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.
