Vector2 operator * (Transform2D right) is using Transform2D.inverse() instead of Transform2D.affinity_inverse()

[LuisCarli]

Godot version

4.0 beta10

System information

MacOS 12.6

Issue description

The Vector2 operator * (Transform2D right) gives a similar (but not the same) result to Transform2D.inverse() * Vector2 instead of Transform2D.affinity_inverse() * Vector2. As a result doing Vector2 * Transform2D is not the inverse of doing Transform2D * Vector2.

Transform2D.inverse() does not take in consideration the scale of the transform, Transform2D.affinity_inverse() takes into consideration the scale. The confusion between those two functions is also mentioned here.

Steps to reproduce

Self contained code bellow.

Minimal reproduction project

var xform = Transform2D(PI/2, Vector2(2, 2), 0, Vector2(16, 16))
var from = Vector2(1, 1)

var to = xform * from
var inverse = to * xform
prints(from, inverse) # (1, 1) (4, 4)

to = xform * from
inverse = xform.inverse() * to
prints(from, inverse) # (1, 1) (4.000002, 3.999998)

to = xform * from
inverse = xform.affine_inverse() * to
prints(from, inverse) # (1, 1) (1, 1)

[Calinou]

This may be done by design for performance reasons.

[LuisCarli]

The docs say that Vector2 operator * (Transform2D right) "Inversely transforms (multiplies) the Vector2 by the given Transform2D transformation matrix". But it's doing that without considering scaling.

If this is by design, why doing the transformation without considering scaling is a more relevant use case than with scaling?

[AThousandShips]

Since the documentation has been updated to make it clear it uses inverse this can be said to be complete:

vector * transform is equivalent to transform.inverse() * vector. See Transform2D.inverse.

For transforming by inverse of an affine transformation (e.g. with scaling) transform.affine_inverse() * vector can be used instead. See Transform2D.affine_inverse.