1、班級 姓名： 學(xué)號： 成績： 實驗名稱: 生產(chǎn)者和消費者問題1.實驗?zāi)康模骸吧a(chǎn)者消費者”問題是一個著名的同時性編程問題的集合。通過編寫經(jīng)典的“生產(chǎn)者消費者”問題的實驗，讀者可以進(jìn)一步熟悉Linux 中多線程編程，并且掌握用信號量處理線程間的同步互斥問題。2.實驗內(nèi)容：“生產(chǎn)者消費者”問題描述如下。有一個有限緩沖區(qū)和兩個線程：生產(chǎn)者和消費者。他們分別把產(chǎn)品放入緩沖區(qū)和從緩沖區(qū)中拿走產(chǎn)品。當(dāng)一個生產(chǎn)者在緩沖區(qū)滿時必須等待，當(dāng)一個消費者在緩沖區(qū)空時也必須等待。它們之間的關(guān)系如下圖所示：這里要求用有名管道來模擬有限緩沖區(qū)，用信號量來解決生產(chǎn)者消費者問題中的同步和互斥問題。3.實驗方法：（1）使用信號

2、量解決（2）思考使用條件變量解決4.實驗過程（1）信號量的考慮這里使用3個信號量，其中兩個信號量avail和full分別用于解決生產(chǎn)者和消費者線程之間的同步問題，mutex是用于這兩個線程之間的互斥問題。其中avail初始化為N（有界緩沖區(qū)的空單元數(shù)），mutex 初始化為1，full初始化為0。/*product.c*/#include #include #include #include #include #include #include #include #define FIFO myfifo#define N 5int lock_var;time_t end_time;char bu

3、f_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;/*創(chuàng)建有名管道*/if(mkfifo(FIFO,O_CREAT|O_EXCL)0)&(errno!=EEXIST)printf(cannot create fifoservern);printf(Preparing for

4、 reading bytes.n);memset(buf_r,0,sizeof(buf_r);/*打開管道*/fd=open(FIFO,O_RDWR|O_NONBLOCK,0);if(fd=-1)perror(open);exit(1);/*初始化互斥信號量為1*/ret=sem_init(&mutex,0,1);/*初始化avail信號量為N*/ret=sem_init(&avail,0,N);/*初始化full信號量為0*/ret=sem_init(&full,0,0);if(ret!=0)perror(sem_init);/*創(chuàng)建兩個線程*/ret=pthread_create(&id1

5、,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);/*生產(chǎn)者線程*/void productor(void *arg)int i,nwrite;while(time(NULL) end_time)/*P操作信號量avail和mutex*/se

6、m_wait(&avail);sem_wait(&mutex);/*生產(chǎn)者寫入數(shù)據(jù)*/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操作信號量full和mutex*/sem_post(&full);sem_post(&mutex);sleep(1);/*消費者線程*/void consumer(void *arg)int nolock=0;int ret,

7、nread;while(time(NULL) end_time)/*P操作信號量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操作信號量avail和mutex*/sem_post(&avail);sem_post(&mutex);sleep(1);(2)條件變量的考慮#include #i

8、nclude #define BUFFER_SIZE 4#define OVER (-1)struct producers/定義生產(chǎn)者條件變量結(jié)構(gòu) int bufferBUFFER_SIZE; pthread_mutex_t lock; int readpos, writepos; pthread_cond_t notempty; pthread_cond_t notfull;/初始化緩沖區(qū)void init(struct producers *b) pthread_mutex_init(&b-lock,NULL); pthread_cond_init(&b-notempty,NULL); p

9、thread_cond_init(&b-notfull,NULL); b-readpos=0; b-writepos=0;/在緩沖區(qū)存放一個整數(shù)void put(struct producers *b, int data) pthread_mutex_lock(&b-lock);/當(dāng)緩沖區(qū)為滿時等待 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-

10、writepos=0;/發(fā)送當(dāng)前緩沖區(qū)中有數(shù)據(jù)的信號 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

11、; 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