In C++ given one std::variant type, can one add additional types to make another variant type?

Question

#include <string>
#include <variant>

int main()
{
    using variant_base = std::variant< int, double >;

    using variant_derived_and_flattened = std::variant< std::string, variant_base >;
    // the above does not produce the desired
    // std::variant< int, double, std::string >
    static_assert( std::is_same< variant_base, variant_derived_and_flattened >{} );
}

The static_assert fails.

I assume there is no way to do this and would appreciate a confirmation of this, but perhaps someone knows how to do this?

C++ is a *statically typed* language. You *cannot* modify a type after its declaration. — Jesper Juhl, Jan 07 '20 at 18:12
@JesperJuhl OP specifically asks to make another type, not to modify an existing one. — walnut, Jan 07 '20 at 18:13

score 6 · Answer 1 · answered Jan 07 '20 at 18:11

You can create a meta-function to transform the variant type.

template <class V, class T> struct variant_append_helper;

template <class... A, class T>
struct variant_append_helper<std::variant<A...>, T> {
    using type = std::variant<A..., T>;
};

template <class V, class T>
using variant_append = typename variant_append_helper<V, T>::type;

static_assert(std::is_same<std::variant<int, double, std::string>,
                           variant_append<std::variant<int, double>, std::string>>{});

In C++ given one std::variant type, can one add additional types to make another variant type?

1 Answers1