Mutex可以被多次wait，那么什么情况下Wait一个Mutex才会阻塞?

[asker160]

#include<windows.h>
#include<windef.h>
#include<stdio.h>
#define BUFSIZE 1024
#define NAME "mymutex"
int main(void){
HANDLE hMutex=CreateMutex(NULL,FALSE,NAME);
if(!hMutex || INVALID_HANDLE_VALUE==hMutex){
printf("CreateMutex failed\n");
return 1;
}
ReleaseMutex(hMutex);
ReleaseMutex(hMutex);
WaitForSingleObject(hMutex,1000);
HANDLE h1=OpenMutex(SYNCHRONIZE,TRUE,NAME);
HANDLE h2=OpenMutex(SYNCHRONIZE,TRUE,NAME);
WaitForSingleObject(hMutex,1000);
CloseHandle(hMutex);
WaitForSingleObject(hMutex,1000);
return 0;
}
这个程序不会阻塞，瞬间运行完成。为什么?
我如果要让程序中的某个WaitForSingleObject()阻塞并等待，应该怎么改代码?
谢谢!!!!!

[zgl7903]

看MSDN上的Mutex是用于同步不同线程的，当被其它线程所拥有时会阻塞准备获取信号量的线程
If the mutex object is owned by another thread, the wait function blocks the requesting thread until the owning thread releases the mutex object using the ReleaseMutex function

用Event

[asker160]

补充一下，CreateMutex的时候，Mutex的owner指定为FALSE，当前线程不拥有。那么Mutex的属主是谁呢?

谢谢！！！！！

[asker160]

知道了，我用的不对:

#include<windows.h>
#include<windef.h>
#include<stdio.h>
#define BUFSIZE 1024
#define NAME "mymutex1"
HANDLE hMutex;
VOID* mythread(VOID* ){
printf("子线程等待\n");
WaitForSingleObject(hMutex,INFINITE);
printf("子线程释放\n");
ReleaseMutex(hMutex);
return 0;
}
int main(void){
hMutex=CreateMutex(NULL,TRUE,NAME);
if(!hMutex || INVALID_HANDLE_VALUE==hMutex){
printf("CreateMutex failed:%d\n",GetLastError());
return 1;
}
CreateThread(NULL,0,(LPTHREAD_START_ROUTINE)mythread,0,0,0);
Sleep(100);//使得子线程有时间先运行
ReleaseMutex(hMutex);
printf("主线程释放\n");
BOOL ret=WaitForSingleObject(hMutex,INFINITE);
printf("主线程等待:%d\n",ret);
CloseHandle(hMutex);
return 0;
}

[xxd_qd]

只有一个线程怎么可能阻塞？假若你真的设法把它阻塞了的话，那么再怎么解除？