From: Jacob Fugal <lukfugl@...>
Date: 2006-12-09T09:07:36+09:00
Subject: Re: beginner Q: Kernel#puts, STDOUT, $stdout relation

On 12/8/06, Andreas S <andreas_s@hotmail.com> wrote:
> >From: Daniel Finnie <danfinnie@optonline.net>
> >Reply-To: ruby-talk@ruby-lang.org
> >To: ruby-talk@ruby-lang.org (ruby-talk ML)
> >Subject: Re: beginner Q: Kernel#puts, STDOUT, $stdout relation
> >Date: Sat, 9 Dec 2006 07:28:20 +0900
> >
> >Puts is a method of Kernel.
>
> But, doesn't it depend on what IO object $stdout holds? I redefined $stdout
> 'puts' function, but why it does not affect Kernel's puts method?

It's a tricky relationship, and I'm not quite sure how or why it works
this way, but Kernel#puts does not use STDOUT's (or $stdout, they can
be different) puts. It does use STDOUT/$stdout (I'm not really sure
which) but IIRC Kernel#puts is implemented directly using the write
method. Caveat here though -- the write method is called *twice*, once
for the argument, then again for the newline. So modifying your source
to rewrite "write" instead of "puts":

  class << STDOUT
    def write(*args)
      args[0] = "I say " + args[0] unless args.empty?
      super(args)
    end
  end

  puts "hello"
  STDOUT.puts "hello"
  $stdout.puts "hello"

produces:

  $ ruby test.rb
  I say helloI say
  I say helloI say
  I say helloI say

Jacob Fugal