Re: [9fans] Re: [sources] 20070413: /rc/bin/cpurc.local

["Devon H. O'Dell" <devon.odell@gmail.com>]

2007/4/15, Russ Cox <rsc@swtch.com>:
>         /* cooperative scheduling until the clock ticks */
>         if((p=m->readied) && p->mach==0 && p->state==Ready
>         && runq[Nrq-1].head == nil && runq[Nrq-2].head == nil){
>
> The tests are:
>
>         p=m->readied
>                 some process p was last readied on this cpu (Mach)
>
>         p->mach == 0
>                 p is not running on any cpu right now

Not to be overly annoying, but _is_ it possible that between the
assignment and that test, p->mach is set, indicating another CPU took
over? It just doesn't seem like m->readied is locked, at all; it seems
like another CPU could pick it up in the middle.

>         p->state == Ready
>                 p is still Ready (waiting to run)
>
>         runq[Nrq-1].head == nil && runq[Nrq-2].head == nil
>                 there are no real-time processes waiting to run
>
> If all those succeed, then the code tries to choose
> p to run next.  But it might not -- the next thing that
> happens is
>
>         p = dequeueproc(rq, p);
>
> which can return nil if p has already been grabbed
> or even if there is contention for the runq lock.
> All the accesses in the if condition are just fine --
> they happen without a lock but dequeueproc double-checks
> that p is okay to schedule.
>
> If dequeueproc returns nil, then runproc won't pick
> the readied p after all -- it will fall into the regular
> scheduling loop to find a process.
>
> Russ

Thanks for the low-level explanation. It does really help my understanding.

--dho