C++20 concepts: int not swappable_with int
std::swappable_with<T, U>
checks whether swap
can be invoked (after using std::swap;
) with arguments std::declval<T>()
and std::declval<U>()
. With T
and U
being int
, both arguments are rvalues, which cannot be bound to std::swap
parameters, since these are (non-const) lvalue references.
You wonder that int
cannot be swapped with int
— that's right, you cannot write std::swap(1, -1);
.
Use std::swappable_with<T&,U&>
- swappable with cares about value category, encoded by reference, as well as type.
You are in effect asking if rvalues of type int
can be swapped. And it says "no"; you cannot swap to rvalue int
s.
This might be confusing, but if you do this:
template <class T, class U>
requires std::swappable_with<T,U>
void mySwap(T&& t, U&& u) {
auto temp = std::forward<T>(t);
t = std::forward<U>(u);
u = std::move(temp);
}
it becomes a bit more natural. Here, we use forwarding references, and the l/rvalue categories of the argument are stored along side the bare types in T
and U
respectively.
Note that the above could permit rvalues to be swapped, if objects of that type are swappable_with
each other.