linux理发师多线程问题.doc

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) 多线程编程方法参见附件。）详细代码#include #include #include #include #include #include #include #include #define n 5 time_t end_time; sem_t mutex, customers, barbers; int count = 0; int chair 5 = -1, -1, -1, -1, -1 ; void barber(void *arg) while (time (NULL) 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); 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(&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 (&barbers, 0, 1); for (i =0;i5;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(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(&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) 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_create(&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); pthread_join(id6, NULL); pthread_jo