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Math on Time of Day

 
Wally Hartshorn
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Suppose that you have a time of day in the form of a string, such as "12:15 PM". You want to be able to add some arbitrary amount of minutes to that time to get a new time string. You don't care about the date, only the time (so you don't care whether "11:30 PM" plus 60 minutes rolls over to the next day).

What would be the best way to write a function to handle this? (No, it's not a homework assignment.)

My guess is to pass the time string to the SimpleDateFormat parse() function to get a Date object, then use that to initialize a GregorianCalendar object, then perform the math, then reverse the process to get the new time string, such as "12:45 PM".



I don't expect execution time to be a factor, but the conversions to a Date object, then to a GregorianCalendar object, and then back to a Date object feel somewhat annoying.

Is there a better way to do this?

The only other way I can think of to do it would be to convert the time string into minutes since midnight, do the math, and then convert back. I suspect that would be even more annoying.
 
Campbell Ritchie
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If you really only want hours minutes and seconds, consider a TimeOfDay class with the three fields hours minutes and seconds. You can use the % operator to change the time when you go past midnight. Most simple implementations will record midnight as 00:00:00 rather than 24:00:00.
 
Paul Clapham
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You don't care about the date? Does that mean that you want "1:30 AM" plus 60 minutes to always be "2:30 AM"? You know that in places where daylight saving time is in force, the answer to that is "3:30 AM" on one day each year and "1:30 AM" on one day each year, right?

If your answer to that is that you don't care about that and always want "2:30 AM" for that then your suggestion of parsing the string to a Calendar, doing the arithmetic, and formatting to a new string is still the best idea. Just force the date in the Calendar to be some day in January so that it isn't one of the DST cutover dates.
 
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