From: Ryan Davis <ryand-ruby@...>
Date: 2010-04-02T16:55:01+09:00
Subject: Re: scope question


On Apr 2, 2010, at 00:39 , Harry Kakueki wrote:

> Maybe I'm missing something obvious here, but...
> Does this result surprise anyone else or just me?
> I was expecting a_hash and an_arr to be local to the method.
> Did I just do something stupid?

nope. that's how it works. Every reference (variable, name, whatever you want to call it) to an object (with some exceptions) is a pointer to the actual object. When you call scope_q(h,a,n) you are passing those references into the method and they're modified directly.

> def scope_q(a_hash,an_arr,a_num)
>  a_hash["b"] = 3
>  an_arr[1] = 5
>  a_num += 1
> end

> h = {"a"=>1,"b"=>2,"c"=>3,"d"=>4}
> a = [1,2,3,4]
> n = 7
> 
> scope_q(h,a,n)
> 
> p h  #> {"a"=>1, "b"=>3, "c"=>3, "d"=>4}
> p a  #> [1, 5, 3, 4]
> p n  #> 7

The obvious counter-example is here with a_num/n. Small integers (fixnums) are optimized so they're not a pointer to a 32 bit number and are instead a 30 bit number (2 bits used for flags for special cases). so 7 gets copied in instead of passing a reference to it. As a result, the copy gets modified, not the original value at 'n'.