先看一下这篇文章
https://blog.csdn.net/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;
}
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;
}
这是我们发现票是卖的刚刚好,可是卡在那里了。
因为卖完最后一张票,没有进入if语句,在else语句中没有解锁,特别注意的是,用一次
pthread_mutex_lock(&mutex);
就必须在任何有可能退出的地方进行解锁
pthread_mutex_unlock(&mutex);
改进之后就刚刚好