diffIntTime function Null safety

int diffIntTime(
  1. {required int startTimeA,
  2. required int? endTimeA,
  3. required int timeB}
)

Calculates the difference between timeB and a time or time span defined by a startTimeA and optionally, if it's a span rather than an instance, an endTimeA. The result will be negative and equal to endTimeA-timeB if A is before B, positive and equal to startTimeA-timeB is A is after B.

Situation | Result -------------------+--------- A +--------+ | > 0 B + | -------------------+--------- A +--------+ | = 0 B + | B + | B + | -------------------+--------- A +--------+ | > 0 B + | -------------------+---------

The result is zero if timeB is within the time span defined by the two "A" parameters, even if it's equal to endTimeA. This even though the situation of timeB==endTimeA would not qualify as TimeComparisonResult.overlapping!. This inconsistency is rather hard to avoid, as the infinitesimally small fraction of an instance prior to endTimeA does count as/ overlapping - the difference is infinitely small and can therefore not really be expressed in any way.

Implementation

int diffIntTime(
    {required int startTimeA, required int? endTimeA, required int timeB}) {
  final startDiff = (startTimeA - timeB);
  if (endTimeA == null) {
    return startDiff;
  }

  final endDiff = (endTimeA - timeB);

  // B is within span A if startDiff is negative and endDiff is positive.
  if (0 > startDiff && endDiff > 0) {
    return 0;
  }

  // A is before B -> closest difference will be to the end point.
  if (endDiff <= 0) {
    return endDiff;
  }

  // A must be after B -> closest difference will be to the start point.
  return startDiff;
}