I've seen many answers online such as this one, but they do not seem to work when the parameter pack is of std::size_t.
template <typename ...Ts> struct select_last { using type = typename decltype((std::type_identity<Ts>{}, ...))::type; }; template<std::size_t... N> class Example { private: using type = select_last<int, double>::type; // works using size_t_type = select_last<size_t... N>::type; // doesn't work }; How can I get the last element of a parameter pack of type std::size_t?
The template<std::size_t... N> is based on a non-type template parameter, so you cannot extract the type (or more precisely, there is no sense in trying to extract the type - I can just tell you it is std::size_t!), you may however extract the value, into a static constexpr.
Here is the proposed code:
template<std::size_t... N> struct Last { static constexpr std::size_t val = (N, ...); // take the last }; int main() { std::cout << Last<1, 2, 3, 99>::val; // 99 } If you want, you can actually have it work for any type:
template<auto... N> struct Last { static constexpr auto val = (N, ...); // take the last }; 
constexpr std::array<size_t, sizeof...(N)> vals{N...}. It basically gives you the full control over the pack.
Tscannot be default constructed. You should usestd::declval<Ts>()instead.