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I already have working (ugly) code for this, but I will ask anyway:

I have the time intervals [09:15, 10:00), [21:10, 21:45) during weekdays. Given time t and a number of seconds s, if t is within the intervals, I have to calculate the date and time where t - s would fall into.

  • Example: t = 20130913 21:15, s = 600, t - s falls into 20130913 09:55.
  • Example: t = 20130923 09:16, s = 120, t - s falls into 20130920 21:44.

Is there a way of doing this in C++ cleanly (boost::icl? boost::date_time?)

I have tried boost::icl, it can certainly hold the time ranges in an interval_set<Time> and find which interval a certain Time is in, but if t - s time point does not fall into an interval range, I don't see how I can find the nearest interval before that time point, and how to detect if I have to go back a day or through the whole weekend.

nurettin
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  • What would the result be for `t = 20130915 21:15` and `s = 3300` (55 min)? I suspect `20130914 21:40`. Right? Also, are these two the only intervals that you have, or is it in general a list of intervals? – nickie Sep 27 '13 at 13:07
  • @nickie currently, there are just two intervals within each weekday. Of course clean code would allow more. I forgot to skip weekend in the example (15th was sunday). Examples updated. If 15th was a weekday, your example would have been correct. – nurettin Sep 27 '13 at 13:12
  • In my code, I sum the seconds for all intervals, then do some ugly calculation to find out how many of which intervals I skipped, then calculate the number of days I need to go back, then find out how many weekends I had to skip. – nurettin Sep 27 '13 at 13:30

1 Answers1

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I think the problem is too complicated to allow for a clean solution, at least according to my definition of "clean".

You'll need a container for your (non-overlapping) daily intervals, supporting efficiently the following operations:

  • find in which specific interval a given time belongs,
  • move an interval backwards, in the container, and
  • move to the last interval (in chronological order).

It seems to me that boost::icl::interval_set<Time> is an adequate solution. Your time does not need to keep track of the date, you can have that separately.

Your algorithm will be something like:

let d and t be the date and time portions of your t
let i be the interval where t belongs
loop
   if t-s belongs in i then
      return t-s on day d
   else
      let j be the previous interval from i
      if j does not exist (because i was the first) then
         let j be the last interval
         move d one weekday backwards
      s := s - (t-start(i))
      t := end(j)
      i := j

This is more or less what you say that your code does. I don't think it can be much cleaner.

nickie
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  • This is a very good summary of how to do the job iteratively. I'm currently considering treating intervals that lay outside as real intervals – nurettin Sep 27 '13 at 13:38