Virtual methods and parameters lookup

Question

I have following classes:

Collider (abstract based class)
BoxCollider
MeshCollider
SphereCollider

The last 3 inherit from Collider.

The use case is that it should be possible for each collider to check if it collides with any of the other colliders.

I imagine the call should be in the Collider class and look like:

void Collider::checkCollisionsWithOtherColliders()
{
for(std::vector<Collider*>::iterator it ....)
{
if(this->isCollidingWith(*it){...}
}
}

Now comes the problem how I should define isCollidingWith

I guess I'd need to define it for each pair of colliders, which means each collider type should have a version of this method for each type of collider as argument, e.g. in SphereCollider:

class SphereCollider : public Collider
{

bool isCollidingWith(SphereCollider* c){..}
bool isCollidingWith(BoxCollider* c){..}
bool isCollidingWith(MeshCollider* c){..}

};

To make sure such methods are accessible thru the Collider base class, I'd need to define

class Collider
{
virtual bool isCollidingWith(SphereCollider* c){..}
virtual bool isCollidingWith(BoxCollider* c){..}
.. and so on
};

Also, since a call this->isCollidingWith(*it) takes just a pointer to Collider and NOT a derived class, I'd wish that the appropriate method of those virtual ones listed is taken automatically in runtime. However, this seems to be not the case: since the parameter is just Collider, it looks for a method with just Collider* parameter.

The only solution I can think of is to define in Collider:

bool isCollidingWith(Collider* collider)
{
if (0 != dynamic_cast<SphereCollider*>(collider))
return isCollidingWith(dynamic_cast<SphereCollider*>(collider));
else if (0 != dynamic_cast<BoxCollider*>(collider))
return isCollidingWith(dynamic_cast<BoxCollider*>(collider));
else if ...
}

which is not really elegant. Is there any other (proper) way to handle such situations? Thanks!

Answer 1

This is a case for double dispatch. Won't detail it here since it's an already widely used idiom and even if the wikipedia page become obsolete, a simple google search with the term would yield a good result.