关于经典生成者和消费者问题
- C/C++ code
#include <stdio.h> #include <pthread.h> #define BUFFER_SIZE 16 // 缓冲区数量 // 缓冲区相关数据结构 struct prodcons { int buffer[BUFFER_SIZE]; /* 实际数据存放的数组*/ int readpos; /* 读写指针*/ int writepos; pthread_mutex_t lock; /* 互斥体lock 用于对缓冲区的互斥操作 */ pthread_cond_t notempty; /* 缓冲区非空的条件变量 */ pthread_cond_t notfull; /* 缓冲区未满的条件变量 */ }; //初始化缓冲区结构 void init(struct prodcons *b) { pthread_mutex_init(&b->lock, NULL); pthread_cond_init(&b->notempty, NULL); pthread_cond_init(&b->notfull, NULL); b->readpos = 0; b->writepos = 0; } /* 将产品放入缓冲区,这里是存入一个整数*/ void put(struct prodcons *b, int data) { pthread_mutex_lock(&b->lock); /* 等待缓冲区未满*/ if ((b->writepos + 1) % BUFFER_SIZE == b->readpos) pthread_cond_wait(&b->notfull, &b->lock); /* 写数据,并移动指针 */ b->buffer[b->writepos] = data; b->writepos++; if (b->writepos >= BUFFER_SIZE) b->writepos = 0; /* 设置缓冲区非空的条件变量*/ pthread_cond_signal(&b->notempty); pthread_mutex_unlock(&b->lock); } /* 从缓冲区中取出整数*/ int get(struct prodcons *b) { int data; pthread_mutex_lock(&b->lock); /* 等待缓冲区非空*/ if (b->writepos == b->readpos) pthread_cond_wait(&b->notempty, &b->lock); /* 读数据,移动读指针*/ data = b->buffer[b->readpos]; b->readpos++; if (b->readpos > = BUFFER_SIZE) b->readpos = 0; /* 设置缓冲区未满的条件变量*/ pthread_cond_signal(&b->notfull); pthread_mutex_unlock(&b->lock); return data; } /* 测试:生产者线程将1 到10000 的整数送入缓冲区, 消费者线程从缓冲区中获取整数,两者都打印信息*/ #define OVER ( - 1) struct prodcons buffer; void *producer(void *data) { int n; for (n = 0; n < 10000; n++) { printf("%d --->\n", n); put(&buffer, n); } put(&buffer, OVER); return NULL; } void *consumer(void *data) { int d; while (1) { d = get(&buffer); if (d == OVER) break; printf("--->%d \n", d); } return NULL; } int main(void) { pthread_t th_a, th_b; void *retval; init(&buffer); /* 创建生产者和消费者线程*/ pthread_create(&th_a, NULL, producer, 0); pthread_create(&th_b, NULL, consumer, 0); /* 等待两个线程结束*/ pthread_join(th_a, &retval); pthread_join(th_b, &retval); return 0; }这是网上抄来的觉得几个问题不明白特地问问。
1、极端情况下 put函数中最后1行 pthread_mutex_unlock(&b->lock); 此时put、get函数都应该是竞争状态（即大家都pthread_mutex_lock(&b->lock); ） 应该未必是get函数能拿到，如果是put再次拿到mutex 不是一直是put函数在运行么，更有甚者put函数把缓存都写满从头又开始写？？
2、假设我在put函数中最后2行加入 sleep 10多秒 ，然后放开锁，如果锁被get函数拿到，那么get中的pthread_cond_wait(&b->notempty, &b->lock); 还能获得信号，再放开锁么？
[解决办法]
首先这段代码的实现思路确实是有问题的，楼主说的第1点情况确实是有可能发生。
还有对于条件变量的使用，这里说一下：
- C/C++ code
if (b->writepos == b->readpos) pthread_cond_wait(&b->notempty, &b->lock);
[解决办法]
[Quote=引用:]
首先这段代码的实现思路确实是有问题的，楼主说的第1点情况确实是有可能发生。
还有对于条件变量的使用，这里说一下：
C/C++ code
if (b->writepos == b->readpos)
pthread_cond_wait(&b->notempty, &b->lock);
修改一下 代码就可以实现 连续的 get或者 put，比如连续的get
- C/C++ code
void put(struct prodcons * prod, int data){ pthread_mutex_lock(&prod->lock); /* Wait until buffer is not full */ while ((prod->writepos + 1) % BUFFER_SIZE == prod->readpos) { printf("producer wait for not full\n"); pthread_cond_wait(&prod->notfull, &prod->lock); } /* Write the data and advance write pointer */ prod->buffer[prod->writepos] = data; prod->writepos++; if (prod->writepos >= BUFFER_SIZE) prod->writepos = 0; /*Signal that the buffer is now not empty */ pthread_cond_signal(&prod->notempty); pthread_mutex_unlock(&prod->lock);}/* Read and remove an integer from the buffer */int get(struct prodcons *prod){ int data; pthread_mutex_lock(&prod->lock); /* Wait until buffer is not empty */ while (prod->writepos == prod->readpos) { printf("consumer wait for not empty\n"); pthread_cond_wait(&prod->notempty, &prod->lock); } /* Read the data and advance read pointer */ data = prod->buffer[prod->readpos]; prod->readpos++; if (prod->readpos >= BUFFER_SIZE) prod->readpos = 0; /* Signal that the buffer is now not full */ pthread_cond_signal(&prod->notfull); pthread_mutex_unlock(&prod->lock); return data;}