Re: const correct member variable access thru function?

29 Sep 2006 09:00:58 -0400

The proper way to do it (IMO) would be to use a proxy class that would
know whether it's on the right or on the left side of the assignment op.
It's not as simple as introducing two overloaded functions, like here:

Thanks, I cant believe i did not think of overloading..

because in this case if 'A' is non-const, then the non-const version
will be called if you use it in an expression (even though you're not
trying to assign to it):

That is interseting. So if I wanted to log read and write access to the
variable then overloading would not work.. hmm, I had a go at making a
proxy object emulating the variable, instead of having functions
emulating the variable. The same problem seems to creep up though. That
is, if Z is a non const object and I do a 'int tmp = z.x;' then the non
const assignment operator is called even though this is a "read only"
operation. Anyone got any tips or links to related threads?

Example code (take 2):

 template <class T>
  struct Proxy {

  Proxy(T &x): x_ref_(x) {} //init ref

  //assignment operator..
  Proxy<T>& operator = (const T& rhs) {x_ref_ = rhs; /*cout<<"non-const
lhs";*/ return *this;}

  //conversion operator..
  operator T& () { /*cout<<"non-const rhs";*/ return x_ref_; }
  operator const T& () const { /*cout<<"const rhs";*/ return x_ref_; }

  T &x_ref_;

 struct A{
  int x;

 struct Z {
  int x_;
  Proxy<int> x;

  Z(): x_(0), x(x_) {;}

 void main() {

  int tmp;

  A a = A();
  a.x = 1;
  tmp = a.x;

  const A a_const = A();
  tmp = a_const.x;

  Z z = Z();
  z.x = 1;
  //tmp = z.x; // error C2593: 'operator =' is ambiguous. // ehh!?

  Z z = Z();
  z.x = 1;
  int tmp = z.x; // calls non-const operator=() // calling non const
even though this is read only !?

  const Z z_const = Z();
  int tmp = z_const.x; // calls const operator =

Regards Patrik

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