为什么我不应该为我的类型提供 forward 和 move 版本？

Question

I've used std::swap , std::move and std::forward but I've been recommended not to overload the latter two ones because they already have paramters that can hold any tylpe of value for example a forearding reference. So I should fully-quallify the call to those two functions:

std::move(x);
std::forward<T&&>(x);

And I've read that I can provide my own version of swap and thus I only add a usung declaration for the standard library one then call swap without qualifying it (unqualified lookup):

using std::swap;// in case my class doesn't provide a swap version

swap(x, y); // where x, y are of class type. so don't use std::swap directly

• So why I should not provide a version of std::move and std::forward for my types too?

• Does this mean if I provided versions, there is no benefit because the library's are better like proving a sort function whereas the library's is so far better?

Answer 1

You are only allowed to provide specializations of std functions when the standard permits you to. No such permission is given for std::move or std::forward . Doing so makes your program ill-formed, no diagnostic required.

std::swap you should only specialize (not overload.) when the default implementation has efficiency problems: The default implementation is a basically:

void std::swap( T& lhs, T& rhs ) {
  auto tmp = std::move(lhs);
  lhs = std::move(rhs);
  rhs = std::move(tmp);
}

The most common case is when your type needs to allocate memory when it is "empty". A more efficient swap could avoid such an allocation.

But for most types, this isn't a problem.