生产者消费者问题3

//生产者消费者问题
//使用信号量和互斥量实现
//信号量用于同步：缓冲区有产品，消费者才能取；缓冲区有空位置，生产者才能放入。
//互斥量用于互斥：生产者和消费者不能同时访问缓冲区
//

//缓冲区容量为4
#include <iostream>
#include <Windows.h>

//互斥量
HANDLE g_hMutex;
//信号量
HANDLE g_hSemA; //表示空缓冲区个数
HANDLE g_hSemB; //非空缓冲区个数
//缓冲区 4个
int g_Buffer[4];     

int g_i, g_j;
DWORD WINAPI ProducterThreadFunc(PVOID pM)
{
	int i;
	for(i = 1; i <= 10; i++)
	{
		//等待信号量>0，
		WaitForSingleObject(g_hSemA, INFINITE);//信号量-1
		//等待互斥量被释放（触发）
		WaitForSingleObject(g_hMutex, INFINITE);
		
		g_Buffer[g_i] = i;
		std::cout<<"生产者从缓冲区 "<<g_i<<" 中放入产品："<<g_Buffer[g_i]<<std::endl;
		g_i = (g_i + 1) % 4;
		//释放互斥量
		ReleaseMutex(g_hMutex);
		//信号量+1，非空缓冲区个数+1
		ReleaseSemaphore(g_hSemB, 1, NULL);
	}
	return 0;
}
DWORD WINAPI ConsumerFunc(PVOID pM)
{	
	while (true)
	{
		WaitForSingleObject(g_hSemB, INFINITE);
		//等待互斥量触发
		WaitForSingleObject(g_hMutex,INFINITE);
		
		std::cout<<"    "<<GetCurrentThreadId()<<" 号消费者从缓冲区 "<<g_j<<" 中拿出产品："<<g_Buffer[g_j]<<std::endl;
		
		if(g_Buffer[g_j] == 10)
		{
			ReleaseMutex(g_hMutex);			
			ReleaseSemaphore(g_hSemB, 1, NULL);
			break;
		}
		g_j = (g_j + 1) % 4;
		//释放互斥量
		ReleaseMutex(g_hMutex);
		ReleaseSemaphore(g_hSemA, 1, NULL);		
	}
	return 0;
}
int main()
{
	g_i = 0;
	g_j = 0;

	//设置互斥量为触发状态（释放）,第2个参数FALSE
	g_hMutex = CreateMutex(NULL, FALSE, NULL);

	
	//信号量
	g_hSemA = CreateSemaphore(NULL, 4, 4, NULL);//4个空缓冲区
	g_hSemB = CreateSemaphore(NULL, 0, 4, NULL);
	//线程
	HANDLE hThread[3];
	//生产者线程
	hThread[0] = CreateThread(NULL, 0, ProducterThreadFunc, NULL, 0, NULL);
	//消费者线程
	hThread[1] = CreateThread(NULL, 0, ConsumerFunc, NULL, 0, NULL);
	hThread[2] = CreateThread(NULL, 0, ConsumerFunc, NULL, 0, NULL);
	DWORD xxx = WaitForMultipleObjects(3, hThread, TRUE, INFINITE); 

	std::cout<<std::endl<<"执行完成。\n"<<std::endl;
	CloseHandle(hThread[0]);
	CloseHandle(hThread[1]);

	CloseHandle(g_hSemA);
	CloseHandle(g_hSemB);

	CloseHandle(g_hMutex);
	return 0;
}

　运行结果：

　

参考：http://blog.csdn.net/morewindows/article/details/7577591#reply