哲学家就餐 c语言代码

以下是一个用C语言实现的哲学家就餐问题的代码示例：

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

#define NUM_PHILOSOPHERS 5
#define THINKING 0
#define HUNGRY 1
#define EATING 2

int state[NUM_PHILOSOPHERS];
pthread_mutex_t mutex;
pthread_mutex_t s[NUM_PHILOSOPHERS];

void* philosopher(void* num);
void take_forks(int);
void put_forks(int);
void test(int);
void think();
void eat();

int main() {
    int i;
    pthread_t thread_id[NUM_PHILOSOPHERS];
    
    pthread_mutex_init(&mutex, NULL);
    
    for (i = 0; i < NUM_PHILOSOPHERS; i++)
        pthread_mutex_init(&s[i], NULL);
    
    for (i = 0; i < NUM_PHILOSOPHERS; i++)
        pthread_create(&thread_id[i], NULL, philosopher, (void*)i);
    
    for (i = 0; i < NUM_PHILOSOPHERS; i++)
        pthread_join(thread_id[i], NULL);
    
    for (i = 0; i < NUM_PHILOSOPHERS; i++)
        pthread_mutex_destroy(&s[i]);
    
    pthread_mutex_destroy(&mutex);
    
    return 0;
}

void* philosopher(void* num) {
    int id = (int)num;
    
    while (1) {
        think(id);
        take_forks(id);
        eat(id);
        put_forks(id);
    }
}

void take_forks(int id) {
    pthread_mutex_lock(&mutex);
    state[id] = HUNGRY;
    printf("Philosopher %d is hungry.\n", id);
    test(id);
    pthread_mutex_unlock(&mutex);
    pthread_mutex_lock(&s[id]);
}

void put_forks(int id) {
    pthread_mutex_lock(&mutex);
    state[id] = THINKING;
    printf("Philosopher %d is thinking.\n", id);
    test((id + 4) % NUM_PHILOSOPHERS);
    test((id + 1) % NUM_PHILOSOPHERS);
    pthread_mutex_unlock(&mutex);
}

void test(int id) {
    if (state[id] == HUNGRY && state[(id + 4) % NUM_PHILOSOPHERS] != EATING && state[(id + 1) % NUM_PHILOSOPHERS] != EATING) {
        state[id] = EATING;
        printf("Philosopher %d is eating.\n", id);
        pthread_mutex_unlock(&s[id]);
    }
}

void think(int id) {
    printf("Philosopher %d is thinking.\n", id);
    sleep(rand() % 3 + 1);
}

void eat(int id) {
    printf("Philosopher %d is eating.\n", id);
    sleep(rand() % 3 + 1);
}

这个代码实现了经典的哲学家就餐问题。每个哲学家都被模拟为一个线程，共享的筷子用互斥锁表示。代码使用了以下关键函数：

1. `philosopher`：每个哲学家的主循环，包含思考、取筷子、吃饭和放筷子。
2. `take_forks`：哲学家尝试拿起两边的筷子。
3. `put_forks`：哲学家放下筷子，并尝试唤醒左右两边的哲学家。
4. `test`：检查哲学家是否可以开始吃饭。
5. `think`和`eat`：模拟哲学家思考和吃饭的时间。