erik quanstrom wrote:
How? If there's a stop message already written to /proc/n/ctl. Once
that is done, the process is guaranteed to be in 2 states and those
states only: continue waiting for the I/O, being actually Stopped.
Both of the don't let the scheduler take it to the runqueue.
here's the senerio, i think (works fine on a single processor)
a b
acquire debug lock
sleep complete io
sched
run a bit
syscall
wakeup
But how "run a bit" could possibly happen if after the "stop" message
being sent right after the "complete io" the "b" process goes into
a "Stopped" state?
Thanks,
Roman.