linux理发师多线程问题

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： Linux
2 ．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 <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;
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 : %d\n", 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: %d\n", (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;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("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_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。