一、问题描述
生产者-消费者问题是一个经典的进程同步问题，该问题最早由Dijkstra提出，用以演示他提出的信号量机制。本作业要求设计在同一个进程地址空间内执行的两个线程。生产者线程生产物品，然后将物品放置在一个空缓冲区中供消费者线程消费。消费者线程从缓冲区中获得物品，然后释放缓冲区。当生产者线程生产物品时，如果没有空缓冲区可用，那么生产者线程必须等待消费者线程释放出一个空缓冲区。当消费者线程消费物品时，如果没有满的缓冲区，那么消费者线程将被阻塞，直到新的物品被生产出来。
二、实现代码
#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
//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 (int 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;
}
