Shorten variadic parameter pack to N types-CodePudding

I would like to write a class that takes a size N (> 0) and a variable number of arguments (>= N). It should have a constructor that takes N arguments and a member std::tuple which has the same type:

template <size_t N, typename... Args>
struct Example {
  // A constructor taking N parameters of type Args[N], initializing the member tuple 
  // (e.g. param 1 has type Args[0], param 2 has type Args[1], ..., 
  // param N has type Args[N-1])
  // A tuple with N elements, each corresponding to Args[N]
  // (e.g. std::tuple<Args[0], ..., Args[N-1]>)

  //For instance Example<3, int, float, int, bool> should result in
  constexpr Example(int a, float b, int c): t(a, b, c) {}
  std::tuple<int, float, int> t;
}

In general: Is this possible? If not are there viable alternatives? Why does/ doesn't this work? I'm using C 20.

CodePudding user response：

Here is an implementation using C 20:

#include <cstddef>
#include <utility>
#include <tuple>

template <class... Ts> struct typelist;

template <size_t N, class, class> struct take;

// Takes N elements of the given type pack
template <size_t N, class... Ts>
using take_t = typename take<N, typelist<>, typelist<Ts...>>::type;

template <class Take, class Drop>
struct take<0, Take, Drop>  {
    using type = Take;
};

template <size_t N, class T, class... Ts, class... Us>
requires(N > 0)
struct take<N, typelist<Us...>, typelist<T, Ts...>> {
    using type = typename take<N - 1, typelist<Us..., T>, typelist<Ts...>>::type;
};

template <class Ts> 
struct make_ctor;

template <class... Ts>
struct make_ctor<typelist<Ts...>> {
    constexpr make_ctor(Ts... ts) : tuple(ts...) {}

    std::tuple<Ts...> tuple;
};

template <size_t N, class... Args>
struct Example : 
    make_ctor<take_t<N, Args...>> {
    using make_ctor<take_t<N, Args...>>::make_ctor;
};

int main() {
    Example<3, int, float, int, bool> e(3, 3.14, 4);
}

What we do here is, first we drop the extra template arguments using the take meta-function. However, since type packs are not first class in C , we only get to take a typelist of our desired types. To create a tuple a constructor from a given typelist, we have a helper type called make_ctor. By inheriting its constructor, we get the desired API.

For production, see Barry's answer and use an existing library for the metaprogramming part.

CodePudding user response：

To the extent I understand the question, it simply seems to be asking how to produce a tuple from arguments. Which, using Boost.Mp11, is a short one-liner (as always):

template <size_t N, typename... Args>
using Example = mp_take_c<N, std::tuple<Args...>>;

Rather than Example<3, int, float, int, bool> being some type that has a member tuple<int, float, int> with one constructor, it actually is tuple<int, float int>.

If you, for some reason, specifically need exactly a member tuple and exactly the constructor specified, we can do easily enough:

template <typename... Ts>
struct ExampleImpl {
    std::tuple<Ts...> t;
    constexpr ExampleImpl(Ts... ts) : t(ts...) { }
};

template <size_t N, typename... Args>
using Example = mp_take_c<N, ExampleImpl<Args...>>;

CodePudding user response：

You could make_index_sequence<N> to help with extracting the first N from Args....

Example:

#include <iostream>
#include <tuple>
#include <type_traits>
#include <utility>

// Create a tuple from the sizeof...(I) first elements in the supplied tuple:
template <class... Ts, size_t... I>
constexpr auto extract_from_tuple(std::tuple<Ts...>&& tmp, std::index_sequence<I...>) {
    static_assert(sizeof...(Ts) >= sizeof...(I));
    // make a new tuple:
    return std::make_tuple(std::move(std::get<I>(tmp))...);
}

template <size_t N, class... Args>
struct Example {
    template<class... Ts, std::enable_if_t<sizeof...(Ts)==N, int> = 0>
    constexpr Example(Ts&&... args)
        : t{extract_from_tuple(std::make_tuple(std::forward<Ts>(args)...), 
            std::make_index_sequence<N>{})} {}

    // using the same extract_from_tuple function to deduce the type of `t`
    decltype(extract_from_tuple(std::make_tuple(std::declval<Args>()...),
                                std::make_index_sequence<N>{})) t;
};

int main() {
    Example<3, int, float, double, bool> foo(1, 2.f, 3.);

    static_assert(std::is_same_v<decltype(foo.t), std::tuple<int, float, double>>);
}

CodePudding user response：

Using std::tuple_element and an helper base class, seems to me that you can write something as follows (C 14 is enough)

#include <tuple>
#include <string>

template <typename ...>
struct ExHelper;

template <std::size_t ... Is, typename Tpl>
struct ExHelper<std::index_sequence<Is...>, Tpl>
{
  using t_type = std::tuple<typename std::tuple_element<Is, Tpl>::type...>;

  t_type  t;

  constexpr ExHelper(typename std::tuple_element<Is, Tpl>::type ... args)
    : t{std::move(args)...}
  { }
};

template <std::size_t N, typename... Args>
struct Example : public ExHelper<std::make_index_sequence<N>,
                                 std::tuple<Args...>> {
  static_assert( N <= sizeof...(Args), "!" );

  using ExHelper<std::make_index_sequence<N>,
                 std::tuple<Args...>>::ExHelper;
};

int main ()
{
  Example<3, int, float, int, std::string>  e0{1, 2.0f, 3};

  static_assert( std::is_same<decltype(e0)::t_type,
                              std::tuple<int, float, int>>::value, "!" );
}