Re: Inherited destructor behavior clarification
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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