I want to write a math vector template. I have a class which accepts type and size as template argument, with a lot of math operation methods. Now I want to write specializations where Vector<3> for instance has x, y, z as members which refer to data[0..3] respectively.
The problem is that I don't know how to create a specialization which inherits everything from the default template without creating either a base class or writing everything twice.
What's the most efficient way to do this?
template<class Type, size_t Size>
class Vector {
// stuff
};
template<class T>
class Vector<3,T>: public Vector {
public:
T &x, &y, &z;
Vector(): Vector<>(), x(data[0]), y(data[1]), z(data[2]){}
// and so on
};
Somehow you should be able to derive from default implementation, but you are specializing an instance, so how? it should be a non-specialized version that you can be able to derive from it. So that's simple:
// Add one extra argument to keep non-specialized version!
template<class Type, size_t Size, bool Temp = true>
class Vector {
// stuff
};
// And now our specialized version derive from non-specialized version!
template<class T>
class Vector<T, 3, true>: public Vector<T, 3, false> {
public:
T &x, &y, &z;
Vector(): Vector<>(), x(data[0]), y(data[1]), z(data[2]){}
// and so on
};
Consider making this in a little different way, but goals will be achieved, Add external interface - I mean standalone functions X(), Y(), Z():
template<class T, size_t S>
T& x(Vector<T, S>& obj, typename std::enable_if<(S>=1)>::type* = nullptr)
{
return obj.data[0];
}
template<class T, size_t S>
T& y(Vector<T, S>& obj, typename std::enable_if<(S>=2)>::type* = nullptr)
{
return obj.data[1];
}
template<class T, size_t S>
T& z(Vector<T, S>& obj, typename std::enable_if<(S>=3)>::type* = nullptr)
{
return obj.data[2];
}
There is no big difference between:
Vector<T, 3>& obj
return obj.x();
And
Vector<T, 3>& obj
return x(obj);
As a bonus - this interface works for matching sizes.
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