Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!usc!snorkelwacker.mit.edu!bloom-beacon!eru!hagbard!sunic!mcsun!ukc!mucs!jk
From: jk@cs.man.ac.uk (John Kewley ICL)
Newsgroups: comp.lang.c++
Subject: Re: friend operator +(l,r) vs. operator +(r)
Message-ID: <1939@m1.cs.man.ac.uk>
Date: 23 Nov 90 10:00:34 GMT
References: <11759@hubcap.clemson.edu>
Sender: news@cs.man.ac.uk
Reply-To: jk@cs.man.ac.uk (John Kewley ICL)
Organization: Department of Computer Science, University of Manchester UK
Lines: 56

In article <11759@hubcap.clemson.edu> grimlok@hubcap.clemson.edu (Mike Percy) writes:
>I've tried and tried to get a handle on this question from various
>sources, but haven't been enlightened...
> 
>What semantic/operational and/or stylistic differences are there
>between these two definitions?
> 
>class foo {
>...
>public:
>   foo& operator +(foo& rhs);
>}
> 
>and
> 
>class foo {
>...
>public:
>  friend foo& operator +(foo& lhs, foo&rhs);
>}
> 
>I must be missing something, because to me these effectively are the
>same thing, but the first can operate on this.
> 

If needed, the first definition can be a private or protected function. This
is not possible with a friend.

Implicit type coersions are not permitted on the LHS of the non-friend version.
e.g. 1 + x is not allowed for the member but would be valid for the friend,
assuming a coersion from int -> foo.
 
Note that operator= (and [], (), ->) must be defined using the first method.

The second method should be used when there is a need for the parameters to be
ordered with the object as the second parameter -
operator<< usually needs to be defined using the friend syntax when it is
overloaded with the output function.

A friend can also be used for a function to operate on two objects of
different classes. It would be declared as friend within each class.

If the function + is defined in C (or another language), the friend version
would be neccessary.

In the non-operator case, friends have a different syntax (less object-oriented)
so functions such as print can be called as print(foobar) rather than 
foobar.print().

All this being said it is often unclear whether to declare a function as a
friend or a member. Where there is no over-riding need I prefer members.
--
        J.K.
 
John M. Kewley, ICL, Wenlock Way, West Gorton, Manchester. M12 5DR
Tel:   (+44) 61 223 1301 X2138  Email: jk@r6.cs.man.ac.uk / jk@nw.stl.stc.co.uk