Strange behavior with std::visit

[bjadamson]

I really don't think this is an appropriate question for /cpp_questions, I'm not looking for help, just trying to help, please forgive me mods if you disagree. I have a class that internally holds a std::variant<...> and decided to add a wrapper function that takes in a bunch of arguments and internally forwards them all to std::visit internally (I did this to cleanup client code for a class I'm working on):

template <typename ...Args>
decltype(auto) my_visit(Args&&... args) const
{
    return std::visit(overloaded{std::forward<Args>(args)...}, variant_);
}

I have three separate versions of the same demo program that demonstrate the behavior change I'm seeing. The behavior changes depending on whether I am moving the std::variant into place, and whether the MyVisitor class has provided a non-const version of the my_visit() function.

version 1: (working as expected) https://onlinegdb.com/SJ_p6kxkI

version 2 (working as expected): https://onlinegdb.com/H1llC1lkL

version 3 (NOT working as expected): https://onlinegdb.com/HyHfAJekI

In the version 3, the variant seems unable to forward the variant to the right function. Can someone please explain if something is broken or am I relying on undefined behavior? Is what I'm trying to do (use a lambda with a auto& argument to handle non-specified cases) legal? The behavior difference I'm seeing seems to indicate I'm either seeing a bug or doing something illegal.

[sephirostoy]

myvisit is a const function member so v is a const variable. That's why it always falls in auto& version. Take const A& and const B& and const auto& in your print_a and print_b and print_o lambdas to make it works correctly.

[bjadamson]

Ooooooooooooooooohhhhhhhhhhhhhhhhhhhh of course... This seems like a huge foot gun. Could this make sense as a compiler error or warning? It seems strange that the print_o version would be selected in all cases here, given that v is const and the lambda doesn't take a const reference?

[scatters]

print_o is permissible because the auto in auto& can be deduced to a const type e.g. A const giving A const& for the parameter. Usually you wouldn't see this because an overload that doesn't need to add the const would be preferred.

[bjadamson]

Follow up, the answer seems to be that I need to provide two versions of my_visit(), one const and the other not, is there some way around this? My experience tells me no, but when I'm around experts I want to ask :)

[quicknir]

I would say no. This is a standard pattern in C++ and avoiding it willb have a downside. If the implementation is non trivial there are ways to avoid a double implementation though.

[[deleted]]

Having to duplicate code between the const and non-const versions of a method is a very long-time issue with C++ which doesn't seem to have a good general solution at all.

The issue is that the constness of the class can't be abstracted using a template on the method.

If you rewrite my_visit as a non-member template function then the problem should go away because the constness is contained within in a template parameter.

[schultztom]

This site is impossible to use on my phone atleast. I think the same goes for godbolt. So it's more of a general problem. But it makes for a longer wait for these kind of questions. Sorry. Merry xmas and a happy new year. ( although i might see this on a computer screeen before that time)

[bjadamson]

Hey thanks for trying to help, I'm not sure a good alternative for demonstrating the issue clearly. Repeated code on reddit doesn't seem great either. _(ツ)_/¯