From: "Dmitry Adamushko" <dmitry.adamushko@gmail.com>
To: vatsa@linux.vnet.ibm.com
Cc: "Gautham R Shenoy" <ego@in.ibm.com>,
"Ingo Molnar" <mingo@elte.hu>,
npiggin@suse.de, linux-kernel@vger.kernel.org,
"Srivatsa Vaddagiri" <vatsa@in.ibm.com>
Subject: Re: [PATCH] sched: Improve readability in update_cpu_load() code
Date: Thu, 15 May 2008 17:31:52 +0200 [thread overview]
Message-ID: <b647ffbd0805150831p1b61437cjbd47697d5dd72f01@mail.gmail.com> (raw)
In-Reply-To: <b647ffbd0805150815t523a6242xd78e85b3f4c5045b@mail.gmail.com>
2008/5/15 Dmitry Adamushko <dmitry.adamushko@gmail.com>:
> 2008/5/15 Srivatsa Vaddagiri <vatsa@linux.vnet.ibm.com>:
>> On Thu, May 15, 2008 at 06:34:59PM +0530, Gautham R Shenoy wrote:
>>> Author: Gautham R Shenoy <ego@in.ibm.com>
>>> Date: Thu May 15 17:55:49 2008 +0530
>>>
>>> sched: Improve readability in update_cpu_load() code
>>>
>>> Currently the cpu_load[i] is calculated as:
>>> this_rq->cpu_load[i] = (old_load*(scale-1) + new_load) >> i;
>>>
>>> However, since scale = 2^i, this can be simplified as:
>>> this_rq->cpu_load[i] = old_load + ((new_load - old_load) >> i);
>>>
>>> Makes it easier to read.
>>> Signed-off-by: Gautham R Shenoy <ego@in.ibm.com>
>>>
>>> diff --git a/kernel/sched.c b/kernel/sched.c
>>> index 2d7d8f1..e1a6985 100644
>>> --- a/kernel/sched.c
>>> +++ b/kernel/sched.c
>>> @@ -2921,7 +2921,7 @@ static void update_cpu_load(struct rq *this_rq)
>>> */
>>> if (new_load > old_load)
>>> new_load += scale-1;
>>> - this_rq->cpu_load[i] = (old_load*(scale-1) + new_load) >> i;
>>> + this_rq->cpu_load[i] = old_load + ((new_load - old_load) >> i);
>>
>> This wont work when new_load < old_load ..
>>
>> For ex: I tried this prog:
>>
>> #include <stdio.h>
>>
>> main()
>> {
>> unsigned long old_load = 100, new_load = 90, this_load, this_load1;
>> int i = 1, scale = 2 << i;
>>
>> this_load = (old_load*(scale-1) + new_load) >> i;
>> this_load1 = old_load + ((new_load - old_load) >> i);
>
> it should be
>
> this_load2 = 2 * old_load + (new_load >> i) - (old_load >> i);
>
> scale == 2 << i == 1 << (i + 1), so scale >> i = 2.
argh.. sorry, it's (scale = 2 << i) in your test-program which is
wrong. 2^i == (1 << i), so the (2 *) part is redundant, of course.
this_load2 = old_load + (new_load >> i) - (old_load >> i);
--
Best regards,
Dmitry Adamushko
next prev parent reply other threads:[~2008-05-15 15:32 UTC|newest]
Thread overview: 6+ messages / expand[flat|nested] mbox.gz Atom feed top
2008-05-15 13:04 Gautham R Shenoy
2008-05-15 14:52 ` Srivatsa Vaddagiri
2008-05-15 15:02 ` Srivatsa Vaddagiri
2008-05-15 15:15 ` Dmitry Adamushko
2008-05-15 15:31 ` Dmitry Adamushko [this message]
2008-05-15 17:11 ` Gautham R Shenoy
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