寿司店问题 PV操作 pthread表示
寿司店问题。假设一个寿司店有 5 个座位，如果你到达的时候有一个空座位，你可以立刻就坐。但是如果你到达的时候 5 个座位都是满的有人已经就坐，这就意味着这些人都是一起来吃饭的，那么你需要等待所有的人一起离开才能就坐。编写同步原语，实现这个场景的约束。
1 #include <stdio.h> 2 #include <unistd.h> 3 #include <pthread.h> 4 #include <semaphore.h> 5 sem_t mutex; 6 sem_t allowin; 7 int waiting = 0; 8 int eating = 0; 9 int barrier = 0; 10 void *thread_a(void *in) 11 { 12 sem_wait(&mutex); 13 if (barrier) { 14 waiting++; 15 sem_post(&mutex); 16 sem_wait(&allowin); 17 } 18 else { 19 eating++; 20 barrier = (eating == 5); 21 sem_post(&mutex); 22 23 } 24 25 printf("No.%s takes seat and eats (person eating:%d)\n ", (char *)in, eating); 26 sleep(1); 27 sem_wait(&mutex); 28 eating--; 29 if (eating == 0) { 30 int n = (5 < waiting) ? 5 : waiting; 31 waiting -= n; 32 eating += n; 33 barrier = (eating == 5); 34 for (int i = 0; i < n; i++) 35 sem_post(&allowin); 36 } 37 sem_post(&mutex); 38 printf("No.%s leaves\n", (char *)in); 39 pthread_exit((void *)0); 40 } 41 int main() 42 { 43 const int maxn = 10; 44 pthread_t a[maxn]; /* thread id a, b, c*/ 45 char *strs[] = {"a", "b", "c", "d", "e", "f", "g", "h", "i", "j"}; 46 int val; /* used for function return result */ 47 /* init sem1 sem2 to 0 , any thread waits for it will be blocked*/ 48 sem_init(&mutex, 0, 1); 49 sem_init(&allowin, 0, 1); 50 for (int i = 0; i < maxn; i++) 51 pthread_create(&a[i], NULL, thread_a, (void *)strs[i]); 52 for (int i = 0; i < maxn; i++) 53 pthread_join(a[i], (void **)0); 54 sem_destroy(&mutex); 55 sem_destroy(&allowin); 56 57 printf("Main thread is over\n"); 58 return 0; 59 }
