Making a method available depending on a compile-time argument

Question

Is there a way to compile a method or not, depending on template argument ? I'm trying to create a Coordinate class that can handle 2, 3 or more dimensions. I want to provide acces methods as x(), y() and z(), but I would like z() method to be accessible only if dimension is larger than 3. For now (as you can see below), I use a static_assert to prevent use of z() for coordinates of dimension 2.

template<typename DataType, int Dimension>
class Coord
{
private:
    std::array<DataType, Dimension> _data;

public:

    // how to achieve some kind of compile_if()
    DataType& z()
    {
        static_assert(Dimension >= 3, "Trying to access an undefined dimension.");
        return _data[2];
    }
};

What I would like to do is hide existence of z() for dimension 2, so that this

Coord<int, 2> ci2(0,0);
ci2.z() = 3;   // shouldn't compile

doesn't compile without the use of static_assert. I've seen many question around std::enable_if, but for what I understand it is use to enable or disable specific overloads.

Question is : is there a way to make a method available or not depending on a compile-time argument ?

Answer 1

For example, you can declare your function as template and use std::enable_if like this

template<typename DataType, int Dimension>
class Coord
{
private:
    std::array<DataType, Dimension> _data;

public:
    template <class T = DataType>
    typename std::enable_if<Dimension >= 3, T&>::type
    z()
    {
        return _data[2];
    }
};