Re: Virtual calls upon destruction
Jeremy Jurksztowicz wrote:
Nope; by the time operator delete is called, all destructors will be
executed and bptr will point to uninitialized memory.
Hmm... I am curious as to how this occurs? Is this mandated by the
standard (seems likely)? Is there a hidden, behind the scenes operation
that calls T::~T()? In retrospect this seems obvious, given that an
untyped (void) pointer is passed to operator delete. I am imagining
operator delete gets implemented something like this (pardon my
ignorance):
void _impl_delete (_impl_delete_ptr_type * ptr, std::size_t sz)
{
_impl_call_destructor(ptr);
// Calls correct delete operator
_impl_call_delete(
reinterpret_cast<void*>(ptr), sz);
}
Now might such a function be called by a C++ implementation 'behind the
scenes', so to speak?
It sounds like you're confusing operator delete (the memory
deallocation function which can be overridden globally or at the class
level) with the delete operator (which calls the destructor of an
object, then calls operator delete to free the object's memory, and is
not overridable).
Assuming we have a pointer p that points to a T object, this code:
delete p;
gets compiled as if it were written:
p->~T();
T::operator delete(p);
(If T has no operator delete(), then the global operator delete is used
instead.)
You can't change the behavior of the delete operator. The only thing
you can do is try to force users to make extra calls before deleting.
class Base
{
public:
void deleteMe();
virtual void uninitialize() = 0;
private:
static void operator delete(void*); // Thanks to James Kanze for
reminding me of this...
};
void Base::deleteMe()
{
this->uninitialize();
delete this;
}
Now, the only way to delete a pointer to a Base is to used
Base::deleteMe(), which calls the uninitialize() member function and
then deletes the object.
(As James pointed out, a derived class could always make a public
operator delete which could then be used to delete a derived class
object, but I would classify that as a misuse of Base.)
Bob
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