linux理发师多线程问题

1、用多线程同步方法解决睡眠理发师问题（ Sleeping-Barber Problem)1 操作系统： Linux2 程序设计语言：C语言3 设有1个理发师5把椅子（另外还有一把理发椅），几把椅子可用连续存储单元.1技术要求：1）为每个理发师顾客产生一个线程，设计正确的同步算法2）每个顾客进入理发室后，即时显示“ Entered” 及其线程自定义标识，还同时显示理发室共有几名顾客及其所坐的位置。3）至少有10 个顾客，每人理发至少3秒钟。4）多个顾客须共享操作函数代码。提示: (1) 连续存储区可用数组实现。(2) 编译命令可用：gcc -lpthread -o 目标文件名源文件名(3) 多线程

2、编程方法参见附件。）详细代码#include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <errno.h> #include <pthread.h> #include <sys/ipc.h> #include <semaphore.h> #include <fcntl.h> #define n 5 time_t end_time; sem_t mutex, customers, barbers; int count = 0;

3、 int chair 5 = -1, -1, -1, -1, -1 ; void barber(void *arg) while (time (NULL) < end_time | count>0) while (count > 0) sem_wait(&customers); sem_wait(&mutex); count-; printf ("the barber is cutting hair, count is : %dn", count); sem_post(&mutex); sem_post(&barbers);

4、 sleep(3); void customer (void *arg) int i ,id= 0, num=0; while (time (NULL) < end_time) sem_wait(&mutex); if (count< n) count+; num= count % 5; num+; printf("customer entered:the customer %s comes in and sits at %d the chair count is: %dn", (char *)arg, num, count); sem_post(&am

5、p;mutex); sem_post(&customers); sem_wait(&barbers); else sem_post(&mutex); sleep(2); int main (int argc, char *argv) pthread_t id1, id2, id3, id4, id5, id6, id7, id8, id9, id10, id11; int ret= 0; int i; end_time = time (NULL) + 30; sem_init (&mutex, 0, 1); ret = sem_init (&barber

6、s, 0, 1); for (i =0;i<5;i+) chair i = -1; if (0!= ret) perror("sem init"); ret= pthread_create (&id11, NULL, (void*)barber,"id11"); if (0!= ret) perror("create barber"); ret = pthread_create ( &id1,NULL,(void*)customer, "id1"); if (0!= ret) perror(&

7、quot;create customers"); ret = pthread_create(&id2, NULL, (void*)customer, "id2"); if (0!=ret) perror("create customers"); ret = pthread_create(&id3, NULL, (void*)customer, "id3"); if (0!=ret) perror("create customers"); ret = pthread_create(&

8、id4, NULL, (void*)customer, "id4"); if (0!=ret) perror("create customers"); ret = pthread_create(&id5, NULL, (void*)customer, "id5"); if(0!=ret) perror("create custmers"); ret = pthread_create(&id6, NULL, (void*)customer, "id6"); if (0!= ret)

9、 perror("create customers"); ret = pthread_create ( &id7,NULL,(void*)customer, "id7"); if (0!= ret) perror("create customers"); ret = pthread_create(&id8, NULL, (void*)customer, "id8"); if (0!=ret) perror("create customers"); ret = pthread_cr

10、eate(&id9, NULL, (void*)customer, "id9"); if (0!=ret) perror("create customers"); ret = pthread_create(&id10, NULL, (void*)customer, "id10"); if (0!=ret) perror("create customers"); pthread_join(id1, NULL); pthread_join(id2, NULL); pthread_join(id3, NULL); pthread_join(id4, NULL); pthread_join(id5, NULL); pthrea