Hello.

The question are:
Shall access modifiers affect overloading resolution
for derived class ?
Can i define member function with the same name as one access
modifier ajusts ?
Is it possibble to ajust access to one of base class overloded
methods ?

The example is:

struct A {
        int f (short) { printf ("A::f(short)\n"); }
        int f (int) { printf ("A::f(int)\n"); }
};

struct B: public A {
        int f (char) { printf ("B::f(char)\n"); }
        /* using */ A::f;
        int f (long) { printf ("B::f(long)\n"); }
};

main ()
{
        int a;
        char b;
        short e;
        long g;

        B c;

        c.f(a);
        c.f(b);
        c.f(e);
        c.f(g);
}

GNU complains about access modifier and method have the same
name. I have not found corresponding rules in the Jan 96 WP.

Thank you in advance.
--
Alexander N. Krotoff
Research Computer Center
Moscow State University



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Alexander,

It was my understanding that B::f completely hides
A::f even if the names have different signatures.
Maybe someone could help me with a reference as to
where this is written in the std?

I had to change the code to:

main ()
{
        int a;
        char b;
        short e;
        long g;

        B c;

        c.A::f(a);
        c.f(b);
        c.A::f(e);
        c.f(g);
}

to get it to run in Visual C++.

James J. Couball
Enable, Inc.



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> It was my understanding that B::f completely hides
> A::f even if the names have different signatures.
> Maybe someone could help me with a reference as to
> where this is written in the std?

I was wrong, asking the question. I missed one paragraph in the
Section 11. Everything is well defined in the section
"Using declaration" (7.3.3).

B::f do not completely hides A::f.

--
Alexander


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