Convert a tuple to a structure in a template

Question

I am learning OpenGL as an exercise and want to collapse my own math library to get convenient programming using C ++ 11 templates. Solving this problem should not cause runtime - polymorphism.

The main idea is that I want something like this (note that this is clearly invalid C ++ code):

template<class T, int n> //element type is T, size is n
class Vector {
    T v1, v2, ... , vn;
public:
    Vector(T v1, ... , T vn);
    ~Vector() noexcept;
    ...
    // more constructors and stuff here.
}

template<T, n>
Vector<T, n> operator +(Vector<T, n> lhs, Vector<T, n> rhs);

...
// more math functions and operators here...

The problem is that I want to convert these vectors to regular C structures transparently when they are passed as arrays to OpenGL functions. For example, for n == 3I want to convert mine Vector<T, 3>to something like:

template<class T>
struct Vec3 {
    T v1, v2, v3;
}

So that I can make the call:

Vector<float, 3> vertices[1];
vertices[0] = Vector<float, 3>(1.0f, 1.0f, 1.0f);
glBufferData(GL_ARRAY_BUFFER, sizeof(vertices), vertices, GL_STATIC_DRAW);

, Vec3<float>. n == 2, n == 3 n == 4. 3 .

SFINAE : operator T().

// Inside Vector declaration...
public:
    operator typename std::enable_if<n == 3, Vec3<T>>::type();

n == 3, :

    operator typename std::enable_if<n == 2, Vec2<T>>::type();
    operator typename std::enable_if<n == 4, Vec4<T>>::type();

g++ , enable_if ::type typedef 2 4, Vector<float, 3>.

std::array<T, n> , , . , - , std::array<T, n>, Vec3<T>?

std::tuple<class... Types>, . :

• , .
• , .
• - operator Vec3<T>() .
• , sizeof(tuple<float, float, float>) == sizeof(Vec3<float>) , () a tuple<float, float, float> Vec3<float>. , g++ stdlib , .

Answer 1

std::array<T, n> T[n], , , OpenGL.