How to get the class type of an unnamed struct in a typedef inside the unnamed class?

This is a theoritcal question whether it is possible to get the type of the unnamed struct inside its definition.

int main() {
   struct {
         using ThisType = ???? //what do i put here?
    } _; // the instance
}

I tried to extract the class from a member finction pointer but it didnt work

template<typename>
struct get_class;

template<typename Class>
struct get_class<void(Class::*)()>{using type = Class;};

int main(){
struct {
    void getThis(){}
    using ThisType = typename get_class<decltype(getThis)>::type; // error cant know which function to call to determine 
    // i tried to static_cast but I need to know the class type still.
} _;
}

Is it even possible?

I wrote a library to be able to do this in a macro without naming the current type: https://godbolt.org/z/ja77Kzz9q. It works in this context too.

Here's a stripped down version with just the parts you need:

#include <type_traits>

namespace detail {
template<typename Tag>
struct retrieve_type {
    constexpr friend auto friend_function(retrieve_type<Tag>);
    constexpr auto member_function() {
        return friend_function(*this);
    }
};

template<typename Tag, typename Stored>
struct store_type {
    constexpr friend auto friend_function(retrieve_type<Tag>) {
        return std::type_identity<Stored>{};
    }
};
}

int main() {
    struct {
    private:
        struct tag_type;
        auto storer() -> decltype(detail::store_type<tag_type, decltype(this)>{});
    public:
        using ThisType = std::remove_pointer_t<decltype(detail::retrieve_type<tag_type>{}.member_function())::type>;
    } _; // the instance
    static_assert(std::is_same_v<decltype(_)::ThisType, decltype(_)>);
}

This uses template metaprogramming to store the type of this available in the scope of the trailing return type and then retrieves it later.