Date: Thu, 20 Jul 2017 10:15:30 +0200 From: Hans Petter Selasky <hps@selasky.org> To: Sebastian Huber <sebastian.huber@embedded-brains.de>, FreeBSD <freebsd-hackers@freebsd.org> Subject: Re: Relationship between ncallout and callwheelsize Message-ID: <6453f4fa-016f-3d7a-ce03-aa49c7f465e5@selasky.org> In-Reply-To: <8ba43394-41e4-f06c-a8c6-12b773d185c4@embedded-brains.de> References: <8ba43394-41e4-f06c-a8c6-12b773d185c4@embedded-brains.de>
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On 07/20/17 08:59, Sebastian Huber wrote:
> Hello,
>
> I noticed on a LPC4088 running at 96MHz from external SDRAM that the
> callout_process() function needs a lot of time even if no callouts are
> active. The reason is that on a 32-bit architecture with only 13 general
> purpose registers this function needs a lot of load/store operations. So
> I tried to reduce the complexity by using compile time constants. I
> don't understand why the callwheelsize and the timeout(9)
> pre-allocations are related:
>
> ncallout = imin(16 + maxproc + maxfiles, 18508);
> TUNABLE_INT_FETCH("kern.ncallout", &ncallout);
>
> /*
> * Calculate callout wheel size, should be next power of two higher
> * than 'ncallout'.
> */
> callwheelsize = 1 << fls(ncallout);
> callwheelmask = callwheelsize - 1;
>
> The size of the wheel should be related to typical timeout values to
> balance memory size and hash collisions. Why is there a connection to
> the timeout(9) pre-allocations?
>
Hi,
The calloutwheel is simply an optimisation. If you don't need many
callouts, maybe a per-CPU list will be enough for your purpose.
--HPS
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