Take tail of variadic template parameters

Question

Given this type:

template<typename ...As>
struct Base {};

I need to implement a function

template<int i, typename ...As>
constexpr auto Tail() {
   static_assert(i < sizeof...(As), "index out of range");
   return ??;
}

which returns an instance of B using the tail of type parameter list from index i.

For example,

Tail<0, int, float, double>() -> Base<int, float, double>
Tail<1, int, float, double>() -> Base<float, double>
Tail<2, int, float, double>() -> Base<double>
Tail<3, int, float, double>() -> fails with static assert

I know how to get the type at index i :

template <int64_t i, typename T, typename... Ts>
struct type_at
{
    static_assert(i < sizeof...(Ts) + 1, "index out of range");
    typedef typename type_at<i - 1, Ts...>::type type;
};

template <typename T, typename... Ts> struct type_at<0, T, Ts...> {
    typedef T type;
};

but I can't get a working version for the entire problem.

Answer 1

If you don't mind using C++17. Even static_assert is not needed:

template<unsigned i, typename T, typename ...Ts>
constexpr auto Tail() {
    if constexpr (i == 0)
        return Base<T, Ts...>();
    else 
        return Tail<i - 1, Ts...>();
}


int main(){
    Base<int, double, int> a = Tail<0, int, double, int>();    
    Base<double, int> b = Tail<1, int, double, int>();
    Base<int> c = Tail<2, int, double, int>();
    auto d = Tail<3, int, double, int>();
}

Btw, changed int to unsigned to avoid possibility of (nearly) infinite recursion for negative numbers.

Answer 2

For Tail (why is it a function?), the approach can be almost the same as that of type_at. The only thing you have to change is the base case of the recursion:

template <typename ... Ts>
struct types_from<0, Ts...> {
    using type = Base<Ts...>; // we don't like nasty typedef syntax
};