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

#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) < 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);