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.

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