Re: reference member variable question

From:

=?iso-8859-1?q?Erik_Wikstr=F6m?= <eriwik@student.chalmers.se>

Newsgroups:

comp.lang.c++

Date:

3 May 2007 23:11:11 -0700

Message-ID:

<1178259071.308252.207660@u30g2000hsc.googlegroups.com>

On 4 Maj, 02:37, Bart Simpson <123evergr...@terrace.com> wrote:

Salt_Peter wrote:

On May 3, 7:35 pm, Bart Simpson <123evergr...@terrace.com> wrote:

If a class has a member variable that is a reference. What happens to
teh class that is being referenced, when the containing class is destro=

yed?

e.g.

Class A{ };

class A { };

Class B

class B

{

public:

   B(const A& a):m_a(a){}
   A& m_a ;

};

int main()
{
   A a;
   B * b = new B(a);
   delete b ; // is a deleted also at this point ?

};

No, the instance 'a' dies at the end of the scope its in, namely - the
closing brace of int main() in this case. In other words: an instance
of class B does not 'own' the instance of type A. If you require 'a'
to die with the deallocation of *b, you'ld probably want a member of
type A in class B.

class B
{
  A a;
public:
  B() : a() {}
  B( const A& r_a ) : a( r_a ) {}
};

And nothing stops you from declaring and defining a member reference
to private member 'a'.

Class B contains a reference to class A. since a reference IS the object
itself, I dont understand how come A is not destroyed when B is
destroyed - unless some kind of "reference counting" is employed "under
the hood" ?

No, a reference is *not* the object itself, the object and the
reference can have totally different scope (as in your example). A
reference refer to an object, but just like a reference in a book is
not what was referred to a reference in C++ is not the object referred
to. However a reference behaves much like the object itself in that
all operations on the reference will be performed on the object, kind
of like a proxy object.

--
Erik Wikstr=F6m