Get the value of a variant, which itself could be another variant

Question

I have a variant ScalarVar

using ScalarVar = std::variant<int, std::string>;

And a variant Var, which could itself be a ScalarVar or a std::vector of ScalarVars

using Var = std::variant<ScalarVar, std::vector<ScalarVar>>;

I want to make a function template<typename T, typename Variant> T Get(const Variant& var); that when given a variant that does not include inner variants will act just like std::get<T>, i.e it will return the value of T if Variant currently contains a T, or if given a variant that contains other variants, it will recursively get the contained type until a non variant is found, and then return that.

Here is my best attempt at this so far:

#include <iostream>
#include <variant>
#include <string>
#include <typeindex>
#include <vector>
#include <map>

template<typename T, typename... T2>
struct is_variant { static inline constexpr bool value = false; };

template<typename... T>
struct is_variant<std::variant<T...>> { static inline constexpr bool value = true; };

template<typename T, typename Variant>
T Get(const Variant& var) {
    static_assert (is_variant<Variant>::value == true, "Template parameter Variant must be a std::variant");
    auto inner = std::visit([](const auto& i){ return i; }, var);
    if constexpr(is_variant<typeof(inner)>::value) {
        return Get<T>(inner);
    }
    else return inner;
}

int main()
{
    using ScalarVar = std::variant<int, std::string>;
    using Var = std::variant<ScalarVar, std::vector<ScalarVar>>;

    ScalarVar s = 5;
    std::cout << Get<int>(s) << std::endl;

    return 0;
}

This should simply return a T if T is not an std::variant, and return std::get<InnerT> if T is an std::variant that contains type T.

But I get a very complex compilation error from gcc for the line:

std::cout << Get<int>(s) << std::endl

/usr/include/c++/9/variant:1005: error: invalid conversion from ‘std::__success_type<std::__cxx11::basic_string<char> >::type (*)(Get(const Variant&) [with T = int; Variant = std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >]::<lambda(const auto:22&)>&&, const std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)’ {aka ‘std::__cxx11::basic_string<char> (*)(Get(const Variant&) [with T = int; Variant = std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >]::<lambda(const auto:22&)>&&, const std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)’} to ‘int (*)(Get(const Variant&) [with T = int; Variant = std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >]::<lambda(const auto:22&)>&&, const std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)’ [-fpermissive]
 1005 |       { return _Array_type{&__visit_invoke}; }
      |                                           ^
      |                                           |
      |                                           std::__success_type<std::__cxx11::basic_string<char> >::type (*)(Get(const Variant&) [with T = int; Variant = std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >]::<lambda(const auto:22&)>&&, const std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >&) {aka std::__cxx11::basic_string<char> (*)(Get(const Variant&) [with T = int; Variant = std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >]::<lambda(const auto:22&)>&&, const std::variant<int, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)}

How can I get my required behaviour?

Answer 1

The problem is this line:

std::visit([](const auto& i){ return i; }, var);

Calling this directly from main with ScalarVar var = 5; produces a similar message.

Let's see what cppreference says about the return type of std::visit (C++17):

The return type is deduced from the returned expression as if by decltype. The call is ill-formed if the invocation above is not a valid expression of the same type and value category, for all combinations of alternative types of all variants.

This means that you can not call std::visit as above, because it had to return different types (int / std::string) depending on the value inside.

You can use function overloading and use two distinct functions for std::variant and other types:

template <typename T, typename U>
T Get(U const& value)
{
    if constexpr (std::is_same_v<T, U>)
    {
        return value;
    }
    // If you get here, none of the nested variants had a value of type T.
    return T();
}

template <typename T, typename... Args>
T Get(std::variant<Args...> const& var)
{
    return std::visit(
        [](auto const& value) { return Get<T>(value); },
        var
    );
}

Answer 2

You need to distinguish between T, std::variant<...> and everything else when you call std::visit.

template<typename T>
struct Getter {
    T operator()(const T & t) { return t; }

    template <typename ... Ts>
    T operator()(const std::variant<Ts...> & var) { return std::visit(*this, var); }

    template <typename U>
    T operator()(U) { throw std::bad_variant_access(); }
};

template<typename T, typename Variant>
T Get(const Variant& var) {
    Getter<T> getter;
    return getter(var);
}

/* or 
template <typename T>
inline Getter<T> Get;
*/