A doubt on assignment operator.

[Amit Bhatia]

Hi,
I am trying to define the assignment operator for a class called
Cell. I am bit confused between the following two statements:

const Cell& Cell::operator= (const Cell &r)
{
\\ all the stuff

}

and,

Cell& Cell::operator= (const Cell &r)
{
\\ all the stuff

}

Is the first work going to work at all? does it make any sense? I think
it is not going to do what it is intended to do.

thanks,
--a.

[Moritz Beller]

Re: A doubt on assignment operator.

On 26 Sep 2005 11:55:12 -0700
"Amit Bhatia" <amit.bhatia@gm ail.com> wrote:
[color=blue]
> Is the first work going to work at all? [/color]

Try it out for yourself.
[color=blue]
> does it make any sense? [/color]

Well, let's try and figure out what the definition const Cell&
Cell::operator= (const Cell &r) actually means: It says that the
assignment operator shall return a constant reference of type Cell.
It's usually of no great importance whether you have both a const and
non-const assignment operator -- you will bearly use the object
returned by it.

It does however make sense for an (merely used) if-statement such as
if((Cell1=Cell2 ).changeContent ("a"))
when you don't want Cell1's content to be changed.
[color=blue]
> I think it is not going to do what it is intended to do.[/color]

You don't mention what you intend to do, hence I can't tell you. FYI:
Cell& Cell::operator= (const Cell &r) const is certainly rubbish.

best regards / Gruß
Moritz Beller
--
web http://www.4momo.de
mail momo dot beller at t-online dot de
gpg http://gpg.notlong.com

[Victor Bazarov]

Re: A doubt on assignment operator.

Amit Bhatia wrote:[color=blue]
> I am trying to define the assignment operator for a class called
> Cell. I am bit confused between the following two statements:
>
> const Cell& Cell::operator= (const Cell &r)
> {
> \\ all the stuff
>
> }
>
> and,
>
> Cell& Cell::operator= (const Cell &r)
> {
> \\ all the stuff
>
> }
>
> Is the first work going to work at all? does it make any sense? I think
> it is not going to do what it is intended to do.[/color]

The difference between the two is the return value type, right? The
return value type is different only in the const-ness of the object to
which the returned reference points, right? So, why would the first
one *not work*? When you write the expression

someCell = someOtherCell;

it's the same as

someCell.operat or=(someOtherCe ll);

which basically _discards_ the return value. Whatever happens between
passing the argument and returning a value, IOW, whatever happens between
the curly braces of the function, is the same, no? So, how it may not
work if they are doing exactly the same?

V

[Moritz Beller]

Re: A doubt on assignment operator.

On Mon, 26 Sep 2005 21:13:05 +0200
Moritz Beller <momo.beller. No-Spam@t-online.de> wrote:
[color=blue]
> It's usually of no great importance whether you have both a const and[/color]
xx --> or[color=blue]
> non-const assignment operator -- you will bearly use the object[/color]

Who put the "both ... and" in there? ;)

best regards / Gruß
Moritz Beller
--
web http://www.4momo.de
mail momo dot beller at t-online dot de
gpg http://gpg.notlong.com

[Greg]

Re: A doubt on assignment operator.


Victor Bazarov wrote:[color=blue]
> Amit Bhatia wrote:[color=green]
> > I am trying to define the assignment operator for a class called
> > Cell. I am bit confused between the following two statements:
> >
> > const Cell& Cell::operator= (const Cell &r)
> > {
> > \\ all the stuff
> >
> > }
> >
> > and,
> >
> > Cell& Cell::operator= (const Cell &r)
> > {
> > \\ all the stuff
> >
> > }
> >
> > Is the first work going to work at all? does it make any sense? I think
> > it is not going to do what it is intended to do.[/color]
>
> The difference between the two is the return value type, right? The
> return value type is different only in the const-ness of the object to
> which the returned reference points, right? So, why would the first
> one *not work*? When you write the expression
>
> someCell = someOtherCell;
>
> it's the same as
>
> someCell.operat or=(someOtherCe ll);
>
> which basically _discards_ the return value. Whatever happens between
> passing the argument and returning a value, IOW, whatever happens between
> the curly braces of the function, is the same, no? So, how it may not
> work if they are doing exactly the same?
>
> V[/color]

The original poster probably can't tell whether either function works
because they won't compile. The two functions differ only by their
return type - making one an illegal overload of the other if both are
declared.

Greg