最全面的linux信号量解析（5） include, linux, 消费者, 多线程, 生产者

yuyang911220

七．信号量的牛刀小试——生产者与消费者问题
1．问题描述：
有一个长度为N的缓冲池为生产者和消费者所共有，只要缓冲池未满，生产者便可将
消息送入缓冲池；只要缓冲池未空，消费者便可从缓冲池中取走一个消息。生产者往缓冲池
放信息的时候，消费者不可操作缓冲池，反之亦然。
2．使用多线程和信号量解决该经典问题的互斥
#include <pthread.h>
#include <stdio.h>
#include <semaphore.h>
#define BUFF_SIZE 10
char buffer[BUFF_SIZE];
char count; // 缓冲池里的信息数目
sem_t sem_mutex; // 生产者和消费者的互斥锁
sem_t p_sem_mutex; // 空的时候，对消费者不可进
sem_t c_sem_mutex; // 满的时候，对生产者不可进
void * Producer()
{
while(1)
{
sem_wait(&p_sem_mutex); //当缓冲池未满时
sem_wait(&sem_mutex); //等待缓冲池空闲
count++;
sem_post(&sem_mutex);
if(count < BUFF_SIZE)//缓冲池未满
sem_post(&p_sem_mutex);
if(count > 0) //缓冲池不为空
sem_post(&c_sem_mutex);
}
}
void * Consumer()
{
while(1)
{
sem_wait(&c_sem_mutex);//缓冲池未空时
sem_wait(&sem_mutex); //等待缓冲池空闲
count--;
sem_post(&sem_mutex);
if(count > 0)
sem_post(c_sem_nutex);
}
}
int main()
{
pthread_t ptid,ctid;
//initialize the semaphores
sem_init(&empty_sem_mutex,0,1);
sem_init(&full_sem_mutex,0,0);
//creating producer and consumer threads
if(pthread_create(&ptid, NULL,Producer, NULL))
{
printf("\n ERROR creating thread 1");
exit(1);
}
if(pthread_create(&ctid, NULL,Consumer, NULL))
{
printf("\n ERROR creating thread 2");
exit(1);
}
if(pthread_join(ptid, NULL)) /* wait for the producer to finish */
{
printf("\n ERROR joining thread");
exit(1);
}
if(pthread_join(ctid, NULL)) /* wait for consumer to finish */
{
printf("\n ERROR joining thread");
exit(1);
}
sem_destroy(&empty_sem_mutex);
sem_destroy(&full_sem_mutex);
//exit the main thread
pthread_exit(NULL);
return 1;
}