哲学家就餐问题

来源：互联网发布：网络专升本编辑：程序博客网时间：2024/04/30 18:42

五个哲学家围坐在一张圆桌周围，每个哲学家面前都有一盘通心粉。由于通心粉很滑，所以需要两把叉子才能夹住。相邻两个盘子之间放有一把叉子。能为每一个哲学家写一段描述其行为的程序，且绝不会死锁吗？

#include <pthread.h>#include <stdio.h>#define N 5#define LEFT (i+N-1)%N#define RIGHT (i+1)%N#define THINK_TIME 3#define EAT_TIME 2enum { THINKING, HUNGRY, EATING } state[N];pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER, s[N];void test(int i){  if (state[i] == HUNGRY     && state[LEFT] != EATING      && state[RIGHT] != EATING)    {      state[i] = EATING;      pthread_mutex_unlock(&s[i]);     }}void take_forks(int i){  pthread_mutex_lock(&mutex);  state[i] = HUNGRY;  test(i);                  //尝试获取2把叉子  pthread_mutex_unlock(&mutex);  pthread_mutex_lock(&s[i]); //得不到需要就阻塞}void put_forks(int i){  pthread_mutex_lock(&mutex);  state[i] = THINKING;  test(LEFT);              //检查左边的邻居现在可不可以吃  test(RIGHT);             //检查右边的邻居现在可不可以吃  pthread_mutex_unlock(&mutex);}void think(int i){  printf("philosopher %d is thinking...\n", i);  sleep(THINK_TIME);}void eat(int i){  printf("philosopher %d is eating...\n", i);  sleep(EAT_TIME);}void* phi(void* vargp){  int i = *(int*)vargp;  while (1)    {      think(i);      take_forks(i);      eat(i);      put_forks(i);    }  return NULL;}int main(){  int i;  pthread_t tid[N];  for (i = 0; i < N; i++)    pthread_create(&tid[i], NULL, phi, (void*)(&i));  for (i = 0; i < N; i++)    pthread_join(tid[i], NULL);  return 0;}

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