哲學家用餐

Question

所以我試圖解決哲學家就餐問題與一些信號燈..當我編譯它我得到哲學家1-5思考，phisopher 1是餓他叉叉 5和1，哲學家3餓了，然後該程序停在那裏......我不知道這個問題。

#include<stdio.h> 
#include<semaphore.h> 
#include<pthread.h> 

#define N 5 
#define THINKING 0 
#define HUNGRY 1 
#define EATING 2 
#define LEFT (i+4)%N 
#define RIGHT (i+1)%N 


sem_t mutex; 
sem_t S[N]; 
void * philospher(void *num); 
void take_fork(int); 
void put_fork(int); 
void test(int); 

int state[N]; 
int phil_num[N]={0,1,2,3,4}; 

int main() 
{ 
    int i; 
    pthread_t thread_id[N]; 
    sem_init(&mutex,0,1); 
    for(i=0;i<N;i++) 
     sem_init(&S[i],0,1); 
    for(i=0;i<N;i++) 
    { 
     pthread_create(&thread_id[i],NULL,philospher,(void *)i); 
     printf("Philosopher %d is thinking\n",i+1); 
    } 
    for(i=0;i<N;i++) 
     pthread_join(thread_id[i],NULL); 
} 


void *philospher(void *num) 
{ 
    while(1) 
    { 
     int i; 
     i = (int)num; 
     sleep(1); 
     //take_fork(i); 


    sem_wait(&mutex); 
    state[i] = HUNGRY; 
    printf("Philosopher %d is Hungry\n",i+1); 
    //test(i); 


    if (state[i] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING) 
    { 
     sem_wait(&S[LEFT]); 
     sem_wait(&S[RIGHT]); 
     state[i] = EATING; 
     sleep(2); 
     printf("Philosopher %d takes fork %d and %d\n",i+1,LEFT+1,i+1); 
     printf("Philosopher %d is Eating\n",i+1); 
    } 

    sem_post(&mutex); 
    sleep(1); 


     //put_fork(*i); 


    sem_wait(&mutex); 
    state[i] = THINKING; 
    printf("Philosopher %d putting fork %d and %d down\n",i+1,LEFT+1,i+1); 
    printf("Philosopher %d is thinking\n",i+1); 
    sem_post(&S[LEFT]); 
    sem_post(&S[RIGHT]); 
    //test(LEFT); 
    //test(RIGHT); 
    sem_post(&mutex); 

    } 
} 

Answer 1

我認爲這是一種競爭條件。

您在worker函數中指的是state[RIGHT]，但不能保證所有線程都已創建，因此state數組在此時被初始化。