Re: Why can't 'vptr' be static?

Alberto Ganesh Barbati <>
Wed, 5 Nov 2008 08:25:26 CST
irotas ha scritto:

I was reading over the C++ FAQ item here:

In that FAQ, the author explains that the 'vtable' is a static table -
that is, there's one per class that needs a virtual table.

The FAQ warns that this topic is "entirely compiler-dependent". Well,
the warning is a bit of an understatement, as C++ has no concept of
vtable at all, compilers are allowed to use any other means as long as
they provide the correct behaviour.

However, he explains, there is a non-static 'vptr' pointer to the
vtable as a hidden member variable of all classes (that need a virtual

Naively, since all instantiations of a class will have the exact same
'vptr', it seems like it would make sense to have a single static
'vptr' per class.

Of course, this isn't really possible, else they would have done it
already. However, I'm at a loss as to why exactly it isn't possible.

Could someone please enlighten me?

Consider this:

 struct Base
   virtual void f();

 struct Derived
   virtual void f();

 int main()
   Base* pb1 = new Base();
   Base* pb2 = new Derived();
   pb1->f(); // shall call Base::f()
   pb2->f(); // shall call Derived::f(), not Base::f()

How does the line pb1->f() calls Base::f() while pb2->f() calls
Derived::f()? The two lines are perfectly identical from the "static"
point of view, so they must rely on some information stored in a
*non-static* member of the objects in order to determine their
most-derived type. If the hidden vptr were static you would still need
to have a non-static member, so it's much easier to make the vptr itself



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