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From: Christopher Eltschka <celtschk@physik.tu-muenchen.de>
Subject: Re: pointer to member derivation question
Date: 1999/06/26
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Approved: Fergus Henderson <fjh@cs.mu.oz.au>
References: <vgvk8swxj97.fsf@ms.com> <7ktjmc$95f$1@nnrp1.deja.com>
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Jim Hyslop wrote:
> 
> In article <vgvk8swxj97.fsf@ms.com>,
>   Colin Rafferty <craffert@ms.com> wrote:
> > Hi-
> >
> > I have the following code, and it makes sense to me that it should
> > work, but different compilers have different thoughts on the subject.
> >
> >     class Base { /* ... */ };
> >
> >     class Derived : public Base { /* ... */ };
> >
> >     class Holder
> >     {
> >     public:
> >       Derived _derived;
> >     };
> >
> >     void f()
> >     {
> >       Base Holder::* elt = &Holder::_derived;
> >     }
> Er, I hate to point out the obvious, but doesn't elt require an *object*
> to work on, i.e.:
> 
> void f()
> {
>    Holder h;
>    Base Holder::* elt = &h._derived;
> }

No. &Holder::_derived gives a Derived Holder::*.
AFAIK your code wouldn't be legal at all, even if the obvious
conversion were allowed, since &h._derived is of type Derived*,]
not Derived Holder::*.

> 
> BTW, you should avoid using leading underscores in identifiers, since
> they are reserved for compiler and STL usage.

While I agree with the "should", you are not completely right
here. Only identifiers using double underscores and identifiers
beginning with underscore and capital letter are reserved in
every scope; all other identifiers beginning with a single
underscore are only reserved in global scope and in namespace
std.
However, the implementor of your compiler may have the
same misunderstanding, and therefore it's a good idea to
avoid them anyway.
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