linux理發(fā)師多線程問(wèn)題

1、精品文檔用多線程同步方法解決睡眠理發(fā)師問(wèn)題（Sleeping-Barber Problem)1 操作系統(tǒng)：Linux2 程序設(shè)計(jì)語(yǔ)言： C語(yǔ)言3 設(shè)有 1個(gè)理發(fā)師 5把椅子（另外還有一把理發(fā)椅），幾把椅子可用連續(xù)存儲(chǔ)單元 .1技術(shù)要求：1）為每個(gè)理發(fā)師顧客產(chǎn)生一個(gè)線程，設(shè)計(jì)正確的同步算法2）每個(gè)顧客進(jìn)入理發(fā)室后，即時(shí)顯示“ Entered ” 及其線程自定義標(biāo)識(shí)，還同時(shí)顯示理發(fā)室共有幾名顧客及其所坐的位置。3）至少有 10 個(gè)顧客，每人理發(fā)至少 3秒鐘。4）多個(gè)顧客須共享操作函數(shù)代碼。提示 :(1) 連續(xù)存儲(chǔ)區(qū)可用數(shù)組實(shí)現(xiàn)。(2)編譯命令可用：gcc -lpthread -o目標(biāo)文件名源文件名

2、(3) 多線程編程方法參見(jiàn)附件。）詳細(xì)代碼#include #include #include #include #include #include #include #include #define n 5time_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)。1歡迎下載精品文檔sem_wait(&customers);sem_wait(&m

3、utex);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 s

4、its at %d the chair count is:%dn, (char *)arg, num, count);sem_post(&mutex);sem_post(&customers);sem_wait(&barbers);elsesem_post(&mutex);sleep(2);。2歡迎下載精品文檔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;se

5、m_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 = pt

6、hread_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);。3歡迎下載精品文檔if (0!=ret)perror(create customers);ret = pthread_create(&id5

7、, 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, id

8、8);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