用餐哲学家测试死锁

Question

我正在使用信号量在c中实现餐饮哲学家的问题。我的代码工作，虽然我很困惑为什么没有发生死锁。

CODE：

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

sem_t cs[5];
pthread_t phils[5];
void pickup(int i){


    int k = (i+1)%5;
        printf("%d: HUNGRY %d %d \n",i+1,i,k);
        sem_wait(&cs[i]);
        sem_wait(&cs[k]); 
}

void putdown(int i){
    sem_post(&cs[i]);
    sem_post(&cs[(i+1) % 5]);    
    printf("%d: THINK \n",i+1);
}

void eat(int i){
    printf("%d: EAT\n",i+1);
}

void phil(void *i){
    while(1){
        usleep((rand()% 100000) + 100000);
        int n= *(int*)i;
        pickup(n);
        eat(n);
        putdown(n); 

    }
}


int main(){
    int id[5];
    while(1){
        for(int i = 0; i<5;i++){
              sem_init(&cs[i], 0, 1);
              id[i]=i;
              pthread_create (&phils[i], NULL, (void *) &phil, (void *) &id[i]);
        }

        sleep(100);
    }   

}

OUTPUT的块：

2: HUNGRY 1 2 
2: EAT
2: THINK 
4: HUNGRY 3 4 
4: EAT
4: THINK 
5: HUNGRY 4 0 
5: EAT
5: THINK 
1: HUNGRY 0 1 
1: EAT
1: THINK 
3: HUNGRY 2 3 
3: EAT
3: THINK 
5: HUNGRY 4 0 
5: EAT
5: THINK 
4: HUNGRY 3 4 
4: EAT
4: THINK 
2: HUNGRY 1 2 
2: EAT
2: THINK 
1: HUNGRY 0 1 
1: EAT
1: THINK 
3: HUNGRY 2 3 
3: EAT
3: THINK 
2: HUNGRY 1 2 
2: EAT
2: THINK 
5: HUNGRY 4 0 
5: EAT
5: THINK 
4: HUNGRY 3 4 
4: EAT
4: THINK 
1: HUNGRY 0 1 
1: EAT
1: THINK 
2: HUNGRY 1 2 
2: EAT
2: THINK 
3: HUNGRY 2 3 
3: EAT
3: THINK 
4: HUNGRY 3 4 
4: EAT
4: THINK 
1: HUNGRY 0 1 
1: EAT
1: THINK 
5: HUNGRY 4 0 
5: EAT
5: THINK 
2: HUNGRY 1 2 
2: EAT
2: THINK 
3: HUNGRY 2 3 
3: EAT
3: THINK 
4: HUNGRY 3 4 
4: EAT
4: THINK 
1: HUNGRY 0 1 
1: EAT
1: THINK 
5: HUNGRY 4 0 
5: EAT
5: THINK 

我知道这是一件好事，但似乎我的线程一个接一个地执行而不是同时执行，这会破坏线程的点。

提前致谢！