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SubjectRe: [PATCH v2] proc: speedup /proc/stat handling
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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.

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.



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