From: Dave Burt <dave@...>
Date: 2005-05-07T21:44:23+09:00
Subject: Re: object reference handle (like perl's reference to scalar)

"Eric Mahurin" <eric_mahurin@yahoo.com>:
> Where do I get your ref.db?  I saw a link in another post of
> yours and it didn't work.

Sorry, my fault for posting a link that didn't exist yet.
http://www.dave.burt.id.au/ruby/ref.rb is there now.
It may disappear when I become sufficiently ashamed of it.

>
> What does this do?
>
> # silly example as x,y = y,x would be better
> ...

C:\>ruby
require 'ref'
def swap(a,b)
    tmp = a[]
    a[] = b[]
    b[] = tmp
end
x = 1.ref
y = 2.ref
swap(x, y)
p x
p y
x = "hi".ref
y = "world".ref
z = x
swap(x, y)
p x
p y
p z
^Z
2
1
"hi"
"world"
"hi"

> If this doesn't work, I don't see the real use of the way you
> did it.  If on the other hand you use evil.rb/become or replace
> to assign the dereferenced object, you could at least get the
> second case to work (and other non-immediates).  If it does
> work, great.  I want to know how the immediate case is handled.

I'm not being evil. My Ref is just an extra wrapper for an object.

> For now, I'm using replace or become (if available to the
> object) to accomplish the above dereferencing (to an object)
> assignment.  And I still have all the other types of references
> (to a variable, assignable expression, attribute/member,
> get/set methods, etc) in place.  I need to document what I
> have, package it into a gem and upload to rubyforge.
>
> I really don't like the idea of your references looking like
> the underlying object.  This just confuses things.  A reference
> is a reference and you can get/set the object that it
> references.  If you want to get info about the object that it
> references, just get it and query it by getting it from the
> reference.  Why muddy what the reference is?

You're right, there's no real advantage to be gained here. You might as well 
use
  Ref = Struct.new(:deref)
That provides a wrapper for an object - just treat it as if it's name ends 
in ".deref":
  def swap(a, b)
    a.deref, b.deref = b.deref, a.deref
  end

Cheers,
Dave