lerpList method Null safety

List<Shadow>? lerpList(
  1. List<Shadow>? a,
  2. List<Shadow>? b,
  3. double t
)

Linearly interpolate between two lists of shadows.

If the lists differ in length, excess items are lerped with null.

The t argument represents position on the timeline, with 0.0 meaning that the interpolation has not started, returning a (or something equivalent to a), 1.0 meaning that the interpolation has finished, returning b (or something equivalent to b), and values in between meaning that the interpolation is at the relevant point on the timeline between a and b. The interpolation can be extrapolated beyond 0.0 and 1.0, so negative values and values greater than 1.0 are valid (and can easily be generated by curves such as Curves.elasticInOut).

Values for t are usually obtained from an Animation<double>, such as an AnimationController.

Implementation

static List<Shadow>? lerpList(List<Shadow>? a, List<Shadow>? b, double t) {
  assert(t != null);
  if (a == null && b == null) {
    return null;
  }
  a ??= <Shadow>[];
  b ??= <Shadow>[];
  final List<Shadow> result = <Shadow>[];
  final int commonLength = math.min(a.length, b.length);
  for (int i = 0; i < commonLength; i += 1) {
    result.add(Shadow.lerp(a[i], b[i], t)!);
  }
  for (int i = commonLength; i < a.length; i += 1) {
    result.add(a[i].scale(1.0 - t));
  }
  for (int i = commonLength; i < b.length; i += 1) {
    result.add(b[i].scale(t));
  }
  return result;
}