Re: deleting pointers in a list.

From:
"Igor Tandetnik" <itandetnik@mvps.org>
Newsgroups:
microsoft.public.vc.language
Date:
Thu, 7 Jan 2010 12:02:31 -0500
Message-ID:
<OtSS0s7jKHA.2184@TK2MSFTNGP04.phx.gbl>
bejiz <bruno.julier@wanadoo.fr> wrote:

class linked_list
{
public:
linked_list(): p_begin(0), p_end(0) {};
~linked_list(){};


You'd probably want to call clear() from destructor.

void clear( linked_list A)


Why does clear() take a parameter? It's a method of linked_list - =
shouldn't it work on the instance it's called on?

Also, it takes its parameter by value, meaning it works on a copy of =
linked_list. But it's a shallow copy - you now have two instances of =
linked_list whose p_begin and p_end members point to the same set of =
nodes. You delete those nodes and update p_begin and p_end in the =
temporary copy of linked_list, but the original remains unchanged: its =
p_begin and p_end members are now dangling pointers (pointing to data =
that's already been deallocated).

{
while(p_begin!=0)
{
                        Node* p_zap = new Node(0,0);
                        p_zap = A.p_begin;


You allocate a new instance of Node, make p_zap point to it - and then =
immediately make it point to something else. Thus, you leak that Node(0, =
0).

                        cout << " delete : " << p_zap->data << endl;
                        p_begin = p_begin->p_next;
                        delete p_zap;
                }


You probably want to set p_end = NULL here, otherwise it still points =
to a now-deleted last node.

void push_back( const int& a)


No need to pass int by reference, just make it

void push_back(int a)

--
With best wishes,
    Igor Tandetnik

With sufficient thrust, pigs fly just fine. However, this is not =
necessarily a good idea. It is hard to be sure where they are going to =
land, and it could be dangerous sitting under them as they fly overhead. =
-- RFC 1925

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