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.
来源:https://stackoverflow.com/questions/18264829/universal-reference-vs-const-reference-priority