Messages in this thread | | | Subject | Re: [PATCH v2] proc: speedup /proc/stat handling | From | Eric Dumazet <> | Date | Mon, 30 Jan 2012 10:25:55 +0100 |
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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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