effective way to select last parameter of variadic template

Question

I know how to select first parameter of variadic template:

template< class...Args> struct select_first;
template< class A, class ...Args> struct          


        

Answer 1

Sorry for being a bit late to the party, but I just ran across the same problem , looked up for an answer, didn't like what I see here and realised it can be done using a tuple. Please see C++11 implementation below. Note: one can also get access to an Nth type of a variadic template this way. (The example doesn't check that N exceeds the number of variadic arguments , however the check can be done with SFINAE technique (enable_if) for instance) Is that an acceptable answer or I'm missing anything in the question?

#include <tuple>
#include <iostream>

struct A
{
    char ch = 'a';
};
struct B
{
    char ch = 'b';
};
struct C
{
    char ch = 'c';
};


template <typename... Types>
struct SomeVariadic {

    using TypesTuple = std::tuple<Types...>;

    using LastType = typename std::tuple_element<sizeof...(Types)-1, TypesTuple>::type;

    template <int N>
    using NthType = typename std::tuple_element<N, TypesTuple>::type;
};



int main(int argc, char* argv[]) {

    SomeVariadic<A,B,C>::LastType l;

    std::cout << SomeVariadic<A,B,C>::LastType().ch << " "
            << SomeVariadic<A,B,C>::NthType<1>().ch<< std::endl;
}

Answer 2

Same approach as last time, O(logN) instantiation depth. Using only one overload, so it should consume less resources.

Warning: it currently removes references from the tuple types. Note: Removed the reference from pack::declval. I think it still works in every case.

indices trick in O(log(N)) instantiations, by Xeo; modified to use std::size_t instead of unsigned

    #include <cstddef>

    // using aliases for cleaner syntax
    template<class T> using Invoke = typename T::type;

    template<std::size_t...> struct seq{ using type = seq; };

    template<class S1, class S2> struct concat;

    template<std::size_t... I1, std::size_t... I2>
    struct concat<seq<I1...>, seq<I2...>>
      : seq<I1..., (sizeof...(I1)+I2)...>{};

    template<class S1, class S2>
    using Concat = Invoke<concat<S1, S2>>;

    template<std::size_t N> struct gen_seq;
    template<std::size_t N> using GenSeq = Invoke<gen_seq<N>>;

    template<std::size_t N>
    struct gen_seq : Concat<GenSeq<N/2>, GenSeq<N - N/2>>{};

    template<> struct gen_seq<0> : seq<>{};
    template<> struct gen_seq<1> : seq<0>{};

---

Today, I realized there's a different, simpler and probably faster (compilation time) solution to get the nth type of a tuple (basically an implementation of std::tuple_element). Even though it's a direct solution of another question, I'll also post it here for completeness.

namespace detail
{
    template<std::size_t>
    struct Any
    {
        template<class T> Any(T&&) {}
    };

    template<typename T>
    struct wrapper {};

    template<std::size_t... Is>
    struct get_nth_helper
    {
        template<typename T>
        static T deduce(Any<Is>..., wrapper<T>, ...);
    };

    template<std::size_t... Is, typename... Ts>
    auto deduce_seq(seq<Is...>, wrapper<Ts>... pp)
    -> decltype( get_nth_helper<Is...>::deduce(pp...) );
}

#include <tuple>

template<std::size_t n, class Tuple>
struct tuple_element;

template<std::size_t n, class... Ts>
struct tuple_element<n, std::tuple<Ts...>>
{
    using type = decltype( detail::deduce_seq(gen_seq<n>{},
                                              detail::wrapper<Ts>()...) );
};

Helper for last element:

template<typename Tuple>
struct tuple_last_element;

template<typename... Ts>
struct tuple_last_element<std::tuple<Ts...>>
{
    using type = typename tuple_element<sizeof...(Ts)-1,
                                        std::tuple<Ts...>> :: type;
};

Usage example:

#include <iostream>
#include <type_traits>
int main()
{
    std::tuple<int, bool, char const&> t{42, true, 'c'};

    tuple_last_element<decltype(t)>::type x = 'c'; // it's a reference

    static_assert(std::is_same<decltype(x), char const&>{}, "!");
}

---

Original version:

#include <tuple>
#include <type_traits>

namespace detail
{
    template<typename Seq, typename... TT>
    struct get_last_helper;

    template<std::size_t... II, typename... TT>
    struct get_last_helper< seq<II...>, TT... >
    {
        template<std::size_t I, std::size_t L, typename T>
        struct pack {};
        template<typename T, std::size_t L>
        struct pack<L, L, T>
        {
            T declval();
        };

        // this needs simplification..
        template<typename... TTpacked>
        struct exp : TTpacked...
        {
            static auto declval_helper()
                -> decltype(std::declval<exp>().declval());
            using type = decltype(declval_helper());
        };

        using type = typename exp<pack<II, sizeof...(TT)-1, TT>...>::type;
    };
}

template< typename Tuple >
struct get_last;

template< typename... TT >
struct get_last<std::tuple<TT...>>
{
    template<std::size_t... II>
    static seq<II...> helper(seq<II...>);
    using seq_t = decltype(helper(gen_seq<sizeof...(TT)>()));

    using type = typename detail::get_last_helper<seq_t, TT...>::type;
};


int main()
{
    using test_type = std::tuple<int, double, bool, char>;

    static_assert(std::is_same<char, get_last<test_type>::type>::value, "!");
    // fails:
    static_assert(std::is_same<int, get_last<test_type>::type>::value, "!");
}

Answer 3

With C++17, the cleanest way is

template<typename T>
struct tag
{
    using type = T;
};

template<typename... Ts>
struct select_last
{
    // Use a fold-expression to fold the comma operator over the parameter pack.
    using type = typename decltype((tag<Ts>{}, ...))::type;
};

with O(1) instantiation depth.

Answer 4

template <class... Args>
struct select_last;

template <typename T>
struct select_last<T>
{
     using type = T;
};

template <class T, class... Args>
struct select_last<T, Args...>
{
    using type = typename select_last<Args...>::type;
};

Answer 5

The following is another lean C++17 approach which also uses a fold-expression; but avoids an ad-hoc class proxy, by using std::enable_if:

template <typename ...Ts>
struct select_last
{
  using type = typename decltype((std::enable_if<true,Ts>{}, ...))::type;
};

template <typename ...Ts>
using select_last_t = typename select_last<Ts...>::type;

static_assert(std::is_same_v<char, select_last_t<int,double,char>>);

In C++20 std::type_identity offers a more readable approach:

// C++20
template <typename ...Ts>
struct select_last
{
  using type = typename decltype((std::type_identity<Ts>{}, ...))::type;
};

Answer 6

If you are willing to strip references blindly from your type list (which is quite often the case: either you know they are references, or you don't care), you can do this with little machinery outside of std. Basically stuff the data into a tuple or tie, then use std::get<sizeof...(X)-1>( tuple or tie ) to extract the last element.

You can do this in a pure-type context using std::declval< std::tuple<Args...> >() and decltype, and possibly std::remove_reference.

As an example, suppose you have a variardic set of arguments, and you want to return the last argument ignoring the rest:

#define RETURNS(x) ->decltype(x) { return (x); }

template<typename ...Args>
auto get_last( Args&&... args )
  RETURNS( std::get< sizeof...(Args)-1 >( std::tie(std::forward<Args>(args)...) ) )

we can then use this in another function:

template<typename ...Args>
void foo( Args&&... args ) {
  auto&& last = get_last(std::forward<Args>(args)...);
}