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  • 生产者消费者问题

    
    
    

    http://wenku.baidu.com/view/b602a1beb9d528ea81c77986.html?re=view
    #include<stdio.h> #include<malloc.h> #include<pthread.h> #include<semaphore.h> #define BUFFER_SIZE 30 #define OVER (-1) struct Product { int tid; int data; }; struct producers { //定义生产者条件变量结构 struct Product buffer[BUFFER_SIZE];//缓冲区 sem_t sem_read; // 读信号量 sem_t sem_write; // 写信号量 pthread_mutex_t wlock; // 缓冲区写锁 pthread_mutex_t rlock; // 缓冲区读锁 pthread_mutex_t lock; // thread_count的读写锁 int readpos , writepos;//读写位置 }; struct producers buffer; int thread_count = 30; //存活生产者计数 //用于在线程内部标识线程ID int ids[30] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30}; int count = 0; // 计数消费产品数量 int pcount = 0; // 计数生产产品数量 void init() { //初始化相关锁和变量以及信号量 buffer.readpos = 0; buffer.writepos = 0; //写信号量比缓冲区小1,防止缓冲区满和缓冲区空分不清 sem_init(&buffer.sem_write, 0, BUFFER_SIZE-1); sem_init(&buffer.sem_read, 0, 0); pthread_mutex_init(&buffer.wlock, NULL); pthread_mutex_init(&buffer.rlock, NULL); pthread_mutex_init(&buffer.lock, NULL); } void put(int tid, int data) { //缓冲区中放入一个数据 sem_wait(&buffer.sem_write); //生产前先加锁,已防止其他线程同时生产 pthread_mutex_lock(&buffer.wlock); buffer.buffer[buffer.writepos].tid = tid; buffer.buffer[buffer.writepos].data = data; buffer.writepos ++; ++ pcount; if( buffer.writepos >= BUFFER_SIZE ) buffer.writepos = 0; pthread_mutex_unlock(&buffer.wlock); sem_post(&buffer.sem_read); } //读数据并移走 struct Product * get(int tid) { struct Product * produce = NULL; //消费前先上锁,以防止其他线程同时消费 pthread_mutex_lock(&buffer.rlock); // 如果生产者线程没有全部退出,或者缓冲区内仍有产品,则说明可以尝试去获取产品 if(thread_count > 0 || buffer.readpos != buffer.writepos) { //从信号量的值减去一个“1”,但它永远会先等待该信号量为一个非零值才开始做减法 sem_wait( &buffer.sem_read ); int pos = buffer.readpos; // 在已有产品中迭代,查找适合的产品 while(pos != buffer.writepos) { int id = buffer.buffer[pos].tid; if(id >10 && id <=20 && tid == id) { //如果产品是的生产者线程号 10<id<=20 则可以供和它的线程号相同的消费者线程消费 produce = (struct Product *)malloc(sizeof(struct Product)); produce->tid = id; produce->data = buffer.buffer[pos].data; break; } else if(id <= 10 && (id%2 == tid%2)) { //如果产品是的生产者线程号 <=10 则可以供和它的线程号奇偶性相同的消费者消费 produce = (struct Product *)malloc(sizeof(struct Product)); produce->tid = id; produce->data = buffer.buffer[pos].data; break; } else if(id > 20) { //如果产品是的生产者线程号>20则可以供任何消费者消费 produce = (struct Product *)malloc(sizeof(struct Product)); produce->tid = id; produce->data = buffer.buffer[pos].data; break; } pos = (pos+1)%BUFFER_SIZE; } if( produce ) { //如果取得了产品,消费计数+1,将在队头的元素覆盖到被取走的位置 ++ count; buffer.buffer[pos].tid = buffer.buffer[buffer.readpos].tid; buffer.buffer[pos].data = buffer.buffer[buffer.readpos].data; ++ buffer.readpos; } if( buffer.readpos >= BUFFER_SIZE ) buffer.readpos = 0; //如果取得了产品则释放一个缓冲区可写的信号量,否则释放一个可读的信号量 if( produce ) sem_post(&buffer.sem_write); else sem_post(&buffer.sem_read); pthread_mutex_unlock(&buffer.rlock); }else pthread_mutex_unlock(&buffer.rlock); return produce; } void *producer(void *data) { //每个线程循环生产30个产品 int tid = *((int *)data); int n; for(n = 1; n <= 30; n++) { printf("producer %d product %d ", tid, n); put(tid, n); } // 每退出一个生产者线程后,thread_count 减1 pthread_mutex_lock(&buffer.lock); -- thread_count; pthread_mutex_unlock(&buffer.lock); return NULL; } void *consumer(void * data) { int tid = *((int *)data); struct Product *d = NULL; while(1) { d = get(tid); if( d ) { printf("consumer %d consum %d from Producer %d ", tid, d->data,d->tid); free(d); } pthread_mutex_lock(&buffer.lock); // 当所有生产者线程已退出,且缓冲区内已没有该线程可消费的产品时,退出该线程 if(d == NULL && thread_count == 0 ) { pthread_mutex_unlock(&buffer.lock); break; }else pthread_mutex_unlock(&buffer.lock); } return NULL; } int main(void) { pthread_t th_a[30], th_b[30]; void *retval; init(&buffer); int i; for(i = 0; i<30; ++i) { // 创建生产者和消费者 pthread_create(&th_a[i], NULL, producer, &ids[i]); pthread_create(&th_b[i], NULL, consumer, &ids[i]); } for(i = 0; i<30; ++i) { // 将线程加入到主线程组 pthread_join(th_a[i], &retval); pthread_join(th_b[i], &retval); } printf("product %d products ", pcount); printf("consume %d products ", count); return 0; }
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  • 原文地址:https://www.cnblogs.com/xs-yqz/p/5055971.html
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