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SubjectRe: [PATCH v2] proc: speedup /proc/stat handling
Date
Eric Dumazet wrote, on 01/30/12 10:25:
> Le lundi 30 janvier 2012 à 09:06 +0100, Jörg-Volker Peetz a écrit :
>> Eric Dumazet wrote, on 01/25/12 01:26:
>>> Le mercredi 25 janvier 2012 à 09:18 +0900, KAMEZAWA Hiroyuki a écrit :
>>>
>>>> BTW, what is the reason of this change ?
>>>>
>>>>> - unsigned size = 4096 * (1 + num_possible_cpus() / 32);
>>>>> + unsigned size = 1024 + 128 * num_possible_cpus();
>>>>
>>>> I think size of buffer is affected by the number of online cpus.
>>>> (Maybe 128 is enough but please add comment why 128 ?)
>>>>
>>>
>>> There is no change, as 4096/32 is 128 bytes per cpu.
>>>
>>
>> Wrong math, only num_possible_cpus() is divided by 32. Thus,
>>
>> - unsigned size = 4096 * (1 + num_possible_cpus() / 32);
>> + unsigned size = 4096 + 128 * num_possible_cpus();
>>
>> <snip>
>
>
> It is good math, once you take the time to think a bit about it.
>
> The original question was about the 128 * num_possible_cpus()
>
> 4096/32 is 128 as I said.
>
> The 4096 -> 1024 is just taking into account fact that once you do the
> correct computations, you dont need initial 4096 value, and 1024 is more
> than enough.

Thanks for your explanation, that makes sense now. At first, I was mislead by
your comment "There is no change" which was apparently related to the second
term in the formula.

>
> Example on a dual core machine :
>
> # dmesg|grep nr_irq
> [ 0.000000] nr_irqs_gsi: 40
> [ 0.000000] NR_IRQS:2304 nr_irqs:712 16
>
> size = 1024 + 2*128 + 2*712 = 2704 bytes (rounded to 4096 by kmalloc())
>
> # wc -c /proc/stat
> 1767 /proc/stat
>
> Problem with original math was that for a machine with 16 cpus or a
> machine with 1 cpu, we ended with the same 4096 value. That was a real
> problem.
>
> If we instead use "unsigned size = 4096 + 128 * num_possible_cpus();" as
> you suggest, we would always allocate 2 pages of memory, this is not
> needed at all for typical 1/2/4 way machines.
>
--
Best regards,
Jörg-Volker.

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