Re: Canonical assignment operator
On 25 Mai, 19:18, Ucayaly Fish <UcayalyF...@gmail.com> wrote:
Rvalue References 101 (https://www.boostpro.com/trac/wiki/BoostCon09/
RValue101) suggest the following form
-- code --
T& operator=(T x)
{
swap(*this, x);
return *this;
}
-- code --
of assingment operator as the best in most of the cases. Does I
understand it correctly that:
a) this form assumes that a free function swap exists in the same
namespace type T is defined?
b) this swap function should be 'friend' for type T in order to be
efficient?
It probably is a friend. It doesn't really matter as long as it's a
special swap that can handle the type T efficiently. You could also
provide a swap function as non-static member and an inline free
function swap that delegates to the member swap function:
class Foo {
.....
public:
.....
void swap(Foo&);
Foo& operator=(Foo t) {swap(t); return *this;}
.....
};
inline void swap(Foo & x, Foo & y) {x.swap(y);}
Cheers!
SG
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