1. I expected to see the date in seconds since time epoch, but result is
variable name
# execlineb -Pc 'backtick D { date "+%s" } echo $D'
$D
Normal behaviour, since there's no shell to interpret $D as the
contents of variable D. Try using "importas D D" before the echo:
it will read the value of D and substitute $D with this value, so
echo will print the value. Yeah, execline is annoying like that, it's
just a habit to take.
Also, you generally want "backtick -n", to chomp the newline at
the end of your input.
---
2. When I use emptyenv within an execlineb script, I have a "defunct"
zombie process
89685 3 S< 0:00.01 |-- s6-supervise base:time-srv
3020 - S<s 0:00.03 | `-- /usr/local/sbin/ntpd -c /etc/ntp.conf
-N -g -u ntpd --nofork
3601 - Z< 0:00.00 | `-- <defunct>
The time server script is
#!/usr/local/bin/execlineb -P
emptyenv
multidefine -d " " "base time ntpd /usr/local/sbin/ntpd" { JAIL SERVICE
USER PROGRAM }
background { echo Starting service $SERVICE using $PROGRAM on $JAIL
under user $USER }
fdmove 2 1
redirfd -w 1 /m/base:time/fifo
$PROGRAM -c /etc/ntp.conf -N -g -u $USER --nofork
removing emptyenv, prevents the zombie from being created. Is this normal?
The zombie is the echo program in your background block, since it's a
direct child of your run script and there's nothing that reaps it
after it's forked (fdmove, redirfd, ntpd - those programs don't expect
to inherit a child). So the zombie is expected. To prevent that, use
"background -d", which will doublefork your echo program, so it will
be reparented to pid 1 which will reap it properly.
The anomaly is that you *don't* have that zombie without emptyenv;
my first guess is that there's something in your environment that
changes
the behaviour of ntpd and makes it reap the zombie somehow.
---
3. Is it normal/standard/good practice to include a dependency in a
bundle. For example, I have a "time" bundle whose contents are
time-srv. time-srv starts the ntpd service, and has as a dependency
time-log.
Using "s6-rc -u change time", everything behaves as documented, ie
starts "time" which starts time-log, then time-srv. However
# s6-rc -v 9 -d change base:time
s6-rc: info: bringing selected services down
s6-rc: info: processing service base:time-srv: stopping
s6-rc: info: service base:time-srv stopped successfully
# Starting logging service time for base with user s6log folder
/var/log/time
and the time-log continues running.
If you only have time-srv in your 'time' bundle, then time-srv and
time are equivalent. Telling s6-rc to bring down time will do the
exact same thing as telling it to bring down time-srv. time-log is
not impacted. So the behaviour is expected.
If you want "s6-rc -d change time" to also bring down time-log, then
yes, you should add time-log to the time bundle. Then 'time' will
address both time-srv and time-log.
y 6 seconds # This is time-srv
up (pid 85131) 6 seconds # This is time-log,so it
has been restarted
If you're using a manually created named pipe to transmit data
from time-srv to time-log, that pipe will close when time-srv exits,
and your logger will get EOF and probably exit, which is why it
stopped; but time-log's supervisor has received no instruction that
it should stop, so it will restart it. This is also expected.
The simplest way of achieving the behaviour you want is s6-rc's
integrated pipeline feature. Get rid of your named pipe and of your
stdout (for time-srv) and stdin (for time-log) redirections; get rid
of your time bundle definition. Then declare time-log as a consumer
for time-srv and time-srv as a producer for time-log. In the
time-log source definition directory, write 'time' into the
pipeline-name file. Then recompile your database.
This will automatically create a pipe between time-srv and time-log;
the pipe will be held open so it won't close even if one of the
processes exits; and it will automatically create a 'time' bundle
that contains both time-srv and time-log.
You're on the right track. :)
--
Laurent
