Question

I have a template base class like this:

template<typename T, std::size_t Size>
class VectorT
{
public:
    typedef T data_type;
}

and a few specialised derived classes:

template<typename T>
class Vector2d : public VectorT<T, 2U>
{ // some specialised functions }

template<typename T>
class Vector3d : public VectorT<T, 3U>
{  // some other specialised functions }

and these work fine. Howerver, I have a few free-standing functions for operators. For example:

template<typename T, size_t Size>
VectorT<T, Size> operator*(T lhs, const VectorT<T, Size>& rhs)
{
    ... 
}

Unfortunately these do not work for my derived classes, because they return a VectorT<T, Size> instead of a Vector2d<T>.

So I tried with

template<V>
V operator*(typename V::data_type lhs, const V& rhs)
{
    ... 
}

and this works fine, however it can lead to ambiguities because it sucks in anything else with a data_type member.

How can I get around this: how can I write type safe functions that only work with my vector base, or any derivatives of?

I am trying to get around having to re-declare and redefine the operators again for the subclasses.

Answer 1

You could add yet another base class, one that is independent of the template parameters, and use SFINAE to disable calls for types other than derived from such base:

struct VectorBase {};

template< typename T, std::size_t Size >
class VectorT : public VectorBase { ... }

template< typename V >
typename boost::enable_if< boost::is_base_of< VectorBase, V >, V >::type
operator*( V lhs, V const& rhs ){ ... }

Note that is_base_of< X, X > is always true, so this function will work for one more type than required, namely the base class VectorBase.

If you are using a compiler that implements TR1, you can replace boost:: for std:: in both places where is used.