On Thu Jan 2 17:54:50 EST 2014, [email protected] wrote:
> 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.
>
the code says you can set alarm = 0 without holding the alarms lock.
ulong
procalarm(ulong time)
{
Proc **l, *f;
ulong when, old;
if(up->alarm)
old = tk2ms(up->alarm - sys->ticks);
else
old = 0;
if(time == 0) {
>> up->alarm = 0;
>> return old;
- erik
