CODEWITHSUNDEEP
c++c++20variant
user997112
user997112

Reputation: 30635

Is it possible to group structs for use with std::variant

I'm designing a class to represent data from 10+ sources using std::variant. Each source may have 10 different messages, so the variant will represent 100 different underlying structs.

Understandably I'd like to declare the std::variant without listing 100 types in the declaration:

std::variant<Type1, Type2, Type3............................ Type100>

I'd prefer to sub-group the 10 structs per 'source' and the variant consist of the groups:

std::variant<Source1Types, Source2Types.... Source10Types>

Is there a way to sub-group structs, for use with std::variant? If not, is there any other way of achieving this?

(I'm using C++20)

Upvotes: 0

Views: 343

Answers (1)

Jarod42
Jarod42

Reputation: 218098

You might do:

using Source0Types = std::variant<Type0, Type1, Type2, Type3, .., Type9>;
using Source1Types = std::variant<Type10, Type11, Type12, Type13, .., Type19>;
// ..

using SourceTypeVariant = std::variant<Source0Types, Source1Types, .., Source9Types>;

The caveat here is that you have two levels of variant (for the visit).

You might create trait to flatten/concatenate the variants:

template <typename ... Vs> struct concatenate;
template <typename ... Vs> using concatenate_t = typename concatenate<Vs...>::type;

template <typename ... Ts> struct concatenate<std::variant<Ts...>>
{
    using type = std::variant<Ts...>;
};

template <typename ... Ts, typename ... Us, typename ... Vs>
struct concatenate<std::variant<Ts...>, std::variant<Us...>, Vs...> :
    concatenate<std::variant<Ts..., Us...>, Vs...>
{
};

And so

using MegaVariant = concatenate_t<Source0Types, Source1Types, .., Source9Types>;

Upvotes: 1

Related Questions