Alexey Starinsky :
How to by a given variant type
using V = std::variant<bool, char, std::string, int, float, double, std::vector<int>>;
declare two variant types
using V1 = std::variant<bool, char, int, float, double>;
using V2 = std::variant<std::string, std::vector<int>>;
where V1 includes all the arithmetic types from V and V2 includes all non-arithmetic types from V?
V can be a parameter of a template class, for example:
template <class V>
struct TheAnswer
{
using V1 = ?;
using V2 = ?;
};
in general the criteria can be a constexpr variable like this:
template <class T>
constexpr bool filter;
Quentin :
If for whatever reason you don't want to use Barry's short and reasonable answer, here is one that is neither (thanks @xskxzr for removing the awkward "bootstrap" specialization, and to @max66 for warning me against the empty variant corner case):
namespace detail {
template <class V>
struct convert_empty_variant {
using type = V;
};
template <>
struct convert_empty_variant<std::variant<>> {
using type = std::variant<std::monostate>;
};
template <class V>
using convert_empty_variant_t = typename convert_empty_variant<V>::type;
template <class V1, class V2, template <class> class Predicate, class V>
struct split_variant;
template <class V1, class V2, template <class> class Predicate>
struct split_variant<V1, V2, Predicate, std::variant<>> {
using matching = convert_empty_variant_t<V1>;
using non_matching = convert_empty_variant_t<V2>;
};
template <class... V1s, class... V2s, template <class> class Predicate, class Head, class... Tail>
struct split_variant<std::variant<V1s...>, std::variant<V2s...>, Predicate, std::variant<Head, Tail...>>
: std::conditional_t<
Predicate<Head>::value,
split_variant<std::variant<V1s..., Head>, std::variant<V2s...>, Predicate, std::variant<Tail...>>,
split_variant<std::variant<V1s...>, std::variant<V2s..., Head>, Predicate, std::variant<Tail...>>
> { };
}
template <class V, template <class> class Predicate>
using split_variant = detail::split_variant<std::variant<>, std::variant<>, Predicate, V>;