Re: Inherited destructor behavior clarification

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 3 Mar 2011 06:09:23 CST
Message-ID:
<07ca8c0e-7d8b-4c2c-bec7-0f4e5af88789@s4g2000vbq.googlegroups.com>
On Mar 2, 11:23 pm, Kevin Brandon <kjbron...@gmail.com> wrote:

On Mar 2, 5:52 pm, James Kanze <james.ka...@gmail.com> wrote:


      [...]

So, in the second example, declaring "base"'s destructor virtual, but
deleting one of the most derived types through a pointer to an
arbitrary type in inheritance hierarchy, I believe, should not trigger
ub.


Once a function is virtual, it's always virtual. Even if the
function is a destructor (which formally doesn't have a name).

class base {
public:
 virtual ~base() { }
};

class secondBase : public base {
public:
};

class thirdBase : public secondBase {
public:
};

class derived : public thirdBase {
public:
};

void foo() {
 thirdBase* p = new derived;
 delete p;
}

or should all types declare destructors as virtual, particularly
thirdBase?


They all have virtual destructors, so you don't have to worry.

--
James Kanze

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