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To: Paul Flinders <paul@dawa.demon.co.uk>
Cc: Paul Flinders <P.Flinders@ftel.co.uk>, Jeff Mcadams <jeffm@iglou.com>,
        Rik van Riel <riel@conectiva.com.br>,
        John Kodis <kodis@mail630.gsfc.nasa.gov>,
        "Richard B. Johnson" <root@chaos.analogic.com>,
        linux-kernel@vger.kernel.org, bug-bash@gnu.org
Subject: Re: binfmt_script and ^M
In-Reply-To: <20010305095512.A30787@tux.gsfc.nasa.gov>
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	<jeae70m97e.fsf@hawking.suse.de> <3AA3EEDF.D0547D4@dawa.demon.co.uk>
X-Yow: Oh my GOD -- the SUN just fell into YANKEE STADIUM!!
From: Andreas Schwab <schwab@suse.de>
Date: 06 Mar 2001 11:41:09 +0100
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Paul Flinders <paul@dawa.demon.co.uk> writes:

|> Andreas Schwab wrote:
|> 
|> > This [isspace('\r') == 1] has no significance here.  The right thing to
|> 
|> > look at is $IFS, which does not contain \r by default.  The shell only splits
|> 
|> > words by "IFS whitespace", and the kernel should be consistent with it:
|> >
|> > $ echo -e 'ls foo\r' | sh
|> > ls: foo: No such file or directory
|> 
|> The problem with that argument is that #!<interpreter> can be applied
|> to more than just shells which understand $IFS, so which environment
|> variable does the kernel pick?

The kernel should use the same default value of IFS as the Bourne shell,
ie. the same value you'll get with /bin/sh -c 'echo "$IFS"'.  This is
independent of any settings in the environment.

|> It's a difficult one - logically white space should terminate the interpreter

No, IFS-whitespace delimits arguments in the Bourne shell.

Andreas.

-- 
Andreas Schwab                                  "And now for something
SuSE Labs                                        completely different."
Andreas.Schwab@suse.de
SuSE GmbH, Schanz?ckerstr. 10, D-90443 N?rnberg
Key fingerprint = 58CA 54C7 6D53 942B 1756  01D3 44D5 214B 8276 4ED5
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