Why the 30-second pause after executing this script?

[Jilles Tjoelker]

On Mon, Apr 12, 2004 at 05:04:34PM -0400, Alex Vasylenko wrote:
> Kirk Strauser wrote:
> >to attach to a jail to execute /etc/rc.shutdown.  I've noticed that this
> >works as expected, unless its output is being piped into another program, 
> >in
> >which case 'sh' waits about 30 seconds after the 'exit 0' line is executed
> >before closing its end of the pipe.  For example:

> /etc/rc.shutdown kills the _shell_ of the watchdog timer instead of the 
> timer itself, so the timer keeps going. Please try the patch attached.
> -	sleep $rcshutdown_timeout && ( 
> +	exec sleep $rcshutdown_timeout && ( 

This means that the part after && is never executed, invalidating the
whole purpose of the watchdog!

It would be easiest if sh had a builtin similar to the alarm(3) C
function, but I can think of complicated setups with trap and wait.

Another option is to close the stdin/stdout/stderr in the watchdog (this
doesn't seem very satisfying though):

diff -u -r1.25 rc.shutdown
--- etc/rc.shutdown     8 Jul 2003 02:52:14 -0000       1.25
+++ etc/rc.shutdown     12 Apr 2004 20:55:51 -0000
@@ -66,7 +66,7 @@
 _rcshutdown_watchdog=
 if [ -n "$rcshutdown_timeout" ]; then
 	debug "Initiating watchdog timer."
-	sleep $rcshutdown_timeout && (
+	exec 0<>/dev/tty 1>&0 2>&1 && sleep $rcshutdown_timeout && (
 		_msg="$rcshutdown_timeout second watchdog" \
 		    " timeout expired. Shutdown terminated."
 		logger -t rc.shutdown "$_msg"

Note that I haven't tested anything of this.

Also note that the behaviour of 'time cmd1 | cmd2' differs quite a lot
between shells.

-- 
Jilles Tjoelker