Re: Using assignment operator when rhs is temporary object.

From:

"Vladimir Grigoriev" <vlad.moscow@mail.ru>

Newsgroups:

microsoft.public.vc.language

Date:

Mon, 5 Oct 2009 19:33:11 +0400

Message-ID:

<eULhPEdRKHA.1372@TK2MSFTNGP02.phx.gbl>

I am sorry. I did not specify the reference symbol in the definition by
mistake .
All works.

Vladimir Grigoriev

"Vladimir Grigoriev" <vlad.moscow@mail.ru> wrote in message
news:eTnyx7cRKHA.3540@TK2MSFTNGP04.phx.gbl...

Thanks, Ulrich

I have seen that I may not do the following with std::auto_ptr

const std::auto_ptr<int> & f()
{
   static std::auto_ptr<int> ai( new int( 10 ) );
   return ( ai );
}

main()
{
   const std::auto_ptr<int> ai = f();
}

Vladimir Grigoriev

"Ulrich Eckhardt" <eckhardt@satorlaser.com> wrote in message
news:nqdpp6-g7b.ln1@satorlaser.homedns.org...

Vladimir Grigoriev wrote:

The question arised when I tried to trap paths how auto_ptr performs
assignments and creating objects by copy.

Returning an auto_ptr<T> from a function works by bridging the
non-const "copy constructor" with another type. IIRC, the type is called
auto_ptr_ref<T> or auto_ptr<T>::ref. Anyway, what you need is an
additional
constructor taking such a type and an implicit conversion to such a type.
Note that the conversion operator may be a non-const member and that it
can
still be invoked on a temporary.

See also[1] for a template type that adapts the auto_ptr principle to
other
handle types than just pointers. Maybe that code is a bit more readable
than the average C++ stdlib implementation.

Uli

[1] http://lists.boost.org/Archives/boost/2007/11/130127.php

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