Path: utzoo!utgpu!jarvis.csri.toronto.edu!clyde.concordia.ca!uunet!amgraf!cpsolv!rhg
From: rhg@cpsolv.UUCP (Richard H. Gumpertz)
Newsgroups: comp.unix.questions
Subject: use of set in a shell script
Message-ID: <472@cpsolv.UUCP>
Date: 4 Jan 90 21:46:12 GMT
Reply-To: rhg@cpsolv.uucp (Richard H. Gumpertz)
Organization: Computer Problem Solving, Leawood, Kansas
Lines: 39

I recently came across a shell script that, among other things, did the
following:
	OPTIONS=
	DIRS=
	for ARG
	do
		case "$ARG" in
		-*)	OPTIONS="$OPTIONS $ARG";;
		*)	DIRS="$DIRS $ARG";;
		esac
	done
	if test -z "$DIRS"; then
		DIRS="."
	fi
	set $DIRS
	find $@ -type f -exec ...
where the "..." in the find command indicates irrelevant stuff that I have
omitted.

My question is, why do the set in the next to last line?  Isn't the above
roughly equivalent to (but slower than)
	find $DIRS -type f -exec ...
where the set command has been deleted?  The only difference I can see is that
the version with the set command will make one more passing unquoting things,
which could have been avoided using
	set $DIRS
	find "$@" -type f -exec ...
Are there any other subtle differences that I missed?  Is there any way to
avoid the one pass of unquoting things that seems to remain in either case?

Is there a better way to write this whole thing?  (No, perl is not an
acceptable alternative in this case; perl scripts will be read but will not
solve my problem.)

-- 
  ==========================================================================
  | Richard H. Gumpertz    (913) 642-1777 or (816) 891-3561    rhg@CPS.COM |
  | Computer Problem Solving, 8905 Mohawk Lane, Leawood, Kansas 66206-1749 |
  ==========================================================================