From: "Mauricio Fernández" <batsman.geo@...>
Date: 2003-06-12T23:01:46+09:00
Subject: Re: does each work on a copy?

On Thu, Jun 12, 2003 at 10:46:45PM +0900, Rasputin wrote:
> SPACE = 1
> 
> a = Array.new(5)
> b = Array.new(5, SPACE)
> 
> a.each { |el|
>         el = SPACE
> }

You are *not* modifying the elements inside a.
 a.each { |el|    
works like the following:
 a.each { el = corresponding_element_from_a

If you assign to el you'll only change the reference (ie. what the local
is pointing to), not the underlying object.

However the following would work
a.each_with_index { |el,i| a[i] = SPACE }
(it is ugly and inefficient, though)
 
> puts "a = #{a.join("-")}"
> puts "b = #{b.join("-")}"
> 
> ----------------8<----------------
> 
> I'd expect to have a first output line similar to the second,but I get:
> 
> rasputin@shrike rasputin$ ruby /tmp/q
> a = ----
> b = 1-1-1-1-1
> rasputin@shrike rasputin$
> 
> Now, is this because each is returning a copy of the element?
It is giving you a *reference* to the element.
Consider:

>> a = %w{BLA FOO BAZ}
=> ["BLA", "FOO", "BAZ"]
>> a.each{ |x| x.downcase! }
=> ["bla", "foo", "baz"]
>> a
=> ["bla", "foo", "baz"]


> It's the most plausible explanation
> (because if I replace a.each with a.collect! it works as I expected)
> but if so I'm surprised it hasn't bitten me on the ass before.

collect! works because the value of 'el = SPACE' is SPACE, and collect
takes the value of the block in each iteration.

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