Convert static constexpr array to template parameter in C++11

Question

Let's assume we have a constexpr array as follows:

static constexpr unsigned int a[] = {2, 8, ... ,6}; // N element

I'd like to use this array as template parameter pack:

typedef SomeTemplatedStruct<a[0], a[1], ... ,a[N - 1]> tmp;

It is possible to do so in C++14 as it is explained here. However, I failed to convert that code to C++11. Any help is appreciated.

Yes, and at the same time, I am hoping to find a more concise way. — MTMD, Apr 10 '19 at 02:40
Example of [implementation-c14-make-integer-sequence](https://stackoverflow.com/a/32223343/2684539), and linked solution seems concise enough IMO. — Jarod42, Apr 10 '19 at 09:06

Michael Kenzel · Accepted Answer · 2019-04-10T10:36:51.060

If you're already going to roll your own, there's no need to implement an integer sequence helper as we can just use the size of the already expanded pack as index for the next element to take out of the array during recursion:

template <typename A, A& a, typename U = typename std::remove_reference<decltype(a[0])>::type, bool = true, U... unpack>
struct unravel;

template <typename T, int N, T (&a)[N], typename U, U... unpack>
struct unravel<T[N], a, U, false, unpack...>
{
    template <template <U...> class Thingie>
    using into = Thingie<unpack...>;
};

template <typename T, int N, T (&a)[N], typename U, U... unpack>
struct unravel<T[N], a, U, true, unpack...> : unravel<T[N], a, U, (sizeof...(unpack) + 1 < N), unpack..., a[sizeof...(unpack)]> {};

usage:

struct Blub
{
    static constexpr unsigned int a[] = {2, 8, 5, 7, 6};
};

template <unsigned int...>
struct TheThingToMake {};

void test()
{
    typename unravel<decltype(Blub::a), Blub::a>::into<TheThingToMake> blub;
}

live example

Note: this won't work with arrays of size 0, but those are non-standard and, I guess, not really an interesting use-case for this anyways…

score 1 · Answer 2 · answered Apr 10 '19 at 09:09

The best I can imagine, is develop a C++11 substitute for std::index_sequence and std::make_index_sequence and apply the C++14 solution that you've linked.

But if you avoid it, you can use recursion.

Give an home made integer sequence

template <typename T, T...>
struct myIntegerSequence
 { };

and an helper struct mot_h ("make output template helper")

template <typename T, T, std::size_t, typename>
struct mot_h;

// recursive version
template <typename T, std::size_t N, const T(&A)[N], std::size_t Pos, T ... ts>
struct mot_h<const T(&)[N], A, Pos, myIntegerSequence<T, ts...>>
   : mot_h<const T(&)[N], A, Pos+1u, myIntegerSequence<T, ts..., A[Pos]>>
 { };

// ground case
template <typename T, std::size_t N, const T(&A)[N], T ... ts>
struct mot_h<const T(&)[N], A, N, myIntegerSequence<T, ts...>>
 { using type = myIntegerSequence<T, ts...>; };

you can write the following template output

template <typename T, T inp>
using output = typename mot_h<T, inp, 0u,
   myIntegerSequence<
      typename std::remove_const<
         typename std::remove_reference<decltype(inp[0])>::type>::type>
         >::type;

The following is a full compiling C++11 example

#include <type_traits>

template <typename T, T...>
struct myIntegerSequence
 { };

constexpr int input[] = { 2, 3, 5, 7, 11, 13, 17, 19 };

template <std::size_t N, typename T, const T (&A)[N]>
struct foo
 { };

template <typename T, T, std::size_t, typename>
struct mot_h;

template <typename T, std::size_t N, const T(&A)[N], std::size_t Pos, T ... ts>
struct mot_h<const T(&)[N], A, Pos, myIntegerSequence<T, ts...>>
   : mot_h<const T(&)[N], A, Pos+1u, myIntegerSequence<T, ts..., A[Pos]>>
 { };

template <typename T, std::size_t N, const T(&A)[N], T ... ts>
struct mot_h<const T(&)[N], A, N, myIntegerSequence<T, ts...>>
 { using type = myIntegerSequence<T, ts...>; };

template <typename T, T inp>
using output = typename mot_h<T, inp, 0u,
   myIntegerSequence<
      typename std::remove_const<
         typename std::remove_reference<decltype(inp[0])>::type>::type>
         >::type;

int main ()
 {
   using target = myIntegerSequence<int, 2, 3, 5, 7, 11, 13, 17, 19>;

   static_assert( std::is_same<output<decltype((input)), input>,
                               target>::value, "!" );
 }

Convert static constexpr array to template parameter in C++11

2 Answers2