From: Emeka Patrick <lists@...>
Date: 2012-06-21T17:27:01+09:00
Subject: Re: New to Ruby - Scope Question

Hi Stefano & Gennady,

Thanks for writing back and for your explanations.

I just looked over what I wrote and figured out what happened. When I 
was playing around with this in IRB, I accidentally set both x and y to 
have a value of 10 so when I ran the method square with the argument y, 
I got the same results as I got for argument x and got really confused 
about local variable scope.

Thanks again for taking the time to explain, would have walked away with 
some seriously misguided understanding of local variables.

Stefano Crocco wrote in post #1065482:
> On Thursday 21 June 2012 Emeka Patrick wrote
>> "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/.
>
> No, the x being referred to in the "x.to_s" expression is not the one
> you set
> to 10 but the one you gave as argument to the square method.
>
> When you define a method, you create a new scope. This means that local
> variables defined outside the method won't be availlable inside the
> method
> body. Also, local variables defined inside the method body (including
> the
> arguments of the method) won't be availlable outside the method itself.
>
> On the other hand, when you define a method which takes some arguments,
> you
> implicitly introduce a local variable for each argument in the scope of
> the
> method body. When the method is called, this variables will be given the
> values you pass as arguments to the method call.
>
> All this means that your method would work exactly in the same way even
> if you
> called the argument with another name (provided, of course, that you
> change
> all references to it to use the new name). For example:
>
> def square(z)
>   puts "within the method z is " + z.to_s
>   puts z*z
> end
>
> I hope this helps
>
> Stefano

-- 
Posted via http://www.ruby-forum.com/.