用户注册



邮箱:

密码:

用户登录


邮箱:

密码:
记住登录一个月忘记密码?

发表随想


还能输入:200字
云代码 - c代码库

理发师

2013-01-27 作者: TBforest举报

[c]代码库

#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<pthread.h>
#include<semaphore.h>
#include<fcntl.h>
#include<errno.h>
#define n 5   //有五把椅子

//三个信号量: mutex,customer,barbers
sem_t mutex,customers,barbers;
int waiting=0;  //等待的顾客数量
int chair[5];
void * barber();
void * customer ( void *arg );

int main ( int argc,char *argv[] )
{
    //创建10个顾客和1个理发师
    pthread_t Customer_id[10],Barber_id;
    int i;
    sem_init ( &mutex,0,1 );  //互斥信号量初始化1
    sem_init ( &customers,0,0 );//顾客信号量初始化0
    sem_init ( &barbers,0,1 );//理发师信号量初始化1

    for ( i=0; i<5; i++ )
        pthread_create ( &Barber_id,NULL, ( void* ) barber,NULL );
    for ( i=0; i<10; i++ )
        pthread_create ( &Customer_id[i],NULL, ( void* ) customer, ( void* ) ( i+1 ) );
    for ( i=0; i<10; i++ )
        pthread_join ( Customer_id[i],NULL );
    for ( i=0; i<5; i++ )
        pthread_join ( Barber_id,NULL );
    return 0;
}

//创建理发师线程
void * barber()
{
    int i;
    int next;
    //等待顾客,如果没有顾客,理发师睡觉
    sem_wait ( &customers );
    sem_wait ( &mutex );  //等待互斥锁释放
    waiting--;   //等待的顾客减1
    for ( i=0; i<5; i++ )
    {
        if ( chair[i]!=0 )
        {
            next= chair[i];
            chair[i]=0;
            break;
        }
    }
    printf ( "The barber is cutting %dth customer's hair\n",next );
    sleep ( 3 );
    sem_post ( &mutex );
    sem_post ( &barbers );
}

//创建顾客线程
void * customer ( void *arg )
{
    int i;
    sem_wait ( &mutex ); //等待互斥锁释放
    if ( waiting<n )
    {
        waiting++;  //等待的顾客加1
        for ( i=0; i<5; i++ )
        {
            if ( chair[i]==0 )
            {
                chair[i]= ( int ) arg;
                break;
            }
        }

        printf ( "***************************************************\n" );
        printf ( "Entered:Number %d customer comes,and sits at %d chair \n", ( int ) arg, ( i+1 ) );
        printf ( "There are %d customer on the chair\n",waiting );
        printf ( "The customers' location are:" );
        for ( i=0; i<5; i++ )
            printf ( "%d   ",chair[i] );
        printf ( "\n" );

        sleep ( 1 );
        sem_post ( &mutex ); //释放互斥锁
        sem_post ( &customers );
        sem_wait ( &barbers ); //理发师等待
    }
    else
    {
        printf ( "Number %d comes,there are no chairs,the customer %d is leaving\n", ( int ) arg, ( int ) arg );
        sem_post ( &mutex );
    }
}


网友评论    (发表评论)

共1 条评论 1/1页

发表评论:

评论须知:

  • 1、评论每次加2分,每天上限为30;
  • 2、请文明用语,共同创建干净的技术交流环境;
  • 3、若被发现提交非法信息,评论将会被删除,并且给予扣分处理,严重者给予封号处理;
  • 4、请勿发布广告信息或其他无关评论,否则将会删除评论并扣分,严重者给予封号处理。


扫码下载

加载中,请稍后...

输入口令后可复制整站源码

加载中,请稍后...