生产者和消费者问题.doc

精品文档班级 姓名： 学号： 成绩： 实验名称: 生产者和消费者问题1.实验目的：“生产者消费者”问题是一个著名的同时性编程问题的集合。通过编写经典的“生产者消费者”问题的实验，读者可以进一步熟悉Linux 中多线程编程，并且掌握用信号量处理线程间的同步互斥问题。2.实验内容：“生产者消费者”问题描述如下。有一个有限缓冲区和两个线程：生产者和消费者。他们分别把产品放入缓冲区和从缓冲区中拿走产品。当一个生产者在缓冲区满时必须等待，当一个消费者在缓冲区空时也必须等待。它们之间的关系如下图所示：这里要求用有名管道来模拟有限缓冲区，用信号量来解决生产者消费者问题中的同步和互斥问题。3.实验方法：（1）使用信号量解决（2）思考使用条件变量解决4.实验过程（1）信号量的考虑这里使用3个信号量，其中两个信号量avail和full分别用于解决生产者和消费者线程之间的同步问题，mutex是用于这两个线程之间的互斥问题。其中avail初始化为N（有界缓冲区的空单元数），mutex 初始化为1，full初始化为0。/*product.c*/#include #include #include #include #include #include #include #include #define FIFO myfifo#define N 5int lock_var;time_t end_time;char buf_r100;sem_t mutex,full,avail;int fd;void pthread1(void *arg);void pthread2(void *arg);int main(int argc, char *argv)pthread_t id1,id2;pthread_t mon_th_id;int ret;end_time = time(NULL)+30;/*创建有名管道*/if(mkfifo(FIFO,O_CREAT|O_EXCL)0)&(errno!=EEXIST)printf(cannot create fifoservern);printf(Preparing for reading bytes.n);memset(buf_r,0,sizeof(buf_r);/*打开管道*/fd=open(FIFO,O_RDWR|O_NONBLOCK,0);if(fd=-1)perror(open);exit(1);/*初始化互斥信号量为1*/ret=sem_init(&mutex,0,1);/*初始化avail信号量为N*/ret=sem_init(&avail,0,N);/*初始化full信号量为0*/ret=sem_init(&full,0,0);if(ret!=0)perror(sem_init);/*创建两个线程*/ret=pthread_create(&id1,NULL,(void *)productor, NULL);if(ret!=0)perror(pthread cread1);ret=pthread_create(&id2,NULL,(void *)consumer, NULL);if(ret!=0)perror(pthread cread2);pthread_join(id1,NULL);pthread_join(id2,NULL);exit(0);/*生产者线程*/void productor(void *arg)int i,nwrite;while(time(NULL) end_time)/*P操作信号量avail和mutex*/sem_wait(&avail);sem_wait(&mutex);/*生产者写入数据*/if(nwrite=write(fd,hello,5)=-1)if(errno=EAGAIN)printf(The FIFO has not been read yet.Please try latern);elseprintf(write hello to the FIFOn);/*V操作信号量full和mutex*/sem_post(&full);sem_post(&mutex);sleep(1);/*消费者线程*/void consumer(void *arg)int nolock=0;int ret,nread;while(time(NULL) end_time)/*P操作信号量full和mutex*/sem_wait(&full);sem_wait(&mutex);memset(buf_r,0,sizeof(buf_r);if(nread=read(fd,buf_r,100)=-1)if(errno=EAGAIN)printf(no data yetn);printf(read %s from FIFOn,buf_r);/*V操作信号量avail和mutex*/sem_post(&avail);sem_post(&mutex);sleep(1);(2)条件变量的考虑#include #include #define BUFFER_SIZE 4#define OVER (-1)struct producers/定义生产者条件变量结构 int bufferBUFFER_SIZE; pthread_mutex_t lock; int readpos, writepos; pthread_cond_t notempty; pthread_cond_t notfull;/初始化缓冲区void init(struct producers *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;/在缓冲区存放一个整数void put(struct producers *b, int data) pthread_mutex_lock(&b-lock);/当缓冲区为满时等待 while(b-writepos+1)%BUFFER_SIZE=b-readpos) pthread_cond_wait(&b-notfull,&b-lock); b-bufferb-writepos=data; b-writepos+; if(b-writepos=BUFFER_SIZE) b-writepos=0;/发送当前缓冲区中有数据的信号 pthread_cond_signal(&b-notempty); pthread_mutex_unlock(&b-lock);int get(struct producers *b) int data; pthread_mutex_lock(&b-lock); while(b-writepos=b-readpos) pthread_cond_wait(&b-notempty,&b-lock); data=b-bufferb-readpos; b-readpos+; if(b-readpos=BUFFER_SIZE) b-readpos=0; pthread_cond_signal(&b-notfull); pthread_mutex_unlock(&b-lock); return data;struct producers buffer;void *producer(void *data) int n; for(n=0;nn,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(Consumer: - %dn,d); return NULL;int main() pthread_t tha,thb;