Re: C++ Pitfall: const objects do not behave constantly.

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <>
Wed, 26 Sep 2007 08:59:56 GMT
On 2007-09-26 06:27, Kira Yamato wrote:

On 2007-09-25 20:00:23 -0400, Erik Wikstr??m <> said:

On 2007-09-26 00:08, Kira Yamato wrote:

On 2007-09-25 17:32:26 -0400, Erik Wikstr??m <> said:

On 2007-09-25 22:37, Kira Yamato wrote:

So to me, declaring an object 'const' is really just a mechanism to
allow a certain subset of the object's member functions to be called
(those that are declared 'const') and disallow calling the other member
functions, (those that are not declared 'const'). No other conditions
can be assumed.

I would hardly call this a pitfall, and I have yet to meat anyone with a
bit of OO programming skill who fail to grasp the concept of constant
objects (namely that their state are constant/unchangeable).

What is the state of an object? A state of an object is supposed to be
a set of data that determines the behavior of an object. In another
word, knowing the state of an object would dictate what its functions
do given some inputs. Otherwise, what is the point of knowing an
object's state?

So, if you say that declaring an object 'const' should keep its state
constant, then its member functions should yield the same output for
the same input since the object remains in the same state!

However, this is not the case as shown in the example.

You would be right provided that the results (return values and side
effects) of the member functions were totally dependent on the object's
state. In the example above this is not the case, and there is thus no
need to for the results to be the same. Consider the following:

Yes, this is a good point. And this is what I'm saying too. The idea
of keeping the state of an object constant is not guarantee by 'const'

It is not? Exactly which state can be changed in a function marked as const?

#include <iostream>

struct Foo
  int bar(int i) const
    return i;

int main()
  const Foo f;
  std::cout << << "\n";
  std::cout << << "\n";

By your reasoning both calls to bar() should return the same thing, but
that would be totally illogical.

Actually it is logical here. The object remains same state here. So,
everytime I invoke with 1 I get back 1; and everytime I invoke 2 I get
back 2 here. The singleton state of this object dictates that the
function bar be an identity function on its input.

I can actually not see any difference between my example and the one you
gave, I have just moved the source of the randomness from inside the
function to outside. When it comes to the object's state they are equal
(in that the function's results do not depend on the state of the object).

Erik Wikstr??m

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