c - 餐饮哲学家计划C

我正在处理5位哲学家和5条筷子的经典用餐哲学家问题。我的作业是使用1个互斥锁和5个条件。我让它工作了，但我不知道为什么哲学家1从来不吃东西，而是4,3,0吃和2吃两次。这是我的代码（很抱歉，长度）：

//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);
}

最佳答案

您应该#define NUM 5而不是#define NUM 7。这导致您在put_chopsticks()和test()中的模块化算术是错误的，可能还有其他奇怪的事情。

在传递给哲学家线程的int中，您似乎也有一个竞争条件，因为它们都指向i的相同索引地址（并且您也应该在此处收到有关错误类型的编译器警告）。

至此，我已经停止了，因为您在这里遇到的问题太多了，这些问题才刚刚开始。

关于c - 餐饮哲学家计划C，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/23529773/