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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

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.

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

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.";

In reply to Re: Re: Re: Backticks, $?, and Sudo by sauoq
in thread Backticks, $?, and Sudo by ignatz

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