no.. it has to do with the fact that it would be unwise
to make a cluster > 1 page size since we have no guarantee that
all drivers could handle breaking up a DMA if a cluster spanned 2
physical address ranges. (they can handle a chain of discontinuous
mbufs but may assume that a single mbuf will have physically
contiguous data. Now since we cannot span a page boundary,
we should fit in exacly to get as much room as possible
and since (pagesize/3) is too small, the next possibility is (pagesize/2).
If pagesize/3 was big enough, we might have used that..
On Wed, 25 Jul 2001, Zhihui Zhang wrote:
>
> I see. It has something to do with the power-of-two allocator we are
> using inside the kernel.
>
> -Zhihui
>
> On Wed, 25 Jul 2001, Bosko Milekic wrote:
>
> >
> > On Wed, Jul 25, 2001 at 01:51:51PM -0400, Zhihui Zhang wrote:
> > >
> > >
> > > On Tue, 24 Jul 2001, Terry Lambert wrote:
> > >
> > > > Zhihui Zhang wrote:
> > > > > > Hi,
> > > > > > in freebsd can we change the cluster size from 2048
> > > > > > bytes.If yes how can we do that?
> > > > > > do we have to configure in some file?
> > > > >
> > > > > You must be asking why the mbuf cluster size is chosen as 2048, right? It
> > > > > is probably a tradeoff between memory efficient and speed.
> > > >
> > > > Ask yourselves:
> > > >
> > > > "What is the minimum cluster size I would have to have
> > > > to be able to contain the maximum MTU worth of data,
> > > > yet remain an even multiple of sizeof(mbuf) -- 256
> > > > bytes?"
> > >
> > > A dumb question: why even not odd multiple?
> > >
> > > -Zhihui
> >
> > It actually has to do with the fact that 2K is the only size equal to
> > or greater than the maximum MTU worth of data that can be multiplied to a page
> > size without any leftover (in other words, page size modulo 2K is zero).
> >
> > --
> > Bosko Milekic
> > [EMAIL PROTECTED]
> >
> >
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> >
>
>
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