universal reference vs const reference priority?

问题:

When I consider the two following overloads:

template <class... T> void f(const T&... x);
template <class T> void f(const T& x);

I have the guarantee that f(x) will always call the second function and will never lead to an ambiguity. In a sense the second version is universally prioritized compared to the first one for one argument whatever its type is.

Now consider the situation where there is a universal reference and a const reference versions of a function:

template <class T> void f(T&& x);
template <class T> void f(const T& x);

My question is: is their a universal priority between these two functions regardless of the type of x (r-value reference, reference, cv-qualifiers, pointer...) like in the previous case? (and if yes, what is the priority ?)

回答1:

There is not a universal priority between these two functions. They compete equally in the overload resolution algorithm. In general the so-called "universal reference" wins unless const T& is an exact match, and there the const T& wins.

struct A {};

int
main()
{
    f(std::declval<A>());  // calls f<A>(A&&), #1
    f(std::declval<const A>());  // calls f<const A>(const A&&), #1
    f(std::declval<A&>());  // calls f<A&>(A&), #1
    f(std::declval<A&&>());  // calls f<A>(A&&), #1
    f(std::declval<const A&&>());  // calls f<const A>(const A&&), #1
    f(std::declval<const A&>());  // calls f<A>(const A&), #2
}

Good advice is to never overload like this.