How Can I Determine the Collided Side in C++ Collision Detection?

[Peter P.]

My problem is not so much with programming it, but rather determining which side is collided.
And for those who aren't familiar with SDL, SDL_Rect is just a structure that hold an x and y
position along with a width and height ( (x,y) is the top left corner of the rectangle).

So here is how I am planning on going about determining the collided side:
1. I check to see if there is a collision, then
2. I determine the side that collided.

Here is the function I am using for part 1.

bool collision_check (SDL_Rect box1, SDL_Rect box2, bool isOutside) {
	// A collision is determined from the perspective of box1.
	if (isOutside) {
		// box1 must be located somewhere outside box2
		// Checking if the bottom of box1 hits the top of box2
		if (box1.y + box1.h <= box2.y)
			return false;
		// Checking if the top of box1 hits the bottom of box2
		if (box1.y >= box2.y + box2.h)
			return false;
		// Checking if the right side of box1 hits the left side of box2
		if (box1.x + box1.w <= box2.x)
			return false;
		// Checking if the left side fo box1 hits the right side of box2
		if (box1.x >= box2.x + box2.w)
			return false;
	}
	else {
		// box1 must be located somewhere within box2
		// Checking if the top of box1 hits the top of box2
		if (box1.y >= box2.y)
			return false;
		// Checking if the bottom of box1 hits the bottom of box2
		if (box1.y + box1.h <= box2.y + box2.h)
			return false;
		// Checking if the left side of box1 hits the left side of box2
		if (box1.x >= box2.x)
			return false;
		// Checking if the right side of box1 hits the right side of box2
		if (box1.x + box1.w <= box2.x + box2.w)
			return false;
	}
	// If it failed the 4 requirements, it will return true (there was a
	// collision between the boxes.
	return true;
}

For part 2, I was reluctantly planning on doing a pixel by pixel search on all four sides of box1 to
determine which side collided with box2. But I am hoping that i can use a more simple series of if
statements (like part 1), to check for this.

I hope I made sense to anyone reading this. Thanks for your time, and any help is much
appreciated.

-Peter

 

[Mark44]

Can we assume that the sides of the boxes are always vertical, and that the top and bottom are always horizontal? If so, that makes things easier.

Going with that assumption, each point on the top of a box has the same y-coordinate. It's only the x-coordinates that change. For example, suppose that the upper left corner is at (100, 200), and the width is 300 and height is 400.

The corners on the top edge are at (100, 200) and (400, 200). The corners at the bottom edge are (100, 400) and (400, 400). Notice the the top vertices have the same y-coordinate, as do the bottom vertices. Along the left edge, both vertices have the same x-coordinate (100), and both vertices along the right edge have the same x-coordinate (400). It should be a simple matter of checking whether the inside box has any vertices on any of the four edges of the outer box.

 

[chiro]

Are you trying to figure out if a collision happened between two boxes?

If so you will get a maximum of four sides involved in a collision (if they overlap) and at a minimum one side.

The best way IMO to approach this is to return an integer where you use four bits for status: the four bits correspond to a collision of a particular side of a particular box.

You can if you want return the status of not just one box but both boxes resulting in returning collision information for both boxes involved if you want this: this will require a total of eight bits or simply one byte.

To check for a collision, test if the return value is non-zero and if a collision exists simply check the relevant bit for a particular collision.

To implement this you can use a for loop that sets the appropriate bits corresponding to the appropriate check of any pair of two sides (one from each rectangle) where you initialize this return value to 0 and simply set the bit upon an overlap.

 

[Peter P.]

Can we assume that the sides of the boxes are always vertical, and that the top and bottom are always horizontal? If so, that makes things easier.

Going with that assumption, each point on the top of a box has the same y-coordinate. It's only the x-coordinates that change. For example, suppose that the upper left corner is at (100, 200), and the width is 300 and height is 400.

The corners on the top edge are at (100, 200) and (400, 200). The corners at the bottom edge are (100, 400) and (400, 400). Notice the the top vertices have the same y-coordinate, as do the bottom vertices. Along the left edge, both vertices have the same x-coordinate (100), and both vertices along the right edge have the same x-coordinate (400). It should be a simple matter of checking whether the inside box has any vertices on any of the four edges of the outer box.

Yes, all the sides are either vertical or horizontal. And I just thought about using the vertices
to check which side collided in the shower last night as well. :)

Are you trying to figure out if a collision happened between two boxes?

If so you will get a maximum of four sides involved in a collision (if they overlap) and at a minimum one side.

The best way IMO to approach this is to return an integer where you use four bits for status: the four bits correspond to a collision of a particular side of a particular box.

You can if you want return the status of not just one box but both boxes resulting in returning collision information for both boxes involved if you want this: this will require a total of eight bits or simply one byte.

To check for a collision, test if the return value is non-zero and if a collision exists simply check the relevant bit for a particular collision.

To implement this you can use a for loop that sets the appropriate bits corresponding to the appropriate check of any pair of two sides (one from each rectangle) where you initialize this return value to 0 and simply set the bit upon an overlap.

Thanks for the idea of using bits that correspond to different sides and changing them to
represent which side collided with another.