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From: chris@mimsy.UUCP (Chris Torek)
Newsgroups: comp.unix.questions
Subject: Re: Regular Expression delimiters
Message-ID: <16874@mimsy.UUCP>
Date: 12 Apr 89 05:02:29 GMT
References: <993@n8emr.UUCP> <6710@bsu-cs.bsu.edu> <TALE.89Apr11192859@imagine.pawl.rpi.edu>
Distribution: comp
Organization: U of Maryland, Dept. of Computer Science, Coll. Pk., MD 20742
Lines: 45

>In article <6710@bsu-cs.bsu.edu> dhesi@bsu-cs.bsu.edu (Rahul Dhesi)
>suggests something like
>>    pattern="`echo "$ans" | sed -e 's/\//\\\\\\//g'`"

In article <TALE.89Apr11192859@imagine.pawl.rpi.edu> tale@pawl.rpi.edu
(David C Lawrence) writes:
>pattern="`echo $ans | sed 'sX/X\\\/Xg'`"

Actually, sed 'sX/X\\/Xg' suffices (two backslashes, not three).
(I also prefer a comma as the separator: sed 's,/,\\/,g'.)

>a) quoting $ans is only necessary if "-n" could lead it.  If you want
>to allow for that you need to quote it as "`echo \"$ans\" ...`"

This makes little difference.  Quoting $ans is necessary if and only
if it contains special characters (space, tab, newline, and globbing).
If the string is "-n", echo will still break; worse, if it contains
backslashes, it will break on SysV, where echo does escape interpretation
(echo should never have acquired a -n flag either; that job should
have been left to printf(1).)

It is true that if "$ans" is "-n foo", then

	echo $ans

produces

	foo

without a newline, while

	echo "$ans"

produces

	-n foo

(with a newline).  But

	echo "-n"

produces nothing.  There is no workaround for this.
-- 
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7163)
Domain:	chris@mimsy.umd.edu	Path:	uunet!mimsy!chris