一个简单的多线程题目

今天看到一个简单的多线程题目，自己写了一下，功能基本实现，如有问题还望指出交流；

题目：编写一个程序，开启3个线程，这3个线程的ID分别为A、B、C，每个线程将自己的ID在屏幕上打印10遍，要求输出结果必须按ABC的顺序显示；如：ABCABC….依次递推。

// TEST-3.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include 
#include 

CRITICAL_SECTION g_cs; //题目较简单，没有用到临界区
HANDLE g_SemaphoreA,g_SemaphoreB,g_SemaphoreC;
int g_nNum = 0;
bool g_bFlag = true;

unsigned int __stdcall FunA(PVOID pPm)
{
	while(g_bFlag)
	{
		WaitForSingleObject(g_SemaphoreA,INFINITE);
		if (g_nNum ==10)
		{	
			ReleaseSemaphore(g_SemaphoreA,1,NULL);
			ReleaseSemaphore(g_SemaphoreB,1,NULL);
			ReleaseSemaphore(g_SemaphoreC,1,NULL);
			g_bFlag = false;
			break;
		}
		else
			++g_nNum;
		printf("A：第%d次打印，ID%d\r\n",g_nNum,GetCurrentThreadId());
		ReleaseSemaphore(g_SemaphoreB,1,NULL);
		Sleep(10);
	}
	return  0;
}
unsigned int __stdcall FunB(PVOID pPm)
{
	while(g_bFlag)
	{
		WaitForSingleObject(g_SemaphoreB,INFINITE);
		printf("B：第%d次打印，ID%d\r\n",g_nNum,GetCurrentThreadId());
		ReleaseSemaphore(g_SemaphoreC,1,NULL);
		Sleep(10);
	}
	return  0;
}
unsigned int __stdcall FunC(PVOID pPm)
{
	while(g_bFlag)
	{
		WaitForSingleObject(g_SemaphoreC,INFINITE);
		printf("C：第%d次打印，ID%d\r\n",g_nNum,GetCurrentThreadId());
		ReleaseSemaphore(g_SemaphoreA,1,NULL);
		Sleep(10);
	}
	return  0;
}

int _tmain(int argc, _TCHAR* argv[])
{
	g_SemaphoreA = CreateSemaphore(NULL,1,3,NULL);
	g_SemaphoreB = CreateSemaphore(NULL,0,3,NULL);
	g_SemaphoreC = CreateSemaphore(NULL,0,3,NULL);
	InitializeCriticalSectionAndSpinCount(&g_cs,4000);

	HANDLE hThread[3];
	hThread[0]=(HANDLE)_beginthreadex(NULL,0,FunA,NULL,0,NULL);
	hThread[1]=(HANDLE)_beginthreadex(NULL,0,FunB,NULL,0,NULL);
	hThread[2]=(HANDLE)_beginthreadex(NULL,0,FunC,NULL,0,NULL);
	WaitForMultipleObjects(3,hThread,TRUE,INFINITE);
	CloseHandle(g_SemaphoreA);
	CloseHandle(g_SemaphoreB);
	CloseHandle(g_SemaphoreC);
	DeleteCriticalSection(&g_cs);

	for (int i = 0 ;i<3;++i)
	{
		CloseHandle(hThread[i]);
	}
	return 0;
}

运行结果：