I have a Perl script which forks and daemonizes itself. It's run by cron, so in order to not leave a zombie around, I shut down STDIN,STDOUT, and STDERR:
open STDIN, '/dev/null' or die "Can't read /dev/null: $!";
open STDOUT, '>>/dev/null' or die "Can't write to /dev/null: $!";
open STDERR, '>>/dev/null' or die "Can't write to /dev/null: $!";
if (!fork()) {
do_some_fork_stuff();
}
The question I have is: I'd like to restore at least STDOUT after this point (it would be nice to restore the other 2). But what magic symbols do I need to use to re-open STDOUT as what STDOUT used to be?
I know that I could use "/dev/tty" if I was running from a tty (but I'm running from cron and depending on stdout elsewhere). I've also read tricks where you can put STDOUT aside with open SAVEOUT,">&STDOUT"
, but just the act of making this copy doesn't solve the original problem of leaving a zombie around.
I'm looking to see if there's some magic like open STDOUT,"|-"
(which I know isn't it) to open STDOUT the way it's supposed to be opened.
# copy of the file descriptors
open(CPERR, ">&STDERR");
# redirect stderr in to warning file
open(STDERR, ">>xyz.log") || die "Error stderr: $!";
# close the redirected filehandles
close(STDERR) || die "Can't close STDERR: $!";
# restore stdout and stderr
open(STDERR, ">&CPERR") || die "Can't restore stderr: $!";
#I hope this works for you.
#-Hariprasad AJ