Date: Sun, 9 Nov 1997 16:19:08 +0100 (MET) From: Wolfgang Helbig <helbig@Informatik.BA-Stuttgart.DE> To: rkw@dataplex.net (Richard Wackerbarth) Cc: joerg_wunsch@uriah.heep.sax.de, freebsd-hackers@FreeBSD.ORG Subject: Re: Why doesn't /bin/echo use getopt? Message-ID: <199711091519.QAA04460@rvc1.informatik.ba-stuttgart.de> In-Reply-To: <l03110701b08b4c1e60fd@[208.2.87.4]> from Richard Wackerbarth at "Nov 9, 97 05:17:39 am"
next in thread | previous in thread | raw e-mail | index | archive | help
> 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
Want to link to this message? Use this URL: <https://mail-archive.FreeBSD.org/cgi/mid.cgi?199711091519.QAA04460>