From: Gennady Bystritsky <gennady-ruby@...>
Date: 2012-06-21T16:49:22+09:00
Subject: Re: New to Ruby - Scope Question


On Jun 21, 2012, at 12:39 AM, Emeka Patrick wrote:

> x = 10
> y = 5
> 
> def square(x)
>  puts "within the method x is " + x.to_s
>  puts x*x
> end

Here, you define method square but do not invoke it yet. 'x' here is a method parameter that will take the value provided in a method invocation. It does not have anything to do with 'x' in 'x = 10'. 

For example, when you invoke this method as:
  square(25)

x in x.to_s and x*x will be 25. And outer scope x will still be 10 after the method returns.

Gennady.

> 
> 
> In the above when defining the method, the x that is being referenced in
> "x.to_s" and puts x*x is by default the variable x, defined above, with
> a value of 10, correct?  This then means that even though this variable
> isn't defined within the method it can still be called on from outside
> the code. However, if the variable was given a value within the method
> then it wouldn't be available outside of the method, correct?
> 
> Can someone explain why this is so. I guess I don't NEED to know why,
> but I'd like to understand it a bit better if possible. Thanks!
> 
> -- 
> Posted via http://www.ruby-forum.com/.
>