From: Robert Klemme <shortcutter@...>
Date: 2008-04-03T22:29:17+09:00
Subject: Re: nested ruby

2008/4/3, Sunil Chikkegowda <ycsunil@gmail.com>:
> Hello All,
>
>  Take a look at this example:
>
>  def method1
>   puts 'hello'
>   def method2
>     puts 'world'
>   end
>  end
>
>  In IRB, if I say 'method1' it gives 'hello' as output. For
>  'method1.method2' it gives 'hello' in first line and 'world' in second
>  line.
>
>  My question is: Is there any way to print only 'world' if I execute
>  method1.method2

You seem to misunderstand what method1.method2 does: it is not an
access to a nested method definition (because those do not exist in
Ruby) - rather two methods are called one after the other.  If you
only want method2 you can do it:

irb(main):001:0> def method1
irb(main):002:1>  puts 'hello'
irb(main):003:1>  def method2
irb(main):004:2>    puts 'world'
irb(main):005:2>  end
irb(main):006:1> end
=> nil
irb(main):007:0> method2
NameError: undefined local variable or method `method2' for main:Object
        from (irb):7
        from :0
irb(main):008:0> method1
hello
=> nil
irb(main):009:0> method2
world
=> nil
irb(main):010:0>

But in your example there is no point in defining method2 via
executing method1 - at least I cannot see any.

Kind regards

robert

-- 
use.inject do |as, often| as.you_can - without end