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From: Jim Hyslop <jim.hyslop@leitch.com>
Subject: Re: pointer to member derivation question
Date: 1999/06/25
Message-ID: <7ktjmc$95f$1@nnrp1.deja.com>#1/1
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Approved: Valentin Bonnard <bonnard@clipper.ens.fr>
References: <vgvk8swxj97.fsf@ms.com>
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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;
}

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

--
Jim
I ignore all email from recruitment agencies.
Please do not send me email with questions - post
here.


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