生产者消费者问题

标签：操作系统   生产者消费者问题   

C代码：

#include <windows.h>
#include <iostream>
const unsigned short SIZE_OF_BUFFER = 2; //缓冲区长度
unsigned short ProductID = 0; //产品号
unsigned short ConsumeID = 0; //将被消耗的产品号
unsigned short in = 0; //产品进缓冲区时的缓冲区下标
unsigned short out = 0; //产品出缓冲区时的缓冲区下标
int buffer[SIZE_OF_BUFFER]; //缓冲区是个循环队列操作系统课程设计指导书
bool p_ccontinue = true; //控制程序结束
HANDLE Mutex; //用于线程间的互斥
HANDLE FullSemaphore; //当缓冲区满时迫使生产者等待
HANDLE EmptySemaphore; //当缓冲区空时迫使消费者等待
DWORD WINAPI Producer(LPVOID); //生产者线程
DWORD WINAPI Consumer(LPVOID); //消费者线程
int main()
{
	//创建各个互斥信号
	//注意，互斥信号量和同步信号量的定义方法不同，互斥信号量调用的是 CreateMutex 函数，
	//调用的是 CreateSemaphore 函数，函数的返回值都是句柄。
	Mutex = CreateMutex(NULL,FALSE,NULL);
	EmptySemaphore = CreateSemaphore(NULL,SIZE_OF_BUFFER,SIZE_OF_BUFFER,NULL);
	//将上句做如下修改，看看结果会怎样
	//EmptySemaphore = CreateSemaphore(NULL,0,SIZE_OF_BUFFER-1,NULL);
	FullSemaphore = CreateSemaphore(NULL,0,SIZE_OF_BUFFER,NULL);
	//调整下面的数值，可以发现，当生产者个数多于消费者个数时，
	//生产速度快，生产者经常等待消费者；反之，消费者经常等待
	const unsigned short PRODUCERS_COUNT = 3; //生产者的个数
	const unsigned short CONSUMERS_COUNT = 1; //消费者的个数
	//总的线程数
	const unsigned short THREADS_COUNT = PRODUCERS_COUNT+CONSUMERS_COUNT;
	HANDLE hThreads[THREADS_COUNT]; //各线程的 handle
	DWORD producerID[PRODUCERS_COUNT]; //生产者线程的标识符
	DWORD consumerID[CONSUMERS_COUNT]; //消费者线程的标识符
	//创建生产者线程
	for (int i=0;i<PRODUCERS_COUNT;++i){
		hThreads[i]=CreateThread(NULL,0,Producer,NULL,0,&producerID[i]);
		if (hThreads[i]==NULL) return -1;
	}
	//创建消费者线程
	for (i=0;i<CONSUMERS_COUNT;++i){
		hThreads[PRODUCERS_COUNT+i]=CreateThread(NULL,0,Consumer,NULL,0,&consumerID[i]);
		if (hThreads[i]==NULL) return -1;
	}
	while(p_ccontinue){
		if(getchar()){ //按回车后终止程序运行操作系统课程设计指导书
			p_ccontinue = false;
		}
	}
	return 0;
}
//生产一个产品。简单模拟了一下，仅输出新产品的 ID 号
void Produce()
{
	std::cout << std::endl<< "Producing " << ++ProductID << " ... ";
	std::cout << "Succeed" << std::endl;
}
//把新生产的产品放入缓冲区
void Append()
{
	std::cerr << "Appending a product ... ";
	buffer[in] = ProductID;
	in = (in+1)%SIZE_OF_BUFFER;
	std::cerr << "Succeed" << std::endl;
	//输出缓冲区当前的状态
	for (int i=0;i<SIZE_OF_BUFFER;++i){
		std::cout << i <<": " << buffer[i];
		if (i==in) std::cout << " <-- 生产";
		if (i==out) std::cout << " <-- 消费";
		std::cout << std::endl;
	}
}
//从缓冲区中取出一个产品
void Take()
{
	std::cerr << "Taking a product ... ";
	ConsumeID = buffer[out];
	buffer[out] = 0;
	out = (out+1)%SIZE_OF_BUFFER;
	std::cerr << "Succeed" << std::endl;
	//输出缓冲区当前的状态
	for (int i=0;i<SIZE_OF_BUFFER;++i){
		std::cout << i <<": " << buffer[i];
		if (i==in) std::cout << " <-- 生产";
		if (i==out) std::cout << " <-- 消费";
		std::cout << std::endl;
	}
}
//消耗一个产品
void Consume()
{
	std::cout << "Consuming " << ConsumeID << " ... ";
	std::cout << "Succeed" << std::endl;
}
//生产者
DWORD WINAPI Producer(LPVOID lpPara)
{
	while(p_ccontinue){
		WaitForSingleObject(EmptySemaphore,INFINITE); //p(empty);
		WaitForSingleObject(Mutex,INFINITE); //p(mutex);
		Produce();
		Append();
		Sleep(1500);
		ReleaseMutex(Mutex); //V(mutex);
		ReleaseSemaphore(FullSemaphore,1,NULL); //V(full);
	}
	return 0;
}
//消费者
DWORD WINAPI Consumer(LPVOID lpPara)
{
	while(p_ccontinue){
		WaitForSingleObject(FullSemaphore,INFINITE); //P(full);
		WaitForSingleObject(Mutex,INFINITE); //P(mutex);
		Take();
		Consume();
		Sleep(1500);
		ReleaseMutex(Mutex); //V(mutex);
		ReleaseSemaphore(EmptySemaphore,1,NULL); //V(empty);
	}
	return 0;
}

生产者消费者问题,布布扣,bubuko.com

生产者消费者问题

标签：操作系统   生产者消费者问题   

原文地址：http://blog.csdn.net/u012964281/article/details/38254037