From: Brian Candler <b.candler@...>
Date: 2010-08-18T18:36:31+09:00
Subject: Re: question on if / defined?()

Here's a simpler example:

  if false
    x = 123
  end
  puts x   # nil

x is defined, and has value nil. What has happened?

When the code is *parsed* (before it is executed at all, just when Ruby 
is building an in-memory representation of the program), if Ruby sees an 
assignment like x = ... then it marks x as being a local variable, 
reserving a slot for it on the stack.

Whether the assignment is actually executed later or not is irrelevant. 
From this point onwards to the end of the current scope, x is defined, 
and is a local variable.

Why does ruby do this? There is a local variable / method call 
ambiguity. Because you don't declare variables in Ruby, and because a 
method call doesn't need parentheses after it, a bare "x" could be 
either retrieving the value of x, or it could be calling the method x 
with no arguments.

This simple parse-time algorithm decides in advance which you meant, 
without you having to declare it.

    puts x              # undefined local variable or method `x'

    def x
      "Method x"
    end
    puts x              # Method x

    x = 123
    puts x              # 123

At this point it's decided that 'x' is a local variable, although you 
can still force a method call instead:

    puts x              # 123
    puts self.x         # Method x
    puts x()            # Method x
-- 
Posted via http://www.ruby-forum.com/.