pctest.cpp

操作系统中一个生产者消费者问题的在vc++中的简单实现

// PCTest.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"

#include   <windows.h>   
#include   <iostream>   

const   unsigned   short   SIZE_OF_BUFFER   =   10;   //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   g_buffer[SIZE_OF_BUFFER];         //缓冲区是个循环队列   
bool   g_continue   =   true;             //控制程序结束   
HANDLE   g_hMutex;               //用于线程间的互斥   
HANDLE   g_hFullSemaphore;           //当缓冲区满时迫使生产者等待   
HANDLE   g_hEmptySemaphore;           //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   
	g_hMutex   =   CreateMutex(NULL,FALSE,NULL);   
	g_hFullSemaphore   =   CreateSemaphore(NULL,SIZE_OF_BUFFER-1,SIZE_OF_BUFFER-1,NULL);   
	g_hEmptySemaphore   =   CreateSemaphore(NULL,0,SIZE_OF_BUFFER-1,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[PRODUCERS_COUNT];   //各线程的handle   
	DWORD   producerID[CONSUMERS_COUNT];   //生产者线程的标识符   
	DWORD   consumerID[THREADS_COUNT];   //消费者线程的标识符   
    int i;
	//创建生产者线程   
	for   (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(g_continue){   
		if(getchar()){   //按回车后终止程序运行   
			g_continue   =   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::cerr   <<   "Producing   "   <<   ++ProductID   <<   "   ...   ";   
	std::cerr   <<   "Succeed"   <<   std::endl;   
}   

//把新生产的产品放入缓冲区   
void   Append()   
{   
	std::cerr   <<   "Appending   a   product   ...   ";   
	g_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   <<":   "   <<   g_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   =   g_buffer[out];   
	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   <<":   "   <<   g_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::cerr   <<   "Consuming   "   <<   ConsumeID   <<   "   ...   ";   
	std::cerr   <<   "Succeed"   <<   std::endl;   
}   

//生产者   
DWORD     WINAPI   Producer(LPVOID   lpPara)   
{   
	while(g_continue){   
		WaitForSingleObject(g_hFullSemaphore,INFINITE);   
		WaitForSingleObject(g_hMutex,INFINITE);   
		Produce();   
		Append();   
		Sleep(1500);   
		ReleaseMutex(g_hMutex);   
		ReleaseSemaphore(g_hEmptySemaphore,1,NULL);   
	}   
	return   0;   
}   

//消费者   
DWORD     WINAPI   Consumer(LPVOID   lpPara)   
{   
	while(g_continue){   
		WaitForSingleObject(g_hEmptySemaphore,INFINITE);   
		WaitForSingleObject(g_hMutex,INFINITE);   
		Take();   
		Consume();   
		Sleep(1500);   
		ReleaseMutex(g_hMutex);   
		ReleaseSemaphore(g_hFullSemaphore,1,NULL);   
	}   
	return   0;   
}