From: Eric Mahurin <eric_mahurin@...>
Date: 2005-05-07T23:22:19+09:00
Subject: Re: object reference handle (like perl's reference to scalar)

I see you modified my example.  This doesn't do what the
traditional reference/pointer does in other languages.  With a
traditional reference/pointer you do something like this:

C                 Ruby with Eric's reference.rb
--------------    -----------------------------
p = &var          p = ref{:var}
*(p)=value        p[]=value          # changes var
*(p)==var         p[]==var

What you did just appears to create a wrapper around objects to
allow you to make mutable objects out of immutable ones.  I
think this is useful, but not a reference/pointer.

But, I am still taking the idea to create a reference like
this:

x.ref

This will make a reference that uses replace/become to set the
object.  It cannot genernally be used - especially for
immediates.

Here are the various ways my code can create a reference now:

ref{:x} # reference to the variable x
x.ref   # reference to the object in x
ref{"x[1]"} # reference to the assignable expression x[1]
x.ref[1] # reference to the methods [] and []= with 1 as an arg
x.ref("[]",1) # same
x.ref([:"[]",1],[:"[]=",1]) # same - independet get/get
Reference.new(proc{x[1]},proc{|v|x[1]=v}) # explicit get/set

With the exception of the first 2, the rest of the references
will behave the same way - operating on x[1].


--- Dave Burt <dave@burt.id.au> wrote:
> > # 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"




		
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