Value of $? after unsetting non-existent shell variable

["Kevin B. McCarty" <kmccarty@Princeton.EDU>]

Hi all,

I just noticed something interesting.  Let qqqqq represent some shell
variable that isn't defined.  Then, on a system running sid:

bash$ unset qqqqq ; echo $?
0
bash$ zsh
zsh% unset qqqqq ; echo $?
1
zsh% dash
$ unset qqqqq ; echo $?
0
$ posh
$ unset qqqqq ; echo $?
1
$ sh		# sh -> bash
sh-3.00$ unset qqqqq ; echo $?
0

Also, the version of bash (and sh, if pointing to bash) in woody returns
1 instead of 0.

This seems like it could lead to problems in "set -e" maintainer scripts
and other things.  In my case (maybe I'm stupid?), I was happily using
"unset EXIT" in a shell script, only to find that it mysteriously failed
on woody, which prompted this bit of research.  The shell script in
question is part of a Makefile in the cernlib build scripts.  If bash
hadn't changed semantics between woody and sid, I don't know if my
sponsor or I would have caught this, potentially leading to an FTBFS on
some systems.  This is a real-world problem.

So, who's right and who's wrong here?  And if either return value is
acceptable, could it at least be made consistent between different shells?

regards,

-- 
Kevin B. McCarty <kmccarty@princeton.edu>   Physics Department
WWW: http://www.princeton.edu/~kmccarty/    Princeton University
GPG public key ID: 4F83C751                 Princeton, NJ 08544