convert `boost::tuple` to `boost::fusion::tuple`

Question

I need to convert a boost::tuple to the corresponding boost::fusion::tuple. I have figured out the corresponding type.

But I expect there is an inbuilt function to do this. I really don't want to reinvent such things. I have searched in the boost fusion documentation but didn't find any.

Answer 1

c++14 version:

template<std::size_t...Is, class T>
auto to_fusion( std::index_sequence<Is...>, T&& in ) {
  using std::get;
  return boost::fusion::make_tuple( get<Is>(std::forward<T>(in))... );
}
template<class...Ts>
auto to_fusion( boost::tuple<Ts...> in ) {
  return to_fusion( std::make_index_sequence<::boost::tuples::length< boost::tuple<Ts...>>::value>{}, std::move(in) );
}
template<class...Ts>
boost::fusion::tuple<Ts...> to_fusion( std::tuple<Ts...> in ) {
  return to_fusion( std::make_index_sequence<sizeof...(Ts)>{}, std::move(in) );
}

I am unaware of a built-in version.

Add trailing -> decltype(boost::fusion::make_tuple( get<Is>(std::forward<T>(in))... )) in c++11. You also need make_index_sequence, which probably has a boost equivalent.

Live example.

Answer 2

You might use something like:

template <class Tuple>
auto to_fusion(Tuple&& tuple)
{
    std::apply(
        [](auto&&... args){
            return boost::fusion::make_tuple(decltype(args)(args)...);
        },
        std::forward<Tuple>(tuple));
}