From: Rasputin <rasputin@...>
Date: 2003-06-14T08:01:40+09:00
Subject: Re: does each work on a copy?

* Jason Creighton <androflux@remove.to.reply.softhome.net> [030612 19:44]:
> On Thu, 12 Jun 2003 23:14:37 +0900
> Rasputin <rasputin@shrike.mine.nu> wrote:
> 
> > > remember that pretty much everything is a reference and you won't go far 
> > > wrong
> > 
> > I thought so, until this happened!
> > The array was full of references, so I thought el held the
> > reference from the Array.
> > 
> > It seems to hold a copy of the reference...
> 
> No, it doesn't. When you do a:
> 
> a = "Hello world, nice to meet you!".split(' ')
> a.each { |word| word = "lala" }
> p a
> ["Hello", "world,", "nice", "to", "meet", "you!"]
> 
> You're binding word to a different object, not changing the objects themselves. If you did this:

By 'copy of the reference' I meant:

ary = %w( a b c d e)
ary.each { |r| r = "X" }

r is given a copy of the 'reference' (apology for the Perlism),
so when you point (ugh, and the Cism) r at "X" you're updating a
copy of the reference, not the original reference. Is that about right?

And in the below code, 

string2 = "Hi" 
is just pointing string2 to a new object, whereas
string2[0] = "J"
is editing the object through the string2 reference, just
as word.upcase! below is doing, as opposed to the |el| el = SPACE
in my original post.


Sorry for the list noise, think I've got it now.
Thanks again for all replies.


> a = "Hello world, nice to meet you!".split(' ')
> a.each { |word| word.upcase! }
> p a
> ["HELLO", "WORLD,", "NICE", "TO", "MEET", "YOU!"]
> 
> It's the same thing that happens when you do this:
> 
> string1 = "Hello, world!"
> string2 = string1
> string2 = "Hi!"
> p string1
> "Hello, world!"
> 
> You haven't changed the object that string1 points to: You've made a new string object ("Hi!") and bind string2 to that. If you did a:
> 
> string1 = "Hello, world!"
> string2 = string1
> string2[0] = "J"
> p string1
> "Jello, world!"
> 
> Here you call the []= method of the object that string2 points to,(no rebinding of string2) which means that string1 and string2 still point to the same object.
> 
> Jason Creighton
> 

-- 
Rasputin :: Jack of All Trades - Master of Nuns