linux 哲学家进餐问题

仅当哲学家的左右两只筷子均可用时，才允许他拿起筷子进餐

#include<stdio.h>#include<stdlib.h>#include<pthread.h>#include<iostream>#include<unistd.h>#include<semaphore.h>#define N 5#define LEFT i#define RIGHT (i+1)%Nusing namespace std;class Semaphore{   private:      sem_t sem;   public:      Semaphore(int value=1){          sem_init(&sem,0,value);      }      void P(){          sem_wait(&sem);      }      void V(){          sem_post(&sem);      }};Semaphore mutex[N];pthread_t thread[N];pthread_mutex_t mutex1;int id[N];int add=0;enum {think,hungry,eat}pid_state[N];void *check_pid_state(int a){     int i =a;    if(pid_state[i]==hungry&&pid_state[LEFT]!=eat&&pid_state[RIGHT]!=eat){       pid_state[i]=eat;       mutex[i].V();    }}void *Do_take_fork(int param){   int i =param;   pthread_mutex_lock(&mutex1);   pid_state[i]=hungry;   check_pid_state(i);   pthread_mutex_unlock(&mutex1);   mutex[i].P();}void *Do_put_fork(int param){   int i=param;   pthread_mutex_lock(&mutex1);   pid_state[i]=think;   check_pid_state(LEFT);   check_pid_state(RIGHT);   pthread_mutex_unlock(&mutex1);}void *solve(void*param){    int i =*((int*)param);    while(true){       if(add>=30){          cout<<"noodles is over"<<endl;          break;       }       cout<<"philo"<<i<<"  is thinking"<<endl;       sleep(2);       Do_take_fork(i);       add++;       cout<<add<<endl;       cout<<"philo"<<i<<"  is eatting"<<endl;       sleep(1);       Do_put_fork(i);    } }void thread_create(){   int tmp;   for(int i=0; i<N; i++){      tmp=pthread_create(&thread[i],NULL,solve,&id[i]);      if(tmp!=0){          cout<<"thread"<<i<<"error"<<endl;      }   }}void thread_wait(){    for(int i=0; i<N; i++){       pthread_join(thread[i],NULL);    }}int main(){    pthread_mutex_init(&mutex1,NULL);    for(int i=0; i<N; i++){        id[i]=i;    }    thread_create();    thread_wait();    return 0;}

g++ -o  name name.cpp -pthread