Re: reference member variable question
"Bart Simpson" <123evergreen@terrace.com> wrote in message
news:uqGdnQS9Se5846fbnZ2dnUVZ8qDinZ2d@bt.com...
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
destroyed?
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" ?. BTW this is the desired behaviour - I just dont understand
how or why it works though ... and am seeking more of an insight to
explain this (maybe someone has a copy of the language reference)?
I like to call a reference a "pointer on steroids". If it was a pointer
instead,
A*m_a ;
you could see how it's possible for the pointer to go out of scope and leave
what it was pointing to untouched. A pointer going out of scope does not
call the destructor on what the pointer was pointing to, only the pointer
itself.
Same with a reference. A reference points to another variable somewhere.
When the reference goes out of scope, it gets deleted, not what it was
pointing to.
If, in this context, you think of a reference as simply a pointer you don't
have to derefernce (dont' have to use -> but can use . Don't have to use *
but the reference naem itself, etc...) then it should become clearer, as
long as you understand pointers.