Re: Operator overloading and access to private members

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

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

class Integer {
       int i;

public:
          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!