计算两个不同日期两次的时间差 [英] Calculate the difference between two times on two different days

问题描述

我试图确定两次之间的时间差，我将它们表示为无符号整数（以斜线表示），如下所示：

I am trying to determine the time difference between two times, which i represent as unsigned integers (in a sturct) as follows:

unsigned int day;
unsigned int month;
unsigned int year;

unsigned int hour;
unsigned int mins;
unsigned int seconds;

我可以轻松地计算出同一天两次发生的分钟之间的时差：这不是我的确切代码，这只是背后的逻辑。

i can work out the time difference in minutes between two times that occur on the same day easily enough using: This isn't my exact code, this is just the logic behind it.

time1 = hours*3600 + mins*60 +  seconds;
time1 = hours2*3600 + mins2*60 +  seconds2;

    //time2 will always be less than time1

    time_diff_secs = time1_secs - time2_secs;
    time_diff_mins = time_diff_secs / 60;
    time_diff_secs = time_diff_secs % 60;

这会产生以下输出：

Time mayday was issued: 13 Hours 4 Mins 0 Seconds 
Time mayday was recieved: 13 Hours 10 Mins 0 Seconds 
Time between sending and receiving:  6.00Mins

这是正确的，但是当我在不同的日子有两次时，得到的结果是：

which is correct, but when I have two times that are on different days I get this as the result:

Time mayday was issued: 23 Hours 0 Mins 0 Seconds 
Time mayday was recieved: 0 Hours 39 Mins 38 Seconds 
Time between sending and receiving: 71581448.00Mins 

这显然是错误的，我不确定该如何进行从这里开始，实际结果应该是40分钟，而不是7150万。

This is obviously incorrect, I am not sure how to progress from here, the actual result should be 40mins, not 71.5million.

推荐答案

您将出现下溢情况。尝试执行此操作（不管变量是 signed 还是 unsigned ，都可以运行）：

You are getting an underflow. Try this (works regardless of whether the variables are signed or unsigned):

if (time1_secs < time2_secs) {
    // New day. Add 24 hours:
    time_diff_secs = 24*60*60 + time1_secs - time2_secs;
} else {
    time_diff_secs = time1_secs - time2_secs;
}

time_diff_mins = time_diff_secs / 60;
time_diff_secs = time_diff_secs % 60;

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