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 <utility>
#include <type_traits>

template <typename, std::size_t, bool...>
struct bar;

// true case
template <std::size_t ... Is, std::size_t I, bool ... Bs>
struct bar<std::index_sequence<Is...>, I, true, Bs...>
   : public bar<std::index_sequence<Is..., I>, I+1U, Bs...>
 { };

// false case
template <std::size_t ... Is, std::size_t I, bool ... Bs>
struct bar<std::index_sequence<Is...>, I, false, Bs...>
   : public bar<std::index_sequence<Is...>, I+1U, Bs...>
 { };

// end case
template <typename T, std::size_t I>
struct bar<T, I>
 { using type = T; };

template <bool ... Bs>
struct foo : public bar<std::index_sequence<>, 0U, Bs...>
 { };

int main()
 {
   static_assert( std::is_same<typename foo<false, true, true>::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 <tuple>
#include <utility>
#include <type_traits>

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

template <std::size_t I>
struct baz<I, true>
 { using type = std::tuple<std::integral_constant<std::size_t, I>>; };

template <std::size_t I, bool B>
using baz_t = typename baz<I, B>::type;

template <typename, bool...>
struct bar;

template <std::size_t ... Is, bool ... Bs>
struct bar<std::index_sequence<Is...>, Bs...>
 {
   template <std::size_t ... Js>
   constexpr static std::index_sequence<Js...>
      func (std::tuple<std::integral_constant<std::size_t, Js>...> const &);

   using type = decltype(func(std::tuple_cat(baz_t<Is, Bs>{}...)));
 };


template <bool ... Bs>
struct foo : public bar<std::make_index_sequence<sizeof...(Bs)>, Bs...>
 { };

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

Answer 2

This code looks correct and works when compiled as C++17. It uses std::array::begin, which only has been made constexpr in C++17.

A better compilation error can be achieved when using clang, which states:

<source>:23:25: note: non-constexpr function 'begin' cannot be used in a constant expression
    auto it = array.begin();