From: "Jesús Gabriel y Galán" <jgabrielygalan@...>
Date: 2010-04-02T18:24:20+09:00
Subject: Re: scope question

On Fri, Apr 2, 2010 at 10:25 AM, Harry Kakueki <list.push@gmail.com> wrote:
>>
>> a_hash, an_arr and a_num are local to the method indeed. However, a_hash and an_array hold references to hash and array
> object respectively, not copies. So you are changing the objects passed to this method via those references. On the other hand,
> a_num is passed a Fixnum instance which are implemented as immediate objects and hence passed as copy.
>>
>
> def scope_q(an_arr)
>  #an_arr[1] = 5  #This is a reference.
>  an_arr = [6,6] #This is a local an_arr ?
> end
>
> a = [1,2,3,4]
>
> scope_q(a)
>
> p a  #> [1, 2, 3, 4]

To clarify this: variables are references to objects. When you say this:

an_array[1]= 5

you are calling method "[]=" of the object referenced by an_array.
This object happens to be referenced also in the outer scope of the
method scope_q, in your example by variable a. So, if the method "[]="
modifies the object, that modification is seen outside when you
reference the object with variable a. On the other hand, when you do
this:

an_arr = [6,6]

you are not calling any method on the object previously referenced by
an_arr. You are assigning a new object to the local variable an_arr.
This does not affect in any way variable "a" in the outer scope, or
the object referenced by a.

Jesus.