在 Dining Philosophers 上使用 Pthreads 显示死锁

Question

我们有一项任务是展示哲学家就餐问题的僵局。我们已经编写了所有代码并且代码可以编译，但是当运行代码时，其中一位哲学家最终吃掉了。那么这是否意味着实际上并没有发生死锁？

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

#define number_philo 5


pthread_mutex_t cutlery[number_philo];


void* philosopher (void* number)

{

    int my_num = *((int*)number);

    while (1)

    {


        printf("\n Philosopher %d is thinking.\n",my_num);


        pthread_mutex_lock (&cutlery[my_num]);
        printf("\nPhilosopher %d has left cutlery. \n",my_num);
        sleep(3);
        pthread_mutex_lock (&cutlery[(my_num + 1) %  number_philo]);
        printf("\nPhilosopher %d has right cutlery. \n",my_num);



        printf("\n Philosopher %d eating.\n", my_num);


        printf("\n Philosopher %d done.\n", my_num);


        pthread_mutex_unlock (&cutlery[(my_num + 1) % number_philo]);

        pthread_mutex_unlock (&cutlery[my_num]);
        printf("\nPhilosopher %d no longer has cutlery.\n", my_num);
    }

    return NULL;

}


int main ()

{

    int i;

    pthread_t phils[number_philo];

    void* return_val;


    for (i = 0; i < number_philo; i++)

        pthread_mutex_init (&cutlery[i], NULL);


    for (i = 0; i < number_philo; i++)

        pthread_create (&phils[i], NULL, philosopher, &i);


    for (i = 0; i < number_philo; i++)

        pthread_join (phils[i], &return_val);


    return 0;

}

这是输出：output

Answer 1

问题出在这里：

  pthread_create (&phils[i], NULL, philosopher, &i);

您正在将指向同一变量 i 的指针传递给每个线程，而这些线程（和主线程）都在以一种不正当的方式访问 i。你会得到多个哲学家使用相同的数字和一些根本没有使用的数字。

你需要给每个哲学家它自己的变量来读取，或者等待它读取 i，然后再循环并改变 i。前者的例子：

int phil_num[number_philo];

/* ... */

for (i = 0; i < number_philo; i++)
{
    phil_num[i] = i;
    pthread_create (&phils[i], NULL, philosopher, &phil_num[i]);
}

在 Dining Philosophers 上使用 Pthreads 显示死锁

Showing deadlock, using Pthreads, on Dining Philosophers

c

posix

pthreads

dining-philosopher