The default implementation of the declaration with two parameters always returns TRUE. This function is called to release access to the synchronization object owned by the calling thread. The second declaration is provided for synchronization objects such as semaphores that allow more than one access of a controlled resource.

二、产生问题的原因

产生问题的原因如摘要所述：同一个CMutex对象不允许在两个线程中分别调用Lock和Unlock，否则Unlock会失败并返回0。（P.S.这可能并不是唯一会使Unlock返回0的原因，笔者这次是这个原因）

三、例程

1、源码

#include "stdafx.h"
#include "stdlib.h"
#include "afxmt.h"
#include <iostream>
using namespace std;

CMutex g_objLock;

DWORD WINAPI ThrdFun1(LPVOID lpParam)//
{
	int iResult,iResult1,i=0,iTest;
	int iCount=0,iCycle=1;

	for(i=0;i<9999;i++)
	{
		g_objLock.Lock();
		printf("ThrdFun1:%d\n",i);
		g_objLock.Unlock();
		Sleep(1000);
	}

	return 0;
}

DWORD WINAPI ThrdFun2(LPVOID lpParam)//
{
	int iResult,iResult1,i=0,iTest;
	int iCount=0,iCycle=1;
	int iSleep=5;

	/*printf("ThrdFun2:lock and sleep %d sec\n",iSleep);
	g_objLock.Lock();
	Sleep(iSleep*1000);*/
	iResult=g_objLock.Unlock();
	printf("ThrdFun2:unlock ret %d\n",iResult);

	return 0;
}

int _tmain(int argc, _TCHAR* argv[])
{
	int iResult,iResult1,i=0,iTest;
	int iSleep=5;

	printf("start 1\n");
	CreateThread(NULL, 0, ThrdFun1, NULL, 0, NULL);  
	printf("sleep %d sec\n",iSleep);
	Sleep(iSleep*1000);
	
	printf("lock and sleep %d sec\n",iSleep);
	g_objLock.Lock();
	Sleep(iSleep*1000);
	iResult=g_objLock.Unlock();
	printf("unlock ret %d\n",iResult);

	printf("sleep %d sec\n",iSleep);
	Sleep(iSleep*1000);
	
	printf("lock and sleep %d sec\n",iSleep);
	g_objLock.Lock();
	Sleep(iSleep*1000);
	printf("start 2\n");
	CreateThread(NULL, 0, ThrdFun2, NULL, 0, NULL);  

	Sleep(999*1000);

	system("pause");
	return 0;
}