餐飲哲學家計劃C

Question

我正在與5個哲學家和5個筷子的經典餐飲哲學家問題。我的作業是使用1個互斥體和5個條件。我得到它的工作，但我不知道爲什麼哲學家1從不吃，但4,3,0吃，2吃兩次。這裏是我的代碼（遺憾的長度）：

//To compile: gcc -pthread -o monitor monitor.c 
//To execute: ./monitor "an integer as the amount of food (default = 5)" 

#include <pthread.h> 
#include <stdio.h> 
#include <stdlib.h> 
// There are 5 philosophers (0,1,2,3,4) and 5 chopsticks (0,1,2,3,4) 
#define NUM 5 

// States of a philosopher 
#define THINKING 0 
#define HUNGRY 1 
#define EATING 2 

int food = 5; //The total amount of food on the table. 
int monitor = HUNGRY; 
int state[NUM]; //The state of a particular philosopher during the simulation 

//The mutex for monitor variable and the condition resource_ready 
pthread_mutex_t monitor_mutex; 
pthread_cond_t resource_ready; 
//The philosopher thread identifiers 
pthread_t philo[NUM]; 
void *philosopher (void *num); 
void get_chopsticks (int, int, int); 
void test (int); 
void put_chopsticks (int, int, int); 

int main (int argc, char **argv) 
{ 
    if (argc == 2) 
     food = atoi (argv[1]); 
    int i; 

    pthread_mutex_init (&monitor_mutex, NULL); 
    pthread_cond_init (&resource_ready, NULL); 

    //Create a thread for each philosopher 
    for (i = 0; i < NUM; i++) 
     pthread_create (&philo[i], NULL, philosopher, (void *)i); 

    //Wait for the threads to exit 
    for (i = 0; i < NUM; i++) 
     pthread_join (philo[i], NULL); 
    return 0; 
} 
void *philosopher (void *num) 
{ 
    int id; 
    int i, left_chopstick, right_chopstick, f; 

    id = (int)num; //This is the philosopher's id 

    state[id] = THINKING; 
    printf ("Philosopher %d is thinking.\n", id); 

    //Identify the philosopher's right and left chopstick 
    right_chopstick = id; 
    left_chopstick = (id + 1)%NUM; 

    //While there is still food on the table, the loop goes on 
    while (food > 0) 
    { 
     //Get the chopstick 
     get_chopsticks (id, left_chopstick, right_chopstick); 
     sleep(1);  
    } 
    return (NULL); 
} 
void get_chopsticks (int phil, int c1, int c2) 
{ 
    //Lock monitor variable 
    pthread_mutex_lock (&monitor_mutex);  
    state[phil] = monitor;//state[phil] = HUNGRY 
    test(phil); 
    if (state[phil] != EATING) 
    { 
     pthread_cond_wait (&resource_ready, &monitor_mutex); 
     printf ("Philosopher %d is now eating\n", phil); 
     food--; //1 food is eaten 
     if (food == 0) 
     { 
       printf ("There is no more food on the table. Program exit\n"); 
       pthread_mutex_destroy(&monitor_mutex); 
       pthread_cond_destroy(&resource_ready); 
       exit(1); 
      } 
     printf ("There are/is %d food(s) left\n", food); 
     put_chopsticks (phil, c1, c2); 
     sleep(1);  
    } 
    //Unlock monitor variable 
    pthread_mutex_unlock (&monitor_mutex); 
} 

void test (int phil) 
{ 
    //If the philosopher is hungry and the nearest 2 are not eating, he then eats. 
    if ((state[(phil+4)%NUM] != EATING)&& (state[phil] == HUNGRY)) 
    { 
     state[phil] = EATING; 
     pthread_cond_signal (&resource_ready); 
    } 
} 
void put_chopsticks (int phil, int c1, int c2) 
{ 
    state[phil] = THINKING; 
    //Check the 2 nearest philosophers to see if they are eating 
    test ((phil+4)%NUM); 
    test ((phil+1)%NUM); 
} 

Answer 1

的#define NUM 7你應該#define NUM 5代替。這導致您的模塊算術put_chopsticks()和test()是錯誤的，可能是其他奇怪的事情。因爲它們都指向相同的索引地址i（並且你也應該在這裏得到關於錯誤類型的編譯器警告）。在你傳遞給哲學家線程的int中你似乎也有一個競爭條件，因爲它們都指向相同的索引地址i。

我已經停止在這一點，因爲你有太多的問題在這裏，那些只是開始。

Answer 2

右筷子的位置與哲學家的位置相同。所以根據你的代碼，第一哲學家將永遠不會擁有第五哲學家的筷子（這與他們坐在圓桌上的觀點相矛盾）。

所以乾脆試試這個在您的<void *philosopher (void *num)>功能

right_chopstick = (id+1)%NUM ; 
left_chopstick = (id+(NUM-1))%NUM; 

我希望這可以讓你的第一個哲學家吃

&使所有的哲學家吃，嘗試當他們改變自己的狀態，以引進的睡眠時間從思考到食用&反之亦然。