no. the debug qlock is so the process wont exit. once you hold it, the process might be running, might be in the process of exiting might have been already exited or might have been reused.
the key here is that while you hold it. it wont make it to the process freelist when not already there. it wont change from "exiting -> exited". but it wont prevent it from being reused when it was exited already... but... pexit() zeros p->alarm. and p->alarm can not be set to non-zero value again while we hold the alarms qlock in the kproc! so this actually works. -- cinap
