Re: Operator overloading and access to private members

"Bo Persson" <>
Sat, 16 Feb 2008 10:48:48 +0100
pelio wrote:

Christopher dixit:

On Feb 15, 11:28 am, obaqueiro <> wrote:

Hello, reading the Thinking in C++ book, i came into this code

#include <...>...
// code ommited

class Integer {
       int i;

          Integer (int ii) {i = ii}

         const Integer operator+ (const Integer & rv) const {
                       return Integer (i+rv.i); //
------ Isn't rv.i not visible ?? XXX

      const Integer operator= (const Integer & rv){...
          //code ommited}

int main (){
    Integer I(1), J(2), K(3);
   K = I+J;

// This does not compile... of course
// cout << K.i;



This indeed compiles (in GCC 4). What I can't fully understand
why, in the line market with the XXX, the function access rv.i
where i is a private member of the class Integer. Isn't rv.i
supposed to be inaccessible ? if I try to access K.i in main (the
commented code) the compiler does throw a "Integer::i is private"
error. I would expect the same in the other case.

So I think I am missing something, and the specific question
would be why does this happens? is there any special "visibility"
for the members of the parameters provided when overloading an
operator (i.e., is it possible to see all the private members of
the parameters?) Thank you!

private members of a class are certainly visible to the class
itself, otherwise wouldn't they be left in limbo and be
untouchable just taking up space that is never used?

Yes this.privateMember should be but why notThis.privateMember
should ? You did not read the post.

I think he did.

The thing is exactly that notThis is of the same type as this. Members
of a class have access to all other members of the same class, they
are siblings.

This doesn't break encapsulation, because it is the designer of the
class that decides how the class should act. You cannot do this from
the outside.

Bo Persson

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