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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
2、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( sem_wait( count-; printf (the
3、 barber is cutting hair, count is : %dn, count); sem_post( sem_post( sleep(3); void customer (void *arg) int i ,id= 0, num=0; while (time (NULL) end_time) sem_wait( 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, (ch
4、ar *)arg, num, count); sem_post( sem_post( sem_wait( else sem_post( 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 ( ret = sem_init ( for (i =0;i5;i+) chair i = -1; if (0!= ret) per
5、ror(sem init); ret= pthread_create ( if (0!= ret) perror(create barber); ret = pthread_create ( if (0!= ret) perror(create customers); ret = pthread_create( if (0!=ret) perror(create customers); ret = pthread_create( if (0!=ret) perror(create customers); ret = pthread_create( if (0!=ret) perror(crea
6、te customers); ret = pthread_create( if(0!=ret) perror(create custmers); ret = pthread_create( if (0!= ret) perror(create customers); ret = pthread_create ( if (0!= ret) perror(create customers); ret = pthread_create( if (0!=ret) perror(create customers); ret = pthread_create( if (0!=ret) perror(cre
7、ate customers); ret = pthread_create( 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_join(id7, NULL); pthread_join(id8, NULL); pthread_join(id9, NULL); pthread_join(id10, NULL); pthread_join(id11, NULL); exit(0); 运行结果如下
