linux理發(fā)師多線程問題

1、用多線程同步方法解決睡眠理發(fā)師問題(Sleeping-Barber Problem)1操作系統(tǒng)： Linux2程序設(shè)計(jì)語言：C 語言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，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

2、名(3) 多線程編程方法參見附件。)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)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 c

4、omes in and sits 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);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

5、= 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!

6、= 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)perro

7、r(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 cus

8、tomers);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);

9、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(O);運(yùn)行結(jié)果如下CU5tffiflr曲GU5tff

10、Wr 141的祚innd14林5t7thq- CMlr側(cè)1Eustoncr snterDd: the匚ustoracr idl2匚 口 l!5andsitsatJthQchaiITcount15：2tu自ton打Wtiftd: tbi tudtonr ld3andat4thethalr19：3CUftmr entF9E|!：tlN SUfitWMr UMCflies 1ftandat乎thechtlrCWItIS；!custonar antirad! the custonar Ld5 the- &arber iicuttijnq hHlrPcauntcoins jnis : 4-nnr

11、lnt1thachfilrccjnr15!custctwr entered；the custmer ldl tlw hrlw itcutting hair. CQuntcoves in15：4andUt5at1hechairCW14rthp rnar antfirarl: the cnstonar LdS lhe barberis vuttljng halrPcaunE匚DIQS flis : 4-nnriit5nt1thachairVOUflF n:1custowr entered：the custcner ld3 tn bcirtiqr Itcutting何山 count tha-刼 hqis cuttning haire匚DUnt The* barb61 i tut Clhg hdlr , thebarber ishair, touetth hfirbar is cutting hi.countco鹽imIE : 4 is :步1A；215 : 1 1$ ! Bandsitsat1hechairceufiilSr5pig嗣Mg卯心蟲top:（史