Get closest point on 2D perimeter using Bounds

hm. frankly this is not a particularly burdensome task. even doing it with an arbitrarily-oriented bounding box is far from nightmarish. .. or should be, I don't know the code you found.

here's an implementation. I haven't tested it, but it should work whether your point is inside the box, outside the box, or on the edge of the box. it uses sqr$$anonymous$$agnitude instead of magnitude because it's a good practice when you're comparing distances to use the squared version because it may be faster if you're dealing with a bazillion comparisons. in this case, magnitude would probably be more than fine.

 Vector3 p = <the point you're interested in>;
 Bounds  b = <the AABB bounding box>;
 
 // I'll assume the 2D aspect of this is X and Y, and ignore Z.
 
 Vector3[] edgePoints = Vector3[4];  // syntax might be wrong on this. an array of 4 Vector3's.
 
 // point on left edge
 edgePoints[0]   = p;
 edgePoints[0].x = b.min.x;
 
 // point on right edge
 edgePoints[1]   = p;
 edgePoints[1].x = b.max.x;
 
 // point on bottom edge
 edgePoints[2]   = p;
 edgePoints[2].y = b.min.y;
 
 // point on top edge
 edgePoints[3]   = p;
 edgePoints[3].y = b.max.y;
 
 // now find the closest.
 
 float closestSqrDist = -1; // value doesn't matter.
 int   closestIndex   = -1; // value doesn't matter.
 for (int n = 0; n < 4; ++n) {
     float sqrDist = (edgePoints[0] - p).sqr$$anonymous$$agnitude;
     if ((n == 0) || (sqrDist < closestSqrDist)) {
         closestSqrDist = sqrDist;
         closestIndex   = n;
     }
 }
 
 Vector3 closestPointOnEdgeOfBoundingBox = edgePoints[closestIndex];