使用 pthreads、互斥锁和条件变量解决 Dining Philosophers Problem

Question

我正在尝试使用 pthreads、互斥锁和条件变量在 C 中实现哲学家就餐问题。

它需要一个命令行参数来指定程序应该运行多长时间运行。我必须使用睡眠功能来完成这个。
每个哲学家最多可以吃10顿饭。一旦他们吃到 10 顿饭，pthread 应该终止。
在设定的时间结束时，pthreads 需要终止，并且应该打印每个哲学家吃的饭数。

我的输出有一些问题：

让主函数休眠命令行输入的秒数似乎不会使输出不同。
大多数哲学家都在执行大部分程序时挨饿。
当我打印出一个哲学家正在思考或吃饭时，一个 'philosopher 5' 出现了，即使应该只有 0-4 个哲学家。

这是我的代码：

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

//Function declarations
void *pickup_forks(void * philosopher_number);
void *return_forks(void * philosopher_number);
void test(int philosopher_number);
int left_neighbor(int philosopher_number);
int right_neighbor(int philosopher_number);
double think_eat_time(void);
void think(double think_time);
void eat(double eat_time);

//Constants to be used in the program.
#define PHILOSOPHER_NUM 5
#define MAX_MEALS 10
#define MAX_THINK_EAT_SEC 3

//States of philosophers.
enum {THINKING, HUNGRY, EATING} state[PHILOSOPHER_NUM];

//Array to hold the thread identifiers.
pthread_t philos_thread_ids[PHILOSOPHER_NUM];

//Mutex lock.
pthread_mutex_t mutex;

//Condition variables.
pthread_cond_t cond_vars[PHILOSOPHER_NUM];

//Array to hold the number of meals eaten for each philosopher.
int meals_eaten[PHILOSOPHER_NUM];

int main(int argc, char *argv[])
{
    //Ensure correct number of command line arguments.
    if(argc != 2)
    {
        printf("Please ensure that the command line argument 'run_time' is passed.\n");
    }
    else
    {
        //Set command line argument value to variable run_time;
        double run_time = atof(argv[1]);

        //Initialize arrays.
        int i;
        for(i = 0; i < PHILOSOPHER_NUM; i++)
        {
            state[i] = THINKING;
            pthread_cond_init(&cond_vars[i], NULL);
            meals_eaten[i] = 0;
        }


        //Initialize the mutex lock.
        pthread_mutex_init(&mutex, NULL);

        //Join the threads.
        for(i = 0; i < PHILOSOPHER_NUM; i++)
        {
            pthread_join(philos_thread_ids[i], NULL);
        }

        //Create threads for the philosophers.
        for(i = 0; i < PHILOSOPHER_NUM; i++)
        {
            pthread_create(&philos_thread_ids[i], NULL, pickup_forks, (void *)&i);
        }

        sleep(run_time);

        for(i = 0; i < PHILOSOPHER_NUM; i++)
        {
            pthread_cancel(philos_thread_ids[i]);
        }

        //Print the number of meals that each philosopher ate.
        for(i = 0; i < PHILOSOPHER_NUM; i++)
        {
            printf("Philosopher %d: %d meals\n", i, meals_eaten[i]);
        }

    }

    return 0;
}

void *pickup_forks(void * philosopher_number)
{
    int loop_iterations = 0;
    int pnum = *(int *)philosopher_number;

    while(meals_eaten[pnum] < MAX_MEALS)
    {
        printf("Philosoper %d is thinking.\n", pnum);
        think(think_eat_time());

        pthread_mutex_lock(&mutex);
        state[pnum] = HUNGRY;
        test(pnum);

        while(state[pnum] != EATING)
        {
            pthread_cond_wait(&cond_vars[pnum], &mutex);
        }
        pthread_mutex_unlock(&mutex);

        (meals_eaten[pnum])++;

        printf("Philosoper %d is eating meal %d.\n", pnum, meals_eaten[pnum]);
        eat(think_eat_time());

        return_forks((philosopher_number));

        loop_iterations++;
    }
}

void *return_forks(void * philosopher_number)
{
    pthread_mutex_lock(&mutex);
    int pnum = *(int *)philosopher_number;

    state[pnum] = THINKING;

    test(left_neighbor(pnum));
    test(right_neighbor(pnum));

    pthread_mutex_unlock(&mutex);
}

int left_neighbor(int philosopher_number)
{
    return ((philosopher_number + (PHILOSOPHER_NUM - 1)) % 5);
}

int right_neighbor(int philosopher_number)
{
    return ((philosopher_number + 1) % 5);
}

void test(int philosopher_number)
{
    if((state[left_neighbor(philosopher_number)] != EATING) && 
        (state[philosopher_number] == HUNGRY) &&
        (state[right_neighbor(philosopher_number)] != EATING))
    {
        state[philosopher_number] = EATING;
        pthread_cond_signal(&cond_vars[philosopher_number]);
    }
}

double think_eat_time(void)
{
    return ((double)rand() * (MAX_THINK_EAT_SEC - 1)) / (double)RAND_MAX + 1;
}

void think(double think_time)
{
    sleep(think_time);
}

void eat(double eat_time)
{
    sleep(eat_time);
}

这是使用 10 秒的输出：

~$ gcc dining_philos.c -o dp -lpthread
~$ ./dp 10
Philosoper 1 is thinking.
Philosoper 2 is thinking.
Philosoper 3 is thinking.
Philosoper 4 is thinking.
Philosoper 5 is thinking.
Philosoper 2 is eating meal 1.
Philosoper 4 is eating meal 1.
Philosoper 2 is thinking.
Philosoper 4 is thinking.
Philosoper 2 is eating meal 2.
Philosoper 4 is eating meal 4.
Philosoper 4 is thinking.
Philosoper 2 is thinking.
Philosoper 2 is eating meal 3.
Philosoper 4 is eating meal 5.
Philosoper 2 is thinking.
Philosoper 4 is thinking.

Philosopher 0: 2 meals
Philosopher 1: 0 meals
Philosopher 2: 3 meals
Philosopher 3: 0 meals
Philosopher 4: 5 meals

如有任何帮助，我将不胜感激。谢谢！

Answer 1

pthread_create 传递一个指针作为它的最后一个参数。该指针必须在线程运行时有效，而不仅仅是在创建线程时有效。您正在使用指向循环变量 i 的指针，它已经消失了。

也不确定你为什么在 pthread_create 之前调用 pthread_join，但我预计它会返回错误（你没有检查）。

使用 pthreads、互斥锁和条件变量解决 Dining Philosophers Problem

Solve Dining Philosophers Problem Using pthreads, mutex locks, and condition variables

c

mutex

pthreads

condition-variable

dining-philosopher