Re: Why can derived member function not access protected member of a base class object?

From:

Victor Bazarov <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Fri, 26 Sep 2008 08:54:29 -0400

Message-ID:

<gbim25$627$1@news.datemas.de>

blangela wrote:

[..]
Also, I changed the member function to:

void Derived::Method4(Base & B_Param)
{
    Derived * DPtr = dynamic_cast <Derived *> (&B_Param); // this cast
          // creates a Derived class pointer to B_Param, if and only if,
          // B_Param actually is Derived class object, otherwise the
          // pointer will be set to 0

    if (DPtr != 0) // if B_Param is actually a Derived class object
        cout << DPtr->mDerived << endl;
    else
    {
        Base B_obj = B_Param;
        cout << B_obj.mBase << endl;
    }
}

And I still have the same errors:

1>c:\documents and settings\blangela\desktop\polymorphismtest
\derived.cpp(29) : error C2248: 'Base::mBase' : cannot access
protected member declared in class 'Base'
1> c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(17) : see declaration of 'Base::mBase'
1> c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(7) : see declaration of 'Base'

I would have thought that B_obj can only be a Base object now (and not
some subclass of Base), that it would now be allowed?

You changes the run-time behaviour of your program, but not the logic
for that particular statement. The rules of the language do not depend
on the run-time behaviour. The compiler is not going to verify the
run-time conditions when it checks the semantics of an expression.
Access is prohibited, period. It does not matter whether the *actual*
type of the super-object is the same or not because in general it isn't.

V
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