CRTP and overload return type deduction

Question

Consider the following code :

template <class Crtp>
struct Base
{
    const float& get(const short int i) const {return std::get<0>(tuple);}
    const double& get(const int i) const {return std::get<1>(tuple);}
    const long double& get(const unsigned long long int i) const {return std::get<2>(tuple);}
    std::tuple<float, double, long double> tuple;
};

struct Derived
: public Base<Derived>
{
    template <class... Misc, class Return = /*SOMETHING*/>
    const Return& test(Misc&&... misc) const
    {return this->get(std::forward<Misc>(misc)...);} 
};

This is an EXAMPLE code: it does not illustrate something useful, and the problem could be solved using an auto function declaration for instance. I know that and I do not search a workaround concerning this particular example.

My question is : what would be /*SOMETHING*/ in order to get the return type of the correct overload of get depending of the passed Misc types ?

Answer 1

Just using decltype with std::declval (from <type_traits>) should be fine:

class Return = decltype(std::declval<Derived const>().get(std::declval<Misc>()...))