Date: Sun, 11 May 2008 23:10:51 -0700 From: Julian Elischer <julian@elischer.org> To: Elijah Buck <elijah.buck@gmail.com> Cc: freebsd-hackers@freebsd.org Subject: Re: 4bsd fuzzy runq Message-ID: <4827DF6B.6050101@elischer.org> In-Reply-To: <303eddcb0805111943t7c456c79l56f427c526d0b454@mail.gmail.com> References: <303eddcb0805111943t7c456c79l56f427c526d0b454@mail.gmail.com>
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Elijah Buck wrote:
> Hi,
>
> I'm looking at the code for 4bsd fuzzy run queues in kern_switch.c
>
> The relevant bit:
>
> if (fuzz > 1) {
> int count = fuzz;
> int cpu = PCPU_GET(cpuid);
> struct thread *td2;
> td2 = td = TAILQ_FIRST(rqh);
> while (count-- && td2) {
> if (td->td_lastcpu == cpu) {
this should be td2. The bug is mine..
and I wondered why the code commited never had the same result as that
I tested in another tree.
> td = td2;
> break;
> }
> td2 = TAILQ_NEXT(td2, td_runq);
> }
> ...return(td)
> The purpose of this code appears to be to look through the runq to a depth
> defined by fuzz for a thread that was last run on the current cpu.
>
> Here are the cases I see:
> 1.) if (td_lastcpu == cpu) on the first iteration then TAILQ_FIRST(rqh) is
> the selected thread (because of the break).
> 2.) if (td_lastcpu != cpu) on the first iteration then td is never set
> again, and so (td_lastcpu != cpu) is always true. So, again,
> TAILQ_FIRST(rqh) will be selected (because count will reach 0).
>
> Which doesn't seem right (since that's what the else clause does). What am I
> missing here?
>
> Thanks,
> Elijah
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