Re: when to call auto_ptr release?

From:
"Frank Hickman" <fhickman3_NOSP@M_noblesoft.net>
Newsgroups:
microsoft.public.vc.language
Date:
Tue, 19 Feb 2008 05:01:33 -0500
Message-ID:
<Obguo5tcIHA.1204@TK2MSFTNGP03.phx.gbl>
"George" <George@discussions.microsoft.com> wrote in message
news:7F0A4ABC-F3B4-486C-B6B6-4D071558D262@microsoft.com...

Hello everyone,

Auto_ptr is convenient because of it saves our work and provides a
framework
to auto-management life cycle to reduce potential resource leak.

But why do we sometimes need to call release method on auto_ptr to go back
to the style of manual management? Remember, when we call release, we need
to
delete the object instance manually later.

(here is a sample I modified from MSDN auto_ptr sample code)

[Code]
// auto_ptr_release.cpp
// compile with: /EHsc
#include <memory>
#include <iostream>
#include <vector>
using namespace std;

class Int
{
public:
  Int( int i )
  {
     x = i;
     cout << "Constructing " << ( void* )this << " Value: " << x << endl;
  };
  ~Int( ) {
     cout << "Destructing " << ( void* )this << " Value: " << x << endl;
  };

  int x;

};

int main( )
{
  auto_ptr<Int> pi ( new Int( 5 ) );
  Int* pi3 = pi.release ();
  delete pi3;
}
[/Code]

thanks in advance,
George


The only reason I can think of as to why you would want to do this is if
another object was going to take ownership of the object pointed to by the
auto_ptr. If you just needed a pointer to the object for some reason you
would probably call the "get" method rather than "release".

--
============
Frank Hickman
NobleSoft, Inc.
============
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