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Time Arithmetic

Problem

You have a date and time and want to find the date and time of some period in the future or past.

Solution

By using SRFI 19, subtracting, and adding seconds is all you need:

(require (lib "19.ss" "srfi"))

(let ((now (current-seconds))                    ;; using PLT
      (now-srfi-19 (time-second (current-time))) ;; using SRFI-19
      (24-hours-from-now (* 24 60 60)))          ;; in seconds!
  (values (+ now 24-hours-from-now)
          (+ now-srfi-19 24-hours-from-now)))

On the other hand, if what you want is to obtain the difference between two different times, then you could use the procedure time-difference, as in:

(require (lib "19.ss" "srfi"))

(let ((t1 (make-time 'time-utc 0 5000))
      (t2 (make-time 'time-utc 0 2000)))
  (let ((duration (time-difference t1 t2)))
    (values (time-second duration)
            (time-nanosecond duration))))

The example above, as you surely guessed, returns the values 0, and 3000, which represent the nano-seconds, and seconds in the difference of the input times, respectively. However, note that the return value of time-difference is an instance of the time type.

Important note: It is an error if the times passed as arguments to time-difference are of different time types.

Furthermore, if you use care with the time duration types in your program, you can safely add or subtract to a given time directly (instead of adding seconds). In this case, you can use any kind of time as input.

(require (lib "19.ss" "srfi"))

(let ((now (current-time))
      (duration (make-time 'time-duration 0 (* 24 60 60))))
  (values
   (add-duration now duration)        ;; 24 hours from now
   (subtract-duration now duration))) ;; 24 hours in the past

See Also

SRFI 19: Time Data Types and Procedures

Comments here

-- FranciscoSolsona - 05 May 2004