Re: Is destructor automatically be virtual in pure class?
On 25/08/2011 14:57, Goran wrote:
On Aug 25, 1:44 am, linq936<linq...@gmail.com> 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. Counter example: I often use pure virtual classes to often
"interfaces", Java-style. In this case, destructor that is called is
coming from another part of inheritance tree and virtual destructor in
pure virtual base isn't needed.
class D {
public:
virtual void do_sth() {}
... some other things ...
};
int main()
{
Base* p = new D();
p->do_sth();
delete p;
return 0;
}
Here, you have a possible bug: you have a pointer to base, and when
you call delete, compiler will call destructor. But it will not do it
virtually because Base has none (even if you add one to Derived).
Of course Base has a dtor, just because the programmer has not written
it does not mean there isn't one. I think the OP is wondering about
whether the computer generated dtor should not be virtual. I know we
have discussed this many times over the last two decades and some people
have good reasons why this is not the case.
In C++11 you can write (in the base class))
virtual ~Base = default;
and the compiler will generate the dtor for Base just as it would if you
had said nothing, except that the dtor is now virtual.
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