Re: Where is my virtual function???

From:

James Kanze <james.kanze@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 26 Jul 2010 02:50:56 -0700 (PDT)

Message-ID:

<466fc729-46d5-4c1c-9d28-8506d24bd299@5g2000yqz.googlegroups.com>

On Jul 24, 11:10 am, Maxim Rogozhin <max.rogoz...@gmail.com> wrote:

I've created the following class hierarchy in MS Visual Studio:

class B {
};

class D : public B {
public:
    virtual void g();
};

void D::g() {}

and created my object:
D* p = new D;

So far so good. I can see an "B" entry and __vfptr (const
D::'vftable') entry under 'p' entry in "Locals" window, and
I can see an [0x0] entry for function g() under __vfptr.

But if I add a virtual function f() in class B:
class B {
public:
    virtual void f();
};

void B::f() {}

class D : public B {
public:
    virtual void g();
    virtual void f();
};

void D::g() {}
void D::f() {}

then virtual function g() disappears from __vftbl. Moreover
- __vfptr entry has moved under the "B" entry in "Locals"
windows (it was under 'p' entry formerly).

How do you determine this? If it is with the debugger, it may
be just an artifact of the way the debugger displays
information. However...

So my question is where has my function g() got?? And why
__vfptr moves from 'p' to 'B' ?

This is all very implementation dependent, but in general, in
all of the implementations I know, your class D will only have
one vptr. (Multiple inheritance will introduce more.) In the
same way the compiler puts the data in B in front of the data
added by D, it will put the entries for B in the vtable in front
of those for D. The debugger is probably just displaying those
which are valid for both B and D as for B, and those only valid
for D as for D.

--
James Kanze