Re: C2248: cannot access protected member

From:

"Ben Voigt [C++ MVP]" <rbv@nospam.nospam>

Newsgroups:

microsoft.public.vc.language

Date:

Thu, 31 Jan 2008 15:53:53 -0600

Message-ID:

<OSKM0NFZIHA.208@TK2MSFTNGP02.phx.gbl>

"Alex Blekhman" <tkfx.REMOVE@yahoo.com> wrote in message
news:ex7dUDFZIHA.5784@TK2MSFTNGP03.phx.gbl...

"Ben Voigt [C++ MVP]" wrote:

No, I meant what I wrote

void (Y::*pf)() = &X::foo;

which is a fully-bound (non-virtual) member function, calling the exact
function X::foo on any instance of Y.

I don't know if the C++ compiler will accept that syntax, it should
because it is type safe. if Y is a subtype of X, then Y -> unit is a
supertype of X -> unit.

Y can access the base implementation of protected member foo, so it
should be able to create a pointer-to-member that does so.

Actually, I started this thread because the C++ compiler doesn't accept
that syntax. Igor cited relevant part of the Standard that explicitly
forbids it and demonstrated with short example why it forbids such sintax.

Igor's argument applies to

void (X::*pf)() = &X::foo; // inside Y::bar

not what I wrote which is

void (Y::*pf)() = &X::foo; // inside Y::bar

My version, which may or may not be accepted by the compiler, is perfectly
typesafe.

Of course the following will certainly work and provide an identical result:

class Y : X
{
    void fooshim() { X::foo(); }
public:
    void foo() override;
    void bar()
    {
        void (Y::*pf)() = &Y::fooshim;
    }
};

Alex