线程之售票系统pthread_mutex，_lock，_unlock

最新推荐文章于 2025-10-09 09:56:25 发布

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#线程之售票系统pthread_mutex，_lock，_un

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本文通过模拟四个线程同时处理票务销售的过程，探讨了线程同步的问题，并介绍了如何使用互斥锁来解决资源竞争导致的数据不一致性问题。

先看一下这篇文章
https://blog.youkuaiyun.com/csdn_kou/article/details/81148268

四个人同时买票票，引出线程

#include "head.h"
int ticket = 100;
void * route(void *arg)
{
    char *id = (char *)arg;
    while(1)
    {
        if(ticket>0)
        {
            usleep(1000);
            printf("%s sells:%d\n",id,ticket);
            ticket--;
        }
        else
        {
            break;
        }
    }
}

int main()
{
    key_t key = ftok (".",1);

    pthread_t t1,t2,t3,t4;
    pthread_create(&t1,NULL,route,"thread 1");
    pthread_create(&t2,NULL,route,"thread 2");
    pthread_create(&t3,NULL,route,"thread 3");
    pthread_create(&t4,NULL,route,"thread 4");

    pthread_join(t1,NULL);
    pthread_join(t2,NULL);
    pthread_join(t3,NULL);
    pthread_join(t4,NULL);

    return 0;
}

这里写图片描述
四个人抢票，争的票都卖出负数了，这在实际中是不可以的

改进：引入互斥量加锁和解锁概念

#include "head.h"
#include <pthread.h>
#include <unistd.h>

int ticket = 100;
pthread_mutex_t mutex;

void * route(void *arg)
{
    char *id = (char *)arg;

    while(1)
    {
        pthread_mutex_lock(&mutex);
        if(ticket>0)
        {
            usleep(1000);
            printf("%s sells:%d\n",id,ticket);
            ticket--;
            pthread_mutex_unlock(&mutex);
        }
        else
        {
            break;
        }
    }

}

int main()
{
    key_t key = ftok (".",1);

    pthread_t t1,t2,t3,t4;
    pthread_create(&t1,NULL,route,"thread 1");
    pthread_create(&t2,NULL,route,"thread 2");
    pthread_create(&t3,NULL,route,"thread 3");
    pthread_create(&t4,NULL,route,"thread 4");

    pthread_join(t1,NULL);
    pthread_join(t2,NULL);
    pthread_join(t3,NULL);
    pthread_join(t4,NULL);

    return 0;
}