Linux下获取毫秒级时间差
因为要为DTV开发一款简单的T9英文输入法,判断按键之间的间隔就必不可少了
使用Linux的gettimeofday函数可以达到这个目的
其中t1=t_start.tv_sec是公元1970年至今的时间(换算为秒)
t2=t_start.tv_usec是当前秒数下的微妙数
所以将t1*1000+t2/1000可以得到当前的毫秒数
判断两次遥控器按键的敲击间隔就不难了
引用
#include <stdio.h>
#include <sys/time.h>
#include <time.h>
int gettimeofday(struct timeval *tv, struct timezone *tz);
int main(int argc,char * argv[]){
struct timeval t_start,t_end;
long cost_time = 0;
//get start time
gettimeofday(&t_start, NULL);
long start = ((long)t_start.tv_sec)*1000+(long)t_start.tv_usec/1000;
printf("Start time: %ld ms\n", start);
sleep(2);
//get end time
gettimeofday(&t_end, NULL);
long end = ((long)t_end.tv_sec)*1000+(long)t_end.tv_usec/1000;
printf("End time: %ld ms\n", end);
//calculate time slot
cost_time = end - start;
printf("Cost time: %ld ms\n", cost_time);
return 0;
}
提供一些相关的资料:
http://book.chinaunix.net/special/ebook/addisonWesley/APUE2/0201433079/ch06lev1sec10.html
http://www.linuxidc.com/Linux/2010-09/28318.htm
http://blog.sina.com.cn/s/blog_62a9b1bb0100seyz.html
http://lizzy115.blog.163.com/blog/static/36491958201102505054953/