碰撞检测矩形/矩形算法

Question

http://jeffreythompson.org/collision-detection/rect-rect.php

I am trying to figure out the algorithm of the collision detection of the rectangle/rectangle with the rectangles' picture and code. According to Jeff Thompson's collision detection article, valuable r1RightEdge is r1x + r1w.

float r1RightEdge = r1x + r1w;
if (r1RightEdge >= r2x) {
// right edge of r1 is past left edge of r2
}

Is valuable r1RightEdge vertical dot line of the blue rectangle in the picture? If so, why is the valuable r1RightEdge is r1x +r1w instead of r1x+r1h?

Answer 1

Based on the example code

if (r1x + r1w >= r2x &&     // r1 right edge past r2 left
  r1x <= r2x + r2w &&       // r1 left edge past r2 right
  r1y + r1h >= r2y &&       // r1 top edge past r2 bottom
  r1y <= r2y + r2h) {       // r1 bottom edge past r2 top
    return true;
}
return false;

I under stand that x (in you case: r1x) mean horizontal position of the upper left dot of r1 - or R1LeftEdge, it only make sense if we add r1w (which is width) to it as they are both horizontal, the result is the horizontal upper right dot of r1 - or R1RightEdge. "r1x+r1h" doesn't make sense because one is horizontal and the other is vertical.

Answer 2

A generic and simpler solution can be:

// If one rectangle is on left side of other 
if (r1x > r2x + r2w || r2x > r1x + r1w) 
    return false; 

// If one rectangle is above other 
if (r1y > r2y + r2h || r2y > r1y + r1h) 
    return false; 

// If none of the above meet then there will be a intersection
return true;

This will also handle the special case when rectangles intersect but none of the corner of any rectangle lies inside other.