From: Robert Klemme <bob.news@...>
Date: 2004-08-05T04:26:52+09:00
Subject: Re: YAML.load(ARGF)


"Austin Ziegler" <halostatue@gmail.com> schrieb im Newsbeitrag
news:9e7db91104080410463cea4c22@mail.gmail.com...
> On Thu, 5 Aug 2004 02:36:29 +0900, Hans Fugal <hans@fugal.net> wrote:
> > Consider these two short programs:
> >         ARGF.read
> > and
> >         require 'yaml'
> >         YAML.load(ARGF)
> >
> > Then call them like so:
> >
> >         $ ruby foo.rb foo
> >
> > The first program prints the contents of foo and exits. The second
> > program waits for ^D from the terminal, and then tries to parse the YAML
> > and go on with life.
> >
> > I want ARGF to behave like it does in the first example, in the second
> > example. Of course I could do YAML.load(ARGF.read) but that doesn't
> > satisfy my curiousity. :-)
>
> IMO, YAML.load *should* do this with ARGF because the following works:
>
>   f = File.open("foo", "rb")
>   YAML.load(f)
>   f.close

Not so fast: ARGF is quite special.  It's especially *no* an IO.  It just
happens to implement some methods of IO.  IMHO it's not reasonable to invoke
YAML.load(ARGF) because ARGF might draw its data from any number of files -
including stdin.  IMHO it's not very practical to distribute YAML data
across multiple files.  The case is differnt for grep like tools that can
reasonable operate on a multitude of files.

Just my 0.02EUR...

Kind regards

    robert