From: John Feminella <johnf@...>
Date: 2011-02-25T15:17:47+09:00
Subject: Re: parsing rule for this code?

As a rule of thumb, variables exist until you reach the "end" or
closing brace of the innermost block that they're still contained in.
In your case, you have this:

>    def do_something
>        3.times do |x|
>            number = 10
>        end
>        puts number
>    end

The local variable `number` is created with the value 10 and then
immediately discarded. This happens three times, since the block
executes three times.

The goal of the next statement is to print a local variable called
`number`. But there is no such local variable in this block. So Ruby
rightfully complains that you didn't define it. If instead you had
written this:

>    def do_something
>        number = 0 # create `number` in this lexical scope
>        3.times do |x|
>            number = 10
>        end
>        puts number
>    end

then it would work as you probably expect.

~ jf
--
John Feminella
Principal Consultant, BitsBuilder
LI: http://www.linkedin.com/in/johnxf
SO: http://stackoverflow.com/users/75170/



On Thu, Feb 24, 2011 at 22:41, 7stud -- <bbxx789_05ss@yahoo.com> wrote:
> 1)
> class ABC
>    def do_something
>       number = 10
>       puts number
>    end
> end
>
> ABC.new.do_something
>
> --output:--
> 10
>
>
> 2)
> class ABC
>    def do_something
>        3.times do |x|
>            number = 10
>        end
>        puts number
>    end
> end
>
> ABC.new.do_something
>
> --output:--
> Line 6:in `do_something': undefined local variable or method `number'
> for #<ABC:0x401bfaa4> (NameError)
>  from t.rb:10
>
>
> Why does ruby get confused by the setter v. local variable assignment
> when adding a block?
>
> --
> Posted via http://www.ruby-forum.com/.
>
>