Question

Summary

Is there a way to call a class method on a templated type that could be a pointer or a reference without knowing which and not get compiler/linker errors?

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Details

I have a templated QuadTree implementation that can take any of the following non-trivial user-defined types:

//Abstract Base Class
a2de::Shape

//Derived Classes
a2de::Point
a2de::Line
a2de::Rectangle
a2de::Circle
a2de::Ellipse
a2de::Triangle
a2de::Arc
a2de::Spline
a2de::Sector
a2de::Polygon

But they could be a pointer OR a reference as they are all derived from a2de::Shape. So the specializations are declared as:

template class QuadTree<a2de::Shape&>;
//...similar for all derived types as references.

template class QuadTree<a2de::Shape*>;
//...similar for all derived types as pointers

The problem I am having is the ability to call a class method when the indirection (or lack thereof) is unknown and due to the templates, both sets of code are generated:

template<typename T>
bool QuadTree<T>::Add(T& elem) {

    //When elem of type T is expecting a pointer here
    //-> notation fails to compile where T is a reference i.e.:
    //template class QuadTree<a2de::Shape&>
    //with "pointer to reference is illegal"

    if(elem->Intersects(_bounds) == false) return false;

    //...
}

If I change the above line to use the . (dot) notation:

template<typename T>
bool QuadTree<T>::Add(T& elem) {

    //When elem of type T is expecting a reference here
    //. (dot) notation fails to compile where T is a pointer i.e.:
    //template class QuadTree<a2de::Shape*>
    //with "pointer to reference is illegal"

    if(elem.Intersects(_bounds) == false) return false;

    //...

}

If I remove the reference-based types in favor of the pointer-based types (including in the declaration and usage of the Quadtree class) I get the error left of .<function-name> must have class/struct/union.

If I remove the pointer-based type in favor of the reference-based types (including in the declaration and usage of the Quadtree class) I get the aforementioned reference to pointer is illegal again.

compiler: VS2010-SP1

Answer 1

Small overloaded functions can be used to turn reference into pointer:

template<typename T>
T * ptr(T & obj) { return &obj; } //turn reference into pointer!

template<typename T>
T * ptr(T * obj) { return obj; } //obj is already pointer, return it!

Now instead of doing this:

 if(elem->Intersects(_bounds) == false) return false;
 if(elem.Intersects(_bounds) == false) return false;

Do this:

 if( ptr(elem)->Intersects(_bounds) == false) return false;

If elem is a reference, the first overload ptr will be selected, else the second will be selected. Both returns pointer, which means irrespective of what elem is in your code, the expression ptr(elem) will always be a pointer which you can use to invoke the member functions, as shown above.

Since ptr(elem) is pointer, which means checking it for nullptr be good idea:

 if( ptr(elem) && (ptr(elem)->Intersects(_bounds) == false)) return false;