Re: Is destructor automatically be virtual in pure class?
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).
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?
You'll have memory leak if Derived's destructor has memory to free.
Goran.
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