What's the return status of a pipeline???

[Jangling Neck Nipper]

I wrote a filter for lpr some time ago, but it's never worked
properly, and I was wondering if somebody might be able to
help me out.

The printcap entry is:

lp|ps|apple|lwriter|Apple Laserwriter:\
	:lp=/dev/null:if=/usr/local/bin/IF_lwriter:lf=/usr/adm/lprlog:\
	:sd=/usr/spool/lpd/apple:af=/usr/adm/lprlog:\
	:pw#102:pl#78:mx#0:

The filter program IF_lwriter is:

-------------------------------------

#!/bin/csh
#
# [-c] -w<width> -l<length> -i<indent> -n <user> -h <host> <accounting file>
#
# filter used by lpr -P[ps|apple] &c.

setenv LWRITER "LaserWriter Plus"
set BIN = /usr/local/bin

if ("$1" == "-c") then
	set user    = $6
	set host    = $8
	set acct    = $9
else
	set user    = $5
	set host    = $7
	set acct    = $8
endif

#
# Log this print...
#
echo ""                               >> $acct
date                                  >> $acct
echo $user@$host printing to $LWRITER >> $acct

#
# catch signals...
#
onintr quit

#
# DO IT!!!
#
$BIN/psfilter -b $* | $BIN/lwriter -pI "$LWRITER" $* >>& $acct

#
# Say what happened
#
quit:
set printer_status = $status
switch ($printer_status)

case 0:
	echo job was sucessful >> $acct
	breaksw

case 1:
	echo job failed and will be reprinted >> $acct
	breaksw

case 2:
	echo job deleted >> $acct
	breaksw

default:
	echo unknown status $printer_status >> $acct
	breaksw

endsw

exit $printer_status

-------------------------------------

It is the `lwriter' filter that *should* be determining the status,
namely 0, 1 or 2, and I've tried dummy shell scripts and found
that it's always the filter at the *end* of the pipe that determines
the status, so this should be okay, but in fact all I ever get is
status `0', no matter *what* return status I made `psfilter' or
`lwriter' return.  Is there anything fundamentally wrong with
the filter???
-- 
  ,---.'\ 
 (  /@ )/    Nothing is true.        
   /( _/  )  Everything is permitted.
   \,`---'