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From: Wolfgang Helbig <helbig@Informatik.BA-Stuttgart.DE>
Message-Id: <199711091519.QAA04460@rvc1.informatik.ba-stuttgart.de>
Subject: Re: Why doesn't /bin/echo use getopt?
In-Reply-To: <l03110701b08b4c1e60fd@[208.2.87.4]> from Richard Wackerbarth at "Nov 9, 97 05:17:39 am"
To: rkw@dataplex.net (Richard Wackerbarth)
Date: Sun, 9 Nov 1997 16:19:08 +0100 (MET)
Cc: joerg_wunsch@uriah.heep.sax.de, freebsd-hackers@FreeBSD.ORG
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> At 4:50 AM -0600 11/9/97, J Wunsch wrote:
> >As James Raynard wrote:
> >
> >> Because it's supposed to repeat its arguments instead of parse them?
> >> (with the exception of -n, of course).
> >
> >Tough question: how do you echo a "-n\n" then (without using
> >printf(1), of course)?
> 
> echo -n "-n\
> "

Wrong for sh(1). The newline is swallowed by the shell.
>From the manual page:
"A backslash preceding a \n is treated as a line continuation"

$ echo -n "-n
> "

seems to be an answer to this week's question.

Wolfgang