生产者消费者问题

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。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。

#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; }