CODEWITHSUNDEEP
c++variadic-templatestemplate-meta-programming
Regis Portalez
Regis Portalez

Reputation: 4860

Take tail of variadic template parameters

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.

Upvotes: 5

Views: 298

Answers (3)

max66
max66

Reputation: 66210

And now, for something completely different...

Just for fun I propose a C++14 solution that doesn't use recursion but the power of std::tuple_cat().

#include <tuple>
#include <type_traits>

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

template <std::size_t I, std::size_t J, typename A,
          std::enable_if_t<(I <= J), bool> = true>
constexpr std::tuple<Base<A>> Tail_helper3 ();

template <std::size_t I, std::size_t J, typename A,
          std::enable_if_t<(I > J), bool> = true>
constexpr std::tuple<> Tail_helper3 ();

template <typename ... As>
constexpr Base<As...> Tail_helper2 (std::tuple<Base<As>...> const &);

template <std::size_t I, typename ... As, std::size_t ... Is>
constexpr auto Tail_helper1 (std::index_sequence<Is...> const &)
   -> decltype( Tail_helper2(std::tuple_cat(Tail_helper3<I, Is, As>()...)) );

template <std::size_t I, typename ... As>
constexpr auto Tail () 
   -> decltype( Tail_helper1<I, As...>(std::index_sequence_for<As...>{}) )
 {
   static_assert(I < sizeof...(As), "index out of range");

   return {};
 }

int main ()
 {
   static_assert( std::is_same_v<Base<int, double, int>,
                                 decltype(Tail<0u, int, double, int>())> );

   static_assert( std::is_same_v<Base<double, int>,
                                 decltype(Tail<1u, int, double, int>())> );

   static_assert( std::is_same_v<Base<int>,
                                 decltype(Tail<2u, int, double, int>())> );

   // Tail<3u, int, double, int>(); compilation error!
 }

Upvotes: 2

bartop
bartop

Reputation: 10315

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.

Upvotes: 4

LogicStuff
LogicStuff

Reputation: 19607

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
};

Upvotes: 4

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