C++ constexpr : Compute a std array at compile time

Question

I want to convert an \"array\" of bool to a integer sequence. So I need to compute an std::array at compile time.

Here is my code

Answer 1

Is it possible to do otherwise?

About correctness answered JVApen (+1).

A possible alternative is avoid std::array at all and construct the index sequence in a recursive way using template specialization

The following is a full compilable example

#include 
#include 

template 
struct bar;

// true case
template 
struct bar, I, true, Bs...>
   : public bar, I+1U, Bs...>
 { };

// false case
template 
struct bar, I, false, Bs...>
   : public bar, I+1U, Bs...>
 { };

// end case
template 
struct bar
 { using type = T; };

template 
struct foo : public bar, 0U, Bs...>
 { };

int main()
 {
   static_assert( std::is_same::type,
                               std::index_sequence<1U, 2U>>{}, "!" );
 }

If you don't like the recursive solutions, I propose (just for fun) another solution based of std::tuple_cat

#include 
#include 
#include 

template 
struct baz
 { using type = std::tuple<>; };

template 
struct baz
 { using type = std::tuple>; };

template 
using baz_t = typename baz::type;

template 
struct bar;

template 
struct bar, Bs...>
 {
   template 
   constexpr static std::index_sequence
      func (std::tuple...> const &);

   using type = decltype(func(std::tuple_cat(baz_t{}...)));
 };


template 
struct foo : public bar, Bs...>
 { };

int main()
 {
   static_assert( std::is_same::type,
                               std::index_sequence<1U, 2U>>{}, "!" );
 }