Is it possible to store a reference in a std::any?

Question

I was trying some things and came to the following question: Is there a possibility to store references to a value in a std::any?

I tried the following approaches:

#include <any>
#include <iostream>
#include <functional>

auto func_by_pointer(std::any obj)
{
    *std::any_cast<int *>(obj) += 2;
}
auto modify_by_pointer(int &a)
{
    func_by_pointer(std::make_any<int *>(&a));
}

auto func_by_reference_wrapper(std::any obj)
{
    std::any_cast<std::reference_wrapper<int>>(obj).get() -= 2;
}
auto modify_by_reference_wrapper(int &a)
{
    func_by_reference_wrapper(std::make_any<std::reference_wrapper<int>>(a));
}

auto func_by_reference(std::any obj)
{
    std::any_cast<int &>(obj) *= 2;
}
auto modify_by_reference(int &a)
{
    func_by_reference(std::make_any<int &>(a));
}

int main()
{
    auto value = 3;
    std::cout << value << '\n';
    modify_by_pointer(value);
    std::cout << value << '\n';
    modify_by_reference_wrapper(value);
    std::cout << value << '\n';
    modify_by_reference(value);
    std::cout << value << '\n';
}

The result is the following output:

3
5
3
3

Yet, I was expecting it to be:

3
5
3
6

Thus, passing a pointer to value works fine. Passing a std::reference_wrapper to value works fine as well, but passing int& somehow doesn't work. Did I do something wrong in my code, or is it generally not possible to store references inside a std::any?

Answer 1

You cannot store references in std::any because, for a given type T, the constructor std::any(T) stores a value of type std::decay_t<T>, which removes reference qualifiers:

[any.cons]

template<class T>
  any(T&& value);

6. Let VT be decay_­t<T>.

7. Requires: VT shall satisfy the Cpp17CopyConstructible requirements.

8. Effects: Constructs an object of type any that contains an object of type VT direct-initialized with std::forward<T>(value).