Re: Canonical assignment operator

From:

SG <s.gesemann@gmail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Mon, 25 May 2009 20:10:25 CST

Message-ID:

<37bcdb1f-ab91-4894-9e90-62ace3d376bb@h2g2000yqg.googlegroups.com>

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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