Linux 同步理发师问题

快来打我* 2022-04-24 04:28 89阅读 0赞

熟睡的理发师问题描述的是多个进程(线程)之间的通信与同步问题:

有一个理发师的椅子，和n个顾客的椅子  
如果有顾客在椅子上等，那么理发师为他剪发，否则理发师就在自己的椅子上睡觉。  
如果理发师在熟睡，那么顾客会叫醒理发师，否则顾客会看有没有空椅子，有的话，他坐下等，否则，他将离开理发店。

#include <pthread.h>
    #include <stdio.h>
    #include <stdlib.h>
    #include <unistd.h>
    #include <semaphore.h>
    
    #define MAX 10
    #define TRUE 1
    #define p(x) sem_wait(&x)
    #define v(x) sem_post(&x)
    
    sem_t customer,barber;
    int chairs;
    pthread_mutex_t mutex;
    
    void init ( )
    {
        sem_init ( &customer , 0 , 0 );
        sem_init ( &barber , 0 , 1 );
        chairs = MAX;
    }
    
    void* _barber ( )
    {
        while ( TRUE )
        {
            p(customer);
            pthread_mutex_lock(&mutex);
            chairs++;
            pthread_mutex_unlock(&mutex);
            sleep(5);
            printf ( "one work done....\n" );
            v(barber);
        }
    }
    
    void* _customer ( void* arg )
    {
        int *p = (int*) arg;
        int x = *p;
        pthread_mutex_lock(&mutex);
        if ( chairs > 0 )
        {
            chairs--;
            v(customer);
            printf ( "the %dth customer sitting and waiting , %d customers wating...\n" , x , MAX-chairs );
            pthread_mutex_unlock(&mutex);
            p(barber);
        }
        else
        {
           pthread_mutex_unlock(&mutex);
           printf ( "the %dth customer go away!\n" , x );
        }
    }
    
    int main ( )
    {
        int i;
        init ( );
        pthread_t b;
        pthread_t pthreads[MAX*100];
        int cid[MAX*100];
        pthread_create ( &b , NULL , _barber , NULL );
        for ( i = 0 ; i < MAX*100 ; i ++ )
        {
            sleep(1);
            cid[i] = i;
            pthread_create ( &pthreads[i] , NULL , _customer , &cid[i] );
        }
        pthread_join ( b , NULL );
        for ( i = 0 ; i < MAX*100 ; i++ )
            pthread_join ( cid[i] , NULL );
    }