Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!swrinde!sdd.hp.com!apollo!vinoski
From: vinoski@apollo.hp.com (Stephen Vinoski)
Newsgroups: comp.unix.shell
Subject: Re: adjusting string var to fixed length (in ksh)
Keywords: ksh, typeset
Message-ID: <1991Jun13.151112.15311@apollo.hp.com>
Date: 13 Jun 91 15:11:12 GMT
References: <9106111833.AA07449@ucbvax.Berkeley.EDU> <193@atesysv.UUCP>
Sender: netnews@apollo.hp.com (USENET posting account)
Organization: Hewlett-Packard Company, Apollo Division - Chelmsford, MA
Lines: 28
Nntp-Posting-Host: bull.ch.apollo.hp.com

In article <193@atesysv.UUCP> lanzo@atesysv.UUCP (Mark Lanzo) writes:
>"Mark" == Mark Lanzo <lanzo@atesysv.UUCP>    [that's me]
>"Dan"  == Dan_Jacobson@ihlpz.ATT.COM 
>
>From prior posts:
>    Mark> If you are using "ksh" [...]
>    
>    Mark> 	myname=Mark
>    Mark> 	typeset -L10 myname
>    
>Dan>    In my prompt I wanted to put $? (last command exit status) in a fixed
>Dan>    width... (PS1='$? '...) but I can't typeset -L3 ? or \?...
>
>Hmm.  As far as I know, you're out of luck with this.
>You're right, "typeset" doesn't work on special parameters like "$?"; 
>it just complains that "?" isn't a valid identifier.

If your ksh is newer than the 06/86 version, how about this:

  typeset -L3 st=0
  trap 'st=$?' DEBUG
  PS1='$st '


-steve

| Steve Vinoski  (508)256-0176 x5904       | Internet: vinoski@apollo.hp.com  |
| HP Apollo Division, Chelmsford, MA 01824 | UUCP: ...!apollo!vinoski         |