Re: Can i get the class of an object of which i only have the address?
red floyd wrote:
Tom1s # hilidhe wrote:
Virtual functions are achieved via a V-Table. That is, each object has a
hidden pointer in it that points to a static structure that contains
function pointers. (The Standard doesn't necessitate that this is how
it's done, but this *is* how it's done).
Wrong. Virtual functions *may be* achieved by a vtable. Nothing in the
Standard mandates it. Anything code that assumes a vtable is by
definition non-portable.
Hm, you make it sound as though the paragraph is a justification of the
initial "Wrong". However, upon closer examination, it turns out that it
isn't: Tom1s # hilidhe makes it very clear that the standard does not
necessitate a v-table implementation; he just claims that this
implementation happens to be chosen by all C++ implementations. In order to
argue that he is wrong you need to name an implementation that does not use
a v-table implementation.
Also, the standard does not define "portable". In particular, should Tom1s #
hilidhe be correct in the assesment that _all_ implementations use a
v-table, then code assuming that might be portable (in the obvious sense
that it works as documented/expected without change on all platforms). An
example of such code could be this:
template < typename T >
class is_polymorphic {
struct D : public T {
virtual ~D (void ) {}
};
public:
static bool const value = ( sizeof(T) == sizeof(D) );
};
It is based upon the idea that making the destructor virtual would increase
the size of a non-polymorphic class. Your claim that such code is not
portable could be empirically false (at least with regard to this instance
and according to my understanding of "portable" which might be different
from yours). All we can say a priory is that such code relies upon behavior
not determined by the standard.
Best
Kai-Uwe Bux