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  • 生产者-消费者问题...

    有A,B,C,D四个线程共享一个缓冲区,其中A是生产者线程负责向缓冲区送数据,B,C,D是消费者线程需要从缓冲区中取数据。A每送一次数据,B,C,D必须都取一次后A才能再向缓冲区中送数据。要求B,C,D从缓冲区中取数据时必须是互斥进行的。哪位大大会做啊,麻烦给我发信息以后留言给我,谢谢!

    提问者:fineyzhang

    发布于2011-01-25

共1个回答
  • 弱肉强食江湖 丨Lv 1
    #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 (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; }
    +1 2011-01-25 举报
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