linux 哲学家进餐问题 奇数偶数号科学家

规定奇数号哲学家先去拿他左边的筷子，然后再去拿右边的筷子；偶数号哲学家则相反。按此规定，将是1，2号哲学家竞争1号筷子；3，4号哲学家竞争3号筷子。即五位哲学家都先竞争奇数号筷子，获得后，再去竞争偶数号筷子，最后总有一位哲学家能够获得两只筷子而进餐。#include<iostream>#include<stdio.h>#include<stdlib.h>#include<pthread.h>#include<unistd.h>#include<semaphore.h>#define N 5#define LEFT i#define RIGHT (i+1)%Nusing namespace std;sem_t mutex[N];int id[N]={0,1,2,3,4};pthread_t thread[N];void* solve(void*param){    int i=*((int*)param);    while(true){        if(i%2==0){           sem_wait(&mutex[LEFT]);           sem_wait(&mutex[RIGHT]);           cout<<"phlio"<<i<<" is eatting"<<endl;           sleep(1);           sem_post(&mutex[RIGHT]);           sem_post(&mutex[LEFT]);           cout<<"phlio"<<i<<" is over"<<endl;           sleep(1);        }else{           sem_wait(&mutex[RIGHT]);           sem_wait(&mutex[LEFT]);           cout<<"phlio"<<i<<" is eatting"<<endl;           sleep(1);           sem_post(&mutex[LEFT]);           sem_post(&mutex[RIGHT]);           cout<<"phlio"<<i<<" is over"<<endl;           sleep(1);         }    }}int main(){    for(int i=0; i<N; i++){       sem_init(&mutex[i],0,1);       pthread_create(&thread[i],NULL,solve,&id[i]);    }    for(int i=0; i<N; i++){       pthread_join(thread[i],NULL);    }    return 0;}