生產者消費者問題

生產者-消費者問題是一個經典的進程同步問題，該問題最早由Dijkstra提出，用以示範他提出的訊號量機制。在同一個進程地址空間內執行的兩個線程。生產者線程生產物品，然後將物品放置在一個空緩衝區中供消費者線程消費。消費者線程從緩衝區中獲得物品，然後釋放緩衝區。當生產者線程生產物品時，如果沒有空緩衝區可用，那麼生產者線程必須等待消費者線程釋放出一個空緩衝區。當消費者線程消費物品時，如果沒有滿的緩衝區，那麼消費者線程將被阻塞，直到新的物品被生產出來。

#include   <windows.h> #include   <iostream> const   unsigned   short   SIZE_OF_BUFFER   =   10;   //緩衝區長度 unsigned   short   ProductID   =   0;         //產品號 unsigned   short   ConsumeID   =   0;         //將被消耗的產品號 unsigned   short   in   =   0;             //產品進緩衝區時的緩衝區下標 unsigned   short   out   =   0;             //產品出緩衝區時的緩衝區下標 int   g_buffer[SIZE_OF_BUFFER];         //緩衝區是個迴圈隊列 bool   g_continue   =   true;             //控製程序結束 HANDLE   g_hMutex;               //用於線程間的互斥 HANDLE   g_hFullSemaphore;           //當緩衝區滿時迫使生產者等待 HANDLE   g_hEmptySemaphore;           //當緩衝區空時迫使消費者等待 DWORD   WINAPI   Producer(LPVOID);         //生產者線程 DWORD   WINAPI   Consumer(LPVOID);         //消費者線程 int   main() {     //建立各個互斥訊號     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);     //調整下面的數值，可以發現，當生產者個數多於消費者個數時，     //生產速度快，生產者經常等待消費者；反之，消費者經常等待       const   unsigned   short   PRODUCERS_COUNT   =   3;     //生產者的個數     const   unsigned   short   CONSUMERS_COUNT   =   1;     //消費者的個數     //總的線程數     const   unsigned   short   THREADS_COUNT   =   PRODUCERS_COUNT+CONSUMERS_COUNT;     HANDLE   hThreads[PRODUCERS_COUNT];   //各線程的handle     DWORD   producerID[CONSUMERS_COUNT];   //生產者線程的標識符     DWORD   consumerID[THREADS_COUNT];   //消費者線程的標識符     //建立生產者線程     for   (int   i=0;i <PRODUCERS_COUNT;++i){         hThreads[i]=CreateThread(NULL,0,Producer,NULL,0,&producerID[i]);         if   (hThreads[i]==NULL)   return   -1;     }     //建立消費者線程     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()){   //按斷行符號後終止程式運行             g_continue   =   false;         }     }     return   0; } //生產一個產品。簡單類比了一下，僅輸出新產品的ID號 void   Produce() {     std::cerr   < <   "Producing   "   < <   ++ProductID   < <   "   ...   ";     std::cerr   < <   "Succeed "   < <   std::endl; } //把新生產的產品放入緩衝區 void   Append() {     std::cerr   < <   "Appending   a   product   ...   ";     g_buffer[in]   =   ProductID;     in   =   (in+1)%SIZE_OF_BUFFER;     std::cerr   < <   "Succeed "   < <   std::endl;     //輸出緩衝區當前的狀態     for   (int   i=0;i <SIZE_OF_BUFFER;++i){         std::cout   < <   i   < < ":   "   < <   g_buffer[i];         if   (i==in)   std::cout   < <   "   <--   生產 ";         if   (i==out)   std::cout   < <   "   <--   消費 ";         std::cout   < <   std::endl;     } } //從緩衝區中取出一個產品 void   Take() {     std::cerr   < <   "Taking   a   product   ...   ";     ConsumeID   =   g_buffer[out];     out   =   (out+1)%SIZE_OF_BUFFER;     std::cerr   < <   "Succeed "   < <   std::endl;     //輸出緩衝區當前的狀態     for   (int   i=0;i <SIZE_OF_BUFFER;++i){         std::cout   < <   i   < < ":   "   < <   g_buffer[i];         if   (i==in)   std::cout   < <   "   <--   生產 ";         if   (i==out)   std::cout   < <   "   <--   消費 ";         std::cout   < <   std::endl;     } } //消耗一個產品 void   Consume() {     std::cerr   < <   "Consuming   "   < <   ConsumeID   < <   "   ...   ";     std::cerr   < <   "Succeed "   < <   std::endl; } //生產者 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; } //消費者 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; }