I'd like to query the rationale for a behaviour I've experiencing in
monit. I'm testing with the following config:
# Test config start
set daemon 10
check program MyProgram with path "/bin/dash -c 'echo OK!; exit 1'"
every "06 * * * *"
if status != 0 then alert
# Test config end
As expected, monit runs the dash test program at 6 minutes past the
hour. The dash script finishes immediately. However, Monit doesn't
pick up, report or alert on the exit code in a timely manner. Until
the next time Monit is scheduled to run the test script, the dash
script remains as a zombie. But that is an hour later, which is a
long time to wait to be alerted to the script failing.
If the 'every' schedule was "06 0 * * *" then it would seem one
should expect to wait 24 hours before being alerted to the script
I realise the Monit manual explains:
"The asynchronous nature of the program check [...] comes with a
side-effect: when the program has finished executing and is waiting
for Monit to collect the result, it becomes a so-called "zombie"
process [...] the zombie process is removed from the system as soon
as Monit collects the exit status. This means that every "check
program" will be associated with either a running process or a
temporary zombie. This unwanted zombie side-effect will be removed
in a later release of Monit."
That may be so, however why doesn't Monit reap the child and collect
the exit code at the *next poll cycle after the child exits* (i.e.
within 10 seconds of the test script finishing given the 'set daemon
10' line in the test config above) rather than when the program is
next scheduled to be run? Maybe I'm missing something, but the
current behaviour seems to undermine the entire purpose of providing
alerts on program failure (when used in conjunction with cron-style
scheduling). That is the behaviour I'd like to query the rationale
Thanks in advance.
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