Calculate the difference between two times on two

问题:

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;

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;

this produces this output:

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

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 

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

回答1:

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;

回答2:

Another way to do it using the standard C library, the only advantage of this is you do not have to worry about your dates overlapping years, or problems with overlapping month boundaries + leap year nonsense:

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

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


time_t conv(void)
{
   time_t retval=0;
   struct tm tm;
   tm.tm_mday=day;
   tm.tm_mon=month -1;
   tm.tm_year=year - 1900;
   tm.tm_hour=hour;
   tm.tm_min=mins;
   tm.tm_sec=seconds;
   tm.tm_isdst=-1;
   retval=mktime(&tm);
   return retval;
}

int main()
{
   time_t start=0;
   time_t end=0;
   time_t diff=0;
   // assign day, month, year ... for date1
   start=conv();
   // assign day, month, year ... for date2
   end=conv();
   if(start>end)
     diff=start - end;
   else
     diff=end - start;
   printf("seconds difference = %ld\n", diff);
   return 0; 
}

回答3:

Change

time_diff_secs = time1_secs - time2_secs;

to

time_diff_secs = abs(time1_secs - time2_secs) % 86400;

This will force it to be the minimum time difference between both times and will work even if you add days, months, etcetera to the time_diff_secs calculation.