From: Devin Mullins <twifkak@...>
Date: 2005-06-18T09:08:29+09:00
Subject: Re: Why do arrays work this way?

As in many modern languages (Java, most popularly, but I'd imagine 
Python, too), everything's a pointer, so to speak. If you want a copy, 
you should say b=a.dup, instead. Note, however, that not even that's 
perfect.

While this works:

irb(main):001:0> a=[1,2,3,4,5]
=> [1, 2, 3, 4, 5]
irb(main):002:0> b=a.dup
=> [1, 2, 3, 4, 5]
irb(main):003:0> b[2]=7
=> 7
irb(main):004:0> a
=> [1, 2, 3, 4, 5]
irb(main):005:0> b
=> [1, 2, 7, 4, 5]
irb(main):006:0>

There is still a pitfall if the Objects that your Array contains are 
mutable (changeable):

irb(main):006:0> a=["hello","dog","cat"]
=> ["hello", "dog", "cat"]
irb(main):007:0> b=a.dup
=> ["hello", "dog", "cat"]
irb(main):008:0> b[2]="feline"
=> "feline"
irb(main):009:0> b[1].reverse!
=> "god"
irb(main):010:0> a
=> ["hello", "god", "cat"]
irb(main):011:0> b
=> ["hello", "god", "feline"]
irb(main):012:0>

Note that b[2]=... makes the new separate b Array point to a brand new 
String object, while b[1].reverse! modifies the existing String object 
-- the one that both a and b both point to. This is because Array#dup 
(and in general Object#dup) is a shallow copy. In other words, 
Object#dup is not "recursive" on the Objects inside.

Hope that helps, and hope it doesn't kill your buzz. :)

Devin