锁。这是我当前代码的简化版本（只有两个线程，但我想让它可扩展）：
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C和C ++，它实际上是我试图并行化的传统C代码。）
在MSDN上阅读文档，我认为这应该有用。但是，线程0只打印一次，偶尔打印两次，然后程序挂起。这是同步线程的正确方法吗？如果没有，你会推荐什么（MFC中是否真的没有内置的屏障支持？）。
编辑：此解决方案错误，甚至包括Alessandro's fix。例如，请考虑以下情况：
所以问题仍然存在：一个如何安全地确保线程保持同步状态？
答案 0 :(得分：4)
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的标准库和一些随机化）。您应该能够将其放入新的Win32控制台项目（没有预编译的头文件），编译并运行。
#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;
}
在我的机器上运行此程序（双核）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发生变化）。
解决方案的“核心”在于“RandezvousOthers”功能。此函数将阻塞共享信号量（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结构并取消阻塞共享信号量上阻塞的所有线程（如果该线程在其上函数被调用是最后一个调用函数的函数。）
答案 1 :(得分：2)
要使其正常工作，请将CreateEvent的第二个参数设置为TRUE。这将使事件“手动重置”并阻止Waitxxx重置它。
然后在循环的开头放置一个ResetEvent。
答案 2 :(得分：1)
我通过Google搜索“屏障同步窗口”找到了这个SyncTools（下载SyncTools.zip）。它使用一个CriticalSection和一个Event来实现N个线程的屏障。