How to detect std::reference_wrapper in C++ at compile time

Question

Lets say we have some variadic template and need to treat std::reference_wrapper parameters differently. How can we achieve that?

In what way do you want to treat them differently? Print `os << "chicken"` when you print them, while using `os << t` on other types? Treat `reference_wrapper&&` as `T&` while calling `.get()` on them? — Yakk - Adam Nevraumont, Nov 04 '16 at 19:53

score 13 · Accepted Answer · answered Nov 04 '16 at 20:00

You can make a trait to tell if a type is reference_wrapper

template<typename T>
struct is_reference_wrapper : false_type {};

template<typename T>
struct is_reference_wrapper<reference_wrapper<T>> : true_type{};

Then you can use it to disambiguate:

template<typename T>
void do_stuff(T&& t, false_type)
{
    cout << "Normal: " << t << endl;
}

template<typename T>
void do_stuff(T&& ref, true_type)
{
    cout << "Ref: " << ref.get() << endl;
}

template<typename... Ts>
void foo(Ts&&... ts)
{
    [[maybe_unused]] int arr[] = {
        (do_stuff(forward<Ts>(ts), is_reference_wrapper<decay_t<Ts>>{}), 0)...
    };
}

demo

How to detect std::reference_wrapper in C++ at compile time

1 Answers1