Re: Is destructor automatically be virtual in pure class?

From:

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Thu, 25 Aug 2011 05:32:23 -0700 (PDT)

Message-ID:

<j34sm4$td$1@dont-email.me>

On 2011-08-25 01:44, linq936 wrote:

Hi,
The question just came to me and could not find an answer.

Normally we do not declare constructor and destructor in pure class
and compiler generates them, but since the class is pure, compiler
should declare the destructor as virtual, isn't it?

No. You can have a pure class that does not have a virtual destructor. A
pure function just prevents that you can construct an object of this class.

class Base {
    public:
       virtual void do_sth() = 0;
};

class D {
    public:
       virtual void do_sth() {}

     ... some other things ...
};

int main()
{
   Base* p = new D();
   p->do_sth();
   delete p;
   return 0;
}

We write above sort of code a lot, if the destructor generated by
compiler is not virtual, there will be memory leak.

Can you confirm?

I can confirm that you might have a memory leak, if you rely on that
assumption ;-)

Note that pure class types are not necessarily constructed via new
expressions, e.g. we could write

int main()
{
    D d;
    Base* p = &d;
    p->do_sth();
}

In this example there is no need for a virtual destructor.

HTH & Greetings from Bremen,

Daniel Kr?gler

--
      [ See http://www.gotw.ca/resources/clcm.htm for info about ]
      [ comp.lang.c++.moderated. First time posters: Do this! ]