How to avoid code duplication for const versions of functions

Question

  pointer operator->()
  {
    return ptr_+buffer_position_;
  }

  const_pointer operator->() const
  {
    return ptr_+buffer_position_;
  }

So the const overload returns a const pointer (yay) but the two functions do the same thing (boo) how does one avoid code duplication and the ensuing copypasta, while still maintaining const correctness?

(I'm looking at specifically C++98...though a C++11 answer would be pedagogic, as I suspect they did something to solve this problem).

Although I prefer my C++11 answer to the duplicate. – Bathsheba May 22 '14 at 14:37 — Bathsheba, May 22 '14 at 14:37
See also http://stackoverflow.com/q/20533661/103167 – Ben Voigt May 22 '14 at 14:44 — Ben Voigt, May 22 '14 at 14:44

score 0 · Accepted Answer · answered May 22 '14 at 14:40

If you are not using c++11:

You can try this:

class T {
      pointer operator->()
      {
        return const_cast<pointer> (static_cast<const T*>(this)->operator->());
      }

      const_pointer operator->() const
      {
        return ptr_+buffer_position_;
      }
};

How to avoid code duplication for const versions of functions

1 Answers1

Linked