操作系统生产者和消费者问题

1、生产者-消费者问题是一个经典的进程同步问题，已经属于化石级别的了。该问题最早由 Dijkstra提出，用以演示他提出的信号量机制。要求设计在同一个进程地址空间内执行的两个线程。生产者线程生产物品， 然后将物品放置在一个空缓冲区中供消费者线程消费。消费者线程从缓冲区中获得物品，然后释放缓冲区。当生产者线程生产物品时，如果没有空缓冲区可用，那么生产者线程必须等待消费者线程释放出一个空缓冲区。当消费者线程消费物品时，如果没有满的缓冲区， 那么消费者线程将被阻塞，直到新的物品被生产出来。要求设计并实现一个进程，该进程拥有一个生产者线程和一个消费者线程，它们使用N个不同的缓冲区（N为一个自定义的确定的数

2、值，例如 N=32）。需要使用如下信号量： 一个互斥信号量，用以阻止生产者线程和消费者线程同时操作缓冲区列表； 一个信号量，当生产者线程生产出一个物品时可以用它向消费者线程发出信号；一个信号量，消费者线程释放出一个空缓冲区时可以用它向生产者线程发出信号； 看代码吧:/pv操作：生产者与消费者经典问题/author:leaf#include <stdio . h>#include <stdlib . h>#include <unistd . h>#include <pthread . h>#include <semaphore h>#d

3、efine M 32/* 缓冲数目 */#define P(x) sem_wait (&x)#define V(x) sem_post (&x)int in = 0;/*生广者放置广品的位置*/int out = 0 ;/*消者取广品的位置*/int buff M = 0;/*缓冲初始化为0,开始时没有产品*/sem_t empty_sem/*同步信号里，当两了时阻止生广者放广品 */sem_t full_sem ;/*同步信号量，当没产品时阻止消费者消费*/pthread_mutex_t mutex; /*互斥信号量，一次只有一个线程访问缓冲*/*output the buf

4、fer*/void print ()(int i ;for (i = 0 ; i < M; i +)printf ("%d ", buff i );printf ("n");/*producer*/void * producer ()(for (;)(sleep (1);P (empty_sen);pthread_mutex_lock (&mutex););in = in %M;printf ("(+)produce a product. buffer:"buffin = 1 ;print ()；+in ;pthread

5、_mutex_unlock (&mutex);V (full_sem );/*consumer*/void *consumer()for (;)sleep (2);P (full_sem );pthread_mutex_lock (&mutex);out= out %M;printf ("(-)consume a product. buffer:" );buffout = 0 ;print ();+out;pthread_mutex_unlock (&mutex);V (empty_sen); void sem_mutex_init ()(/*sem

6、aphore initialize*/int initl= sem_init (&empty_sem 0 , M);int init2= sem_init (&full_sem , 0, 0);if ( (initl != 0) && (init2 != 0)(printf ("sem init failed n");exit (1);/*mutex initialize*/int init3 = pthread_mutex_init (&mutex, NULL;if (init3 != 0)(printf ("mutex

7、init failed n");exit (1);int main ()pthread_t id1 ;pthread_t id2 ;int i ;int ret ;sem_mutex_init ();/*create the producer thread*/ret = pthread_create (&id1 , NULL producer , NULL;if (ret != 0)printf ("producer creation failed n");exit (1);/*create the consumer thread*/ret = pthre

8、ad_create (&id2 , NULL, consumer, NULL;if (ret != 0)printf ("consumer creation failed n");exit (1);pthread_join (id1 ,NULL;pthread_join (id2 , NULL;exit (0);程序执行结果:|root1 Deal host匚岸 geee-a pv-1 pth r»ad|root1 seal hosteg Jpv(+produce ap roduct.buff«r:l 0 00 0 0 0000000000000

9、0 0 00000000000(+prendu匚* ap roductbuf fvr:l 1 00 0 0 00000000000000 0 00000000000(-Jeonsum? sp roductubuffer:。 1 00 0 0 000000 00000000 0 00000000000(-produce ap roductubuffer:。 1 10 0 0 000000 0* 0000000000000000000(-)con $ urre op rod u ct.buffersO 0 1oooo00000 0000000 0 000000000000(-k)p rod ute

10、 却p rodEt _buff&r：O 0 110 0 00000000000000 0 00000000Q00(4-)p rod ute ap rad u Et Bbufferi-O 0 1110 0OOOOOOOOOOOOD 0 000000000c0(-)ton s urne ap rad u Et BbuffersO 0 0110 000000 O' 0000000 0 00000000000(4-)p rod uce ap rad u Et Bbuffer:0 0 0111000000 O' 0000000 0 00000000000(+)p rod uce

11、ap rad u ct Bbuffer:0 0 0111100000 0* 0000000 0 00000000000(-)con s urre ap rad u ct BbufferrO 0 00 11100000 00000000 0 00000000000(+) p r od u c e ap rad u 匚t: “bu f f b r: 0 0 00 11110000 00000000 0 0000000000Q(+) p rod u ce ap rad u匚七.bu F f e r: 0 0 0Oil：1100000000000 0 0000000000Q(-)can s urre

12、ap rad u 匚t:.bu F f b r: 0 0 00 0 111100000000000 0 00000000000(+) p rod u ce ap rad u ct rbu f f e r: 0 0 00 0 1111100 000000000 0 000000000Q(4) p rod u ce ap r d u e t rbu F f e r: 0 0 00 0 1111110 00000000 0 00000000000(-)LOll S UFTlB 吉p r a d u C t rbu F f 6 r; 0 0 00 0 0 11111000000000000000000

13、ccG(-F)p rod uce ap r n d u C t rbu F f e r:0 Q 00 0 0 1L 1 1 1 100000000000000000c00(Hip rod uce ap rad u Ct rbu Ffer0 00 0 0 11LI111OOOOOODOOO0OOOOO0c0(-)tonsunn& ap rodiict Pbu Ffer:0 0 00 0 0 0I 1 1 1 1 10000000000000000000(-FJproduce ap radud Pbu f f e r; 0 0 00 0 0 0L1111110000000000000000

14、00(+)producs ap roductPbuf ft r:0 0 00 0 0 0III 1 1 1 1 100000000000000000(-)Eonsurre ap rodwetbuffsr:0 0 00 0 0 00 1 1 1 111100000000000000000(+)producfl ap rodwet,buf Rr:Q 0 00 0 0 00 1 1 1 1 1 1 1 10000000000000000(+product ap roduct.byffer:0 0 00 0 0 00 I 1 1 1 I 1 1 1 1 000000000000000(-Jconsuw

15、 fp rcidm匚bi*ffer:0 0 00 0 0 00 0 1 1 1 1 1 1 1 1 000000000000000(+proiduGe opCiduEt Lbuffer:O 0 00 0 0 00 0 1111111 1 1 DDOOOOOOOOO000(Tpr以du匚孕«p rgdmctbuffer:。 0 00 0 0 00 0 1 1 I 1 I 1 1 1 110000000000000(-JconsuHie ap rgductbuffer:0 0 0oooo0 0 0 1 1 1 1 1 1 1 1 i oooooooooo000(d-)p rod uce ap rad u ctbuffer:Q 0 00 0 0 0OOOllllllllllOOOOOOOGOe00(d-)p rod uce ap rad u ct sbuffer