Path: utzoo!utgpu!news-server.csri.toronto.edu!rutgers!mcdchg!tellab5!balr!clrcom!rmartin
From: rmartin@clear.com (Bob Martin)
Newsgroups: comp.lang.c++
Subject: Re: friend operator +(l,r) vs. operator +(r)
Message-ID: <1990Nov22.230835.26713@clear.com>
Date: 22 Nov 90 23:08:35 GMT
References: <11759@hubcap.clemson.edu> <1990Nov21.053431.22340@actrix.co.nz>
Organization: Clear Communications, Inc.
Lines: 54

In article <1990Nov21.053431.22340@actrix.co.nz> Bruce.Hoult@bbs.actrix.gen.nz writes:
>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.
>
>
>The important difference is that the friend version allows implicit
>conversions to be made on both the left hand and right hand arguments
>to +, wheras the member function version allows implicit conversions
>on only the RH argument.

Let me elaborate on this a bit.  Let us say that you have created
a constructor as follows:  foo::foo(int).  This tells the compiler
that it is allowable to convert an int into a foo.

Now lets say that you use the following code:

foo fa,fb;
int ia;
...
fb=ia+fa;  // adding an int to a foo.

If you use the foo::operator+(foo&) method, then the above statement
will produce a compiler error since foo::operator+(foo&) requires that
the left hand side of the + operator must be a foo.  But if you use
the friend operator+(foo&,foo&) then the compiler knows how to convert
ia into a foo before 'adding' fa.


-- 
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