C++ Overload resolution with universal reference function template which can't be changed

Question

Suppose somewhere in my code is a function foo with a universal reference parameter, which I cannot change :

template<typename T>
auto foo(T&& t) { std::cout<<"general version"<<std::endl; }

Now I want to overload foo for a given class A , and make sure that for any qualifier and reference type of A the overload is called. For this I can brute-forcely provide an overload for all possible qualifications (ignore volatile for now):

auto foo(A & a) { std::cout<<"A&"<<std::endl; }
auto foo(A const& a) { std::cout<<"A const&"<<std::endl; }
auto foo(A && a) { std::cout<<"A &&"<<std::endl; }
auto foo(A const&& a) { std::cout<<"A const&&"<<std::endl; }

Demo . This however scales very badly for more parameters.

Alternatively, I can pass by value, which seems to capture as well all the previous cases:

auto foo(A a) { std::cout<<"A"<<std::endl; }

Demo . Now however large object need to be copied (--at least in principle).

Is there an elegant way around these problems?

Remember that I can't change the universal reference function, so SFINAE and the like is no possibility.

Answer 1

Update for C++20 : The answer below remains true for C++11 through C++17, but in C++20 you can do this:

template <typename T>
    requires std::same_as<std::remove_cvref_t<T>, A>
auto foo(T&& t) {
    // since this is more constrained than the generic forwarding reference
    // this one should be preferred for foo(A{})
}

Which you can get using more convenient syntax by creating a named concept:

template <typename T, typename U>
concept DecaysTo = std::same_as<std::decay_t<U>, T>;

// longest form
template <typename T> requires DecaysTo<T, A> void foo(T&&);

// middle form
template <DecaysTo<A> T> void foo(T&&);

// abbreviated form
void foo(DecaysTo<A> auto&&);

---

Honestly, I think you're out of luck here. The typical approaches all fail. You can do...

SFINAE?

template <typename T> auto foo(T&& );
template <typename T,
          typename = only_if_is<T, A>>
auto foo(T&& );

foo(A{}); // error: ambiguous

Write a class that takes an l-or-rvalue reference?

template <typename T> lref_or_ref { ... };
    
template <typename T> auto foo(T&& );
auto foo(lref_or_ref<A> );

foo(A{}); // calls general, it's a better match

The best you could do is introduce a forwarding function using a chooser:

template <int I> struct chooser : chooser<I - 1> { };
template <> struct chooser<0> { };

template <typename T>
auto bar(T&& t, chooser<0> ) {
    // worst-option, general case
    foo(std::forward<T>(t));
}

template <typename T,
          typename = only_if_is<T, A>>
auto bar(T&& t, chooser<1>) {
    // A-specific
}

template <typename T>
auto bar(T&& t) {
    bar(std::forward<T>(t), chooser<20>{});
}

But you mentioned in a comment that this doesn't work for you either.