这是使用信号量解决geeksforgeeks餐饮哲学家问题的解决方案：

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

#define N 5 
#define THINKING 2 
#define HUNGRY 1 
#define EATING 0 
#define LEFT (phnum + 4) % N 
#define RIGHT (phnum + 1) % N 

int state[N]; 
int phil[N] = { 0,1,2,3,4 }; 

sem_t mutex; 
sem_t S[N]; 

void test(int phnum) 
{ 
    if (state[phnum] == HUNGRY 
        && state[LEFT] != EATING 
        && state[RIGHT] != EATING) { 
        // state that eating 
        state[phnum] = EATING; 

        sleep(2); 

        printf("Philosopher %d takes fork %d and %d\n",phnum + 1,LEFT + 1,phnum + 1); 

        printf("Philosopher %d is Eating\n",phnum + 1); 

        // sem_post(&S[phnum]) has no effect 
        // during takefork 
        // used to wake up hungry philosophers 
        // during putfork 
        sem_post(&S[phnum]); 
    } 
} 

// take up chopsticks 
void take_fork(int phnum) 
{ 

    sem_wait(&mutex); 

    // state that hungry 
    state[phnum] = HUNGRY; 

    printf("Philosopher %d is Hungry\n",phnum + 1); 

    // eat if neighbours are not eating 
    test(phnum); 

    sem_post(&mutex); 

    // if unable to eat wait to be signalled 
    sem_wait(&S[phnum]); 

    sleep(1); 
} 

// put down chopsticks 
void put_fork(int phnum) 
{ 

    sem_wait(&mutex); 

    // state that thinking 
    state[phnum] = THINKING; 

    printf("Philosopher %d putting fork %d and %d down\n",phnum + 1); 
    printf("Philosopher %d is thinking\n",phnum + 1); 

    test(LEFT); 
    test(RIGHT); 

    sem_post(&mutex); 
} 

void* philospher(void* num) 
{ 

    while (1) { 

        int* i = num; 

        sleep(1); 

        take_fork(*i); 

        sleep(0); 

        put_fork(*i); 
    } 
} 

int main() 
{ 

    int i; 
    pthread_t thread_id[N]; 

    // initialize the mutexes 
    sem_init(&mutex,1); 

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

        sem_init(&S[i],0); 

    for (i = 0; i < N; i++) { 

        // create philosopher processes 
        pthread_create(&thread_id[i],NULL,philospher,&phil[i]); 

        printf("Philosopher %d is thinking\n",i + 1); 
    } 

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

        pthread_join(thread_id[i],NULL); 
} 

https://www.geeksforgeeks.org/dining-philosopher-problem-using-semaphores/

此代码死锁和死机的可能性很小， 我想更改它，使其极有可能出现死锁，活锁或饥饿， 我怎样才能做到这一点？

还有我如何确保该解决方案在100％的情况下（如果可能）不会出现任何这些问题