Question

Consider:

#include <iostream>

using namespace std;

struct A {
  virtual void f() { cout << "A::f" << endl; }
  virtual void f() const { cout << "A::f const" << endl; }
};

struct B : public A {};

struct C : public A {
   virtual void f() { cout << "C::f" << endl; }
};


int main()
{
   const B b;
   b.f();   // prints "A::f const"

   const C c;
   c.f();
   // Compile-time error: passing ‘const C’ as ‘this’ argument of
   //   ‘virtual void C::f()’ discards qualifiers
}

(I’m using GCC.)

So it seems that the const version of f() gets hidden in C. This makes a lot of sense to me, but is it mandated by the standard?

Answer 1

I will (once more) link this great article :

First, [the compiler] looks in the
immediate scope, in this case the
scope of class C, and makes a list of
all functions it can find that are
named f (regardless of whether they’re
accessible or even take the right
number of parameters). Only if it
doesn’t does it then continue
“outward” into the next enclosing
scope […]

So yes, the const version of f is hidden, and that’s perfectly normal. As pointed out by Simone, you can use a using statement to bring A::f in C scope.