Re: What three steps in overloading the arrow operator?

Bart van Ingen Schenau <>
Mon, 28 Jan 2013 08:09:11 +0000 (UTC)
On Sun, 27 Jan 2013 13:25:54 -0800, fl wrote:


In other words, we want to call the result of evaluating point->action.
The compiler evaluates this code as follows:

1. If point is a pointer to a class object that has a member named
action, then the compiler writes code to call the action member of that

2. Otherwise, if point is an object of a class that defines operator->,
then point->action is the same as point.operator->()->action, That is,
we execute operator->() on point and then repeat these three steps,
using the result of executing operator-> on point.

3. Otherwise, the code is in error."

I do not know the last part of item 2. I do not find any snippet code
using ->display. I do not find any part mentioning 3 steps. Thus, I
cannot solve this myself. Could you help me out? Thanks,

The three steps mentioned in item 2 are the three numbered items I quoted
The operator-> is special, because it's application can be repeated
without this being directly observable in the code.
For example:

#include <iostream>

struct Foo {
  void print() { std::cout << "Hello from Foo" << std::endl; }

struct Bar {
  Foo mFoo;
  Foo* operator->() { return &mFoo; }

struct Baz {
  Bar mBar;
  Bar& operator->() { return mBar; }

int main()
  Foo aFoo, *pFoo = &aFoo;
  Bar aBar;
  Baz aBaz;

  pFoo->print(); /* Item 1: print is a member of Foo */
  aBar->print(); /* Item 2: use operator-> */
  aBaz->print(); /* Item 2 repeated: use operator-> twice */

This code compiles and prints "Hello from Foo" three times.

Bart v Ingen Schenau

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