From: Shannon Fang <xrfang@...>
Date: 2002-11-28T06:54:37+09:00
Subject: Re: call-by-reference problem again

Another problem about reference is that,
if a is an object, let b=a, then b is a copy of the object or a pointer
to a? I encountered the following problem

a=b=c=Array.new
d=e=f=""

In my program, if I modify a, b and c will be affected, in another word,
a, b, c point to the same Array. while I modify d, e and f are not
affected, which means, strings are assigned by value... what will happen
if I write d=e=f=String.new?

Since everything in Ruby is object, I don't understand why String and
Array are different...

Shannon

On Thu, 28 Nov 2002 06:45:26 +0900
Dave Thomas <Dave@PragmaticProgrammer.com> wrote:

> William Djaja Tjokroaminata <billtj@y.glue.umd.edu> writes:
> 
> > Shannon Fang <xrfang@hotmail.com> wrote:
> > > The problem here is, we have seen some Ruby methods, like String.chomp!
> > > and many that end with a !, they can modify parameters in place, how is
> > > it done?? Is it possible to do this in programming, or is it only
> > > available in standard libraries, which may be implemented outside of
> > > ruby (eg. in C)?
> > 
> > Hi,
> > 
> > In general, for the Ruby built-in classes such as String, the methods are
> > indeed implemented in C.  For your own classes, probably it suffices to
> > say that you cannot modify 'self' (such as 'self = something') in Ruby,
> > but it is possible do that through the Ruby C API's.
> 
> That's not really the point. A ! method modifies the state of an
> object, not its value of 'self'. For example
> 
>    class Counter
>      attr_reader :count
> 
>      def initialize(value=0)
>        @count = value
>      end
> 
>      def succ
>        Counter.new(@count.succ)
>      end
> 
>      def succ!
>         @count += 1
>         self
>      end
>     end
> 
> 
>     a = Counter.new
>     p a.count     #=> 0
>     b = a
>     c = a.succ
>     p b.count     #=> 0
>     p c.count     #=> 1
>     c = a.succ!
>     p b.count     #=> 1
>     p c.count     #=> 1
> 
> 
> It's the same for Strings. Mutators change the state of the object:
> the object reference stays the same.
> 
> 
> 
> Dave