哲学家就餐问题

问题

在一个圆桌上坐着五位哲学家，每个哲学家面前有一个碗装有米饭的碗和一个筷子。哲学家的生活包括思考和进餐两个活动。当一个哲学家思考时，他不需要任何资源。当他饿了时，他试图拿起两只相邻的筷子来吃饭。由于碗和筷子不能被共享，因此必须找到一种方法来确保哲学家能够安全地进餐，即不会发生死锁（所有哲学家都在等待筷子）也不会发生饥饿（某些哲学家永远无法拿起筷子）

信号量实现

发生死锁版

#include <pthread.h>
#include <semaphore.h>
#include <vector>
#include <unistd.h>
#include <iostream>

#define N 5
#define P(x) sem_wait(x)
#define V(x) sem_post(x)

sem_t mutex_table;
std::vector<sem_t> mutex_chopsticks(N);


void* T_philosopher(void* arg) {
    int id = *((int*)arg);

    // id哲学家吃饭需要的2根筷子
    int lhs = (id + N - 1) % N;
    int rhs = id % N;

    while (true) {
        P(&mutex_chopsticks[lhs]);
        printf("+ %d by T%d\n", lhs, id);
        P(&mutex_chopsticks[rhs]);
        printf("+ %d by T%d\n", rhs, id);

        // Eat.
        // Philosophers are allowed to eat in parallel.

        printf("- %d by T%d\n", lhs, id);
        printf("- %d by T%d\n", rhs, id);
        V(&mutex_chopsticks[lhs]);
        V(&mutex_chopsticks[rhs]);

        sleep(0.5);
    }
}

int main() {
    for (int i = 0; i < N; i++) {
        sem_init(&mutex_chopsticks[i], 0, 1);
    }
    std::vector<pthread_t> threads(N);
    std::vector<int> ids(N);
    for (int i = 0; i < N; i++) {
        ids[i] = i + 1;
        pthread_create(&threads[i], nullptr, &T_philosopher, &ids[i]);
    }
    for (int i = 0; i < N; i++) {
        pthread_join(threads[i], nullptr);
    }
}

限制人数版

限制最多只能4人同时上桌, 则可以保证至少有一个人可以同时拿起两根筷子

#include <pthread.h>
#include <semaphore.h>
#include <vector>
#include <unistd.h>
#include <iostream>

#define P(x) sem_wait(x)
#define V(x) sem_post(x)

const int N = 5;

sem_t mutex_table;
std::vector<sem_t> mutex_chopsticks(N);


void* T_philosopher(void* arg) {
    int id = *((int*)arg);

    // id哲学家吃饭需要的2根筷子
    int lhs = (id + N - 1) % N;
    int rhs = id % N;

    while (true) {
        // Come to mutex_table
        P(&mutex_table);
        P(&mutex_chopsticks[lhs]);
        printf("+ %d by T%d\n", lhs, id);
        P(&mutex_chopsticks[rhs]);
        printf("+ %d by T%d\n", rhs, id);

        // Eating
        // Philosophers are allowed to eat in parallel.

        printf("- %d by T%d\n", lhs, id);
        printf("- %d by T%d\n", rhs, id);
        V(&mutex_chopsticks[lhs]);
        V(&mutex_chopsticks[rhs]);

        // Leave mutex_table
        V(&mutex_table);
        // Thinking
        sleep(0.5);
    }
}

int main() {
    // 保证任何实际最多只有4个人在桌子上
    // 这样至少有1个人可以拿到2根筷子
    sem_init(&mutex_table, 0, N - 1);

    for (int i = 0; i < N; i++) {
        sem_init(&mutex_chopsticks[i], 0, 1);
    }
    std::vector<pthread_t> threads(N);
    std::vector<int> ids(N);
    for (int i = 0; i < N; i++) {
        ids[i] = i + 1;
        pthread_create(&threads[i], nullptr, &T_philosopher, &ids[i]);
    }
    for (int i = 0; i < N; i++) {
        pthread_join(threads[i], nullptr);
    }
}

规定取筷顺序

规定每个人, 先拿编号小的筷子, 再拿编号大的筷子

#include <pthread.h>
#include <semaphore.h>
#include <vector>
#include <unistd.h>
#include <iostream>

#define P(x) sem_wait(x)
#define V(x) sem_post(x)

const int N = 5;

sem_t mutex_table;
std::vector<sem_t> mutex_chopsticks(N);


void* T_philosopher(void* arg) {
    int id = *((int*)arg);

    // id哲学家吃饭需要的2根筷子
    int lhs = (id + N - 1) % N;
    int rhs = id % N;

    while (true) {
        // 规定每个人, 先拿编号小的筷子, 再拿编号大的筷子
        if (lhs < rhs) {
            P(&mutex_chopsticks[lhs]);
            P(&mutex_chopsticks[rhs]);
        } else {
            P(&mutex_chopsticks[rhs]);
            P(&mutex_chopsticks[lhs]);
        }
        printf("+ %d by T%d\n", lhs, id);
        printf("+ %d by T%d\n", rhs, id);

        // Eating

        printf("- %d by T%d\n", lhs, id);
        printf("- %d by T%d\n", rhs, id);
        V(&mutex_chopsticks[lhs]);
        V(&mutex_chopsticks[rhs]);

        // Thinking
        sleep(0.5);
    }
}

int main() {
    for (int i = 0; i < N; i++) {
        sem_init(&mutex_chopsticks[i], 0, 1);
    }
    std::vector<pthread_t> threads(N);
    std::vector<int> ids(N);
    for (int i = 0; i < N; i++) {
        ids[i] = i + 1;
        pthread_create(&threads[i], nullptr, &T_philosopher, &ids[i]);
    }
    for (int i = 0; i < N; i++) {
        pthread_join(threads[i], nullptr);
    }
}

条件变量实现

#include <pthread.h>
#include <vector>
#include <unistd.h>
#include <iostream>

#define Lock(x) pthread_mutex_lock(x)
#define UnLock(x) pthread_mutex_unlock(x)

const int N = 5;

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
std::vector<int> available(N, true);

void* T_philosopher(void* arg) {
    int id = *((int*)arg);

    // id哲学家吃饭需要的2根筷子
    int lhs = (id + N - 1) % N;
    int rhs = id % N;

    while (true) {
        // Come to mutex_table
        Lock(&mutex);
        while (!(available[lhs] && available[rhs])) {
            pthread_cond_wait(&cond, &mutex);
        }
        printf("+ %d by T%d\n", lhs, id);
        printf("+ %d by T%d\n", rhs, id);
        available[lhs] = available[rhs] = false;
        UnLock(&mutex);

        
        
        // Eating
        sleep(0.5);

        printf("- %d by T%d\n", lhs, id);
        printf("- %d by T%d\n", rhs, id);
        available[lhs] = available[rhs] = true;
        pthread_cond_broadcast(&cond);

        // Thinking
        sleep(0.5);
    }
}

int main() {
    std::vector<pthread_t> threads(N);
    std::vector<int> ids(N);
    for (int i = 0; i < N; i++) {
        ids[i] = i + 1;
        pthread_create(&threads[i], nullptr, &T_philosopher, &ids[i]);
    }
    for (int i = 0; i < N; i++) {
        pthread_join(threads[i], nullptr);
    }
}