using declarations and const overloads

Question

Suppose I have two versions of operator-> (overloaded on const) in a base class. If I say

using Base::operator->;

in a derived class, will I get access to both versions or just the non-const one?

Answer 1

Same business as name hiding. It's all or nothing. Using declarations (7.3.3) bring a name, not a member.

ISO/IEC 14882 (2003), 7.3.3. 1/ A using-declaration introduces a name into the declarative region in which the using-declaration appears. That name is a synonym for the name of some entity declared elsewhere.

I encourage you to read 7.3.3, there are subtle things inside. You cannot using-declare a template, all the members refered by the name you using-declare must be accessible, the names are considerd for overload resolution alongside the names in the block the using declaration is found (ie. they don't hide anything), etc, etc.

Answer 2

You get access to all versions of a method/operator with the same name in that parent.

Answer 3

both. did you try it? (damn this answer is short: ah well, here is example:

#include <iostream>
#include <string>

struct bar
{
  void foo() { std::cout << "non_c:foo()" << std::endl; }
  void foo() const { std::cout << "c:foo()" << std::endl; }
};

class base
{
public:
  bar* operator->() { return &b; }
  bar const* operator->() const { return &b; }

private:
  bar b;  
};

class derived : public base
{
public:
  using base::operator->;
};


int main(void)
{
  const derived d = derived();
  derived e;

  d->foo();
  e->foo();
}