Uhm... $! and $? do not act differently with sudo thrown in the mix. The only real difference is that you are executing a different command (i.e. sudo rather than the command that sudo is executing.)

$> perl sudotest.pl
Password:
sudo: /foo/bar/paco: command not found
$ sudo /foo/bar/paco
ERRNO       =
CHILD_ERROR = 256
[download]

In that case, sudo executed just fine and returned a 1 (which is what sudo does when the command it is given can't be executed.) Remember to shift $?. (256 >> 8 == 1)

Can't exec "/foo/bar/paco": No such file or directory at sudotest.pl l
+ine 8.
$ /foo/bar/paco
ERRNO       = No such file or directory
CHILD_ERROR = -1
Use of uninitialized value in print at sudotest.pl line 12.
[download]

Again, normal behavior. Perl can't execute the command given so it sets $? to -1 and returns the reason in $! (as documented.)

$ ls -laF
ERRNO       = Illegal seek
CHILD_ERROR = 0
total 138
drwxrwxrwx   5 ignatz paco       1024 Jan 16 15:16 ./
drwxr-xr-x  21 ignatz paco       1536 Jan 16 15:34 ../
drwxrwxrwx   2 ignatz paco        512 Dec 18 15:30 CVS/
-rw-r--r--   1 ignatz paco        298 Jan 16 14:44 sudotest.pl
[download]

Once again, this behaved as expected. The command executed successfully, so $? contains 0. In this case, $! is not relevant.

Keep in mind that $! is only meaningful in association with backticks and system when $? == -1.

-sauoq
"My two cents aren't worth a dime.";

So, alas, there's no way to get the result of the command that sudo runs?

()-()
 \"/
  `

Well, as I showed in my other post, sudo does return the exit status of the command it runs.

Unfortunately, the case where the command returns 1 seems to be indistinguishable from the case where sudo returns 1. I guess that little bit of ambiguity could be a big bummer. You might be able to get around it on a case by case basis by using wrappers for your commands.

-sauoq
"My two cents aren't worth a dime.";