/* author: hjjdebug
date: 2023年 09月 20日 星期三 09:33:58 CST
description: 10辆汽车通过承重5辆汽车的桥,处理一个资源争用问题
* 10个线程代表10辆汽车
* 桥上只能承载5辆汽车, 代表最大只能同时有5辆汽车通过
概要:
让10个线程竞争5个资源,用posix 接口, semaphore 来实现
1. sem_init(信号量,num) 初始化资源数
2. 每个线程要 sem_wait(信号量) 申请资源,申请不到会等待
3. 用完要sem_post(信号量), 归还资源
*/
#include
#include
#include
#include
#include
sem_t semPass;
#define MAX_THREAD 10
pthread_t gThread[MAX_THREAD];
#define MAX_LOAD 5
void *func(void *arg)
{
int id = *(int *)arg; //保留下参数
//等待信号量的值
sem_wait(&semPass); //让10个线程竞争5个资源
printf(">>>>>>>>>>>>>> car %d passing >>>>>>>>>>>>>> \n", id);
sleep(2); //这里可以用sleep(rand()%5+1)模拟随即等待1-5秒的时间
printf("============== car %d finished ============== \n", id);
sem_post(&semPass); //资源释放
return NULL;
}
int main()
{
sem_init(&semPass, 0, MAX_LOAD); //初始化资源数(整数)
srand(time(NULL));
for(int i=0; i
//创建线程
pthread_create(&gThread[i], NULL, func, &i); //创建10个线程
//控制一下节奏,否则i变化太快,func拿不到正确参数i,i又变了.
//当然也可以用其它同步手段,例如互斥量等,但usleep 是最简单的了
usleep(1000);
}
for(int i=0; i
//等待线程结束
pthread_join(gThread[i],NULL);
}
return 0;
}
/*
* 运行结果
./semaphore
>>>>>>>>>>>>>> car 0 passing >>>>>>>>>>>>>>
>>>>>>>>>>>>>> car 1 passing >>>>>>>>>>>>>>
>>>>>>>>>>>>>> car 2 passing >>>>>>>>>>>>>>
>>>>>>>>>>>>>> car 3 passing >>>>>>>>>>>>>>
>>>>>>>>>>>>>> car 4 passing >>>>>>>>>>>>>>
============== car 0 finished ==============
>>>>>>>>>>>>>> car 5 passing >>>>>>>>>>>>>>
============== car 1 finished ==============
>>>>>>>>>>>>>> car 6 passing >>>>>>>>>>>>>>
============== car 2 finished ==============
>>>>>>>>>>>>>> car 7 passing >>>>>>>>>>>>>>
============== car 3 finished ==============
>>>>>>>>>>>>>> car 8 passing >>>>>>>>>>>>>>
============== car 4 finished ==============
>>>>>>>>>>>>>> car 9 passing >>>>>>>>>>>>>>
============== car 5 finished ==============
============== car 6 finished ==============
============== car 7 finished ==============
============== car 8 finished ==============
============== car 9 finished ==============
*/