From: Jonah Bloch-Johnson <jsb2142@...>
Date: 2008-08-20T11:38:55+09:00
Subject: Re: Nested Arrays in Method Definitions--a puzzle

Many many many thanks, Adam and Martin! I do think this clears it up.

Quoting Adam Shelly <adam.shelly@gmail.com>:

> On 8/19/08, Jonah Bloch-Johnson <jsb2142@columbia.edu> wrote:
>
>> I am writing because I am somewhat stumped by why the following   
>> discrepency occurs...
>
> Variables and Array entries hold references to objects.
>
> irb(main):001:0> x=[1]
> => [1]
> irb(main):002:0> x.object_id
> => 20774910
>
> irb(main):003:0> y=[];y<<x
> => [[1]]
> irb(main):004:0> y[0].object_id
> => 20774910    #the same object that x referrs to
>
> irb(main):005:0> x=[2]
> => [2]
> irb(main):006:0> x.object_id
> => 21222360   #now x refers to the newly created array
>
> irb(main):008:0> y
> => [[1]]
> irb(main):007:0> y[0].object_id
> => 20774910  #but y still holds a reference to the old one.
>
> irb(main):009:0> y=[x]
> => [[2]]
> irb(main):010:0> y[0].object_id
> => 21222360   #y[0] is now a reference to the same Array as x
>
> irb(main):011:0> x[0]=4
> => 4
> irb(main):013:0> y
> => [[4]]    #so changes to the content of that Array are visible
> through y's reference.
>
>
> hope this helps,
> -Adam
>
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>>        x = [1]
>
> [1] is an Array object, x is a variable referring to that array
>>        y = []
> y referrs to  a 2nd Array
>>        y << x
>
> And now that array contains a reference to
>
>>        x = [2] # (compare this)
>
> Here you are creating an array, labeling it 'x', and storing a
> reference to it in the array labeled y.  Then you reassign the 'x'
> label to yet another array (the one containing 2).
>
>>  end
>>  def havenots
>>        x = [1]
>>        y = []
>>        y << x
>>        x[0] = 2 # (to this)
>>        y
>>  end
>> end
>>
>> I get this in irb:
>>
>> >
>> > > load 'snah.rb'
>> > >
>> >
>> => true
>> >
>> > > a = Warfare.new
>> > >
>> >
>> => #<Warfare:0x33a93c>
>> >
>> > > a.haves
>> > >
>> >
>> => [[1]]
>> >
>> > > a.havenots
>> > >
>> >
>> => [[2]]
>>
>> Why does meddling with the innards of the "x" have retroactive
>> repercussions, whereas redefining it outright does not?
>>
>> Any insight would be greatly appreciated.
>>
>> -Jonah
>>
>>
>>
>
>