关于单片机C语言实现计算任意两个时间点相隔分钟数的问题

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saservice| 中级工程师

楼主 | 2023-9-28 21:00 | 只看该作者回帖奖励

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C语言, 单片机, 单片机C, 单片机C语, 单片机c语言

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#include <stdio.h>



typedef struct Time{

int Y,M,D,h,m,s;

}Time;



Time time1 = {20,7,28,5,30,10};

Time time2 = {20,7,29,6,10,10};



/*>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

*函数名称：Day_Diff

*函数功能：计算任意日期的相隔天数

*输入参数：time1 结构体指针 ， time2结构体指针

*返回参数：天数

* <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<*/

int Day_Diff(Time *time1 , Time *time2)

{

int year2 , month2 , day2;

int year1 , month1 , day1;

month1 = (time1->M + 9)%12;

year1 = time1->Y - month1/10;

day1 = 365 * year1 + year1 / 4 - year1 / 100 + year1 / 400 + (month1 *306 + 5) / 10 + (time1->D - 1);



month2 = (time2->M + 9) % 12;

year2 = time2->Y - month2 / 10;

day2 = 365 * year2 + year2 / 4 - year2 / 100 + year2 / 400 + (month2 *306 + 5) / 10 + (time2->D - 1);

return (day2 - day1);

}



/*>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

*函数名称：TimeSub

*函数功能：计算任意两个时间点相隔分钟数

*输入参数：Day 天数 ， time1 结构体指针 ， time2结构体指针

*返回参数：

* <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<*/

int TimeSub(int Day , Time *time1,Time *time2)

{

long t1,t2,t;

t1 = (time1->h * 3600) + (time1->m * 60) + （time1 -> s）;

t2 = (Day * 24 *3600) + (time2->h * 3600) + (time2->m *60) + (time2 -> s);



t = (t2 - t1) /60;

return t;

}





int main()

{

int num = 0;

int day = 0;

day = Day_Diff(&time1 , &time2);

printf("day = %d\r\n",day);

num = TimeSub(day,&time1 , &time2);

printf("%d\r\n" , num);

return 0;

}