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Date: Tue, 05 Mar 2002 07:29:11 -0800
From: "Martin J. Bligh" <Martin.Bligh@us.ibm.com>
Reply-To: "Martin J. Bligh" <Martin.Bligh@us.ibm.com>
To: Rik van Riel <riel@conectiva.com.br>, Andrea Arcangeli <andrea@suse.de>
cc: "Martin J. Bligh" <Martin.Bligh@us.ibm.com>,
        Daniel Phillips <phillips@bonn-fries.net>,
        Bill Davidsen <davidsen@tmr.com>, Mike Fedyk <mfedyk@matchmail.com>,
        linux-kernel@vger.kernel.org
Subject: Re: 2.4.19pre1aa1
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--On Tuesday, March 05, 2002 9:22 AM -0300 Rik van Riel 
<riel@conectiva.com.br> wrote:

> On Tue, 5 Mar 2002, Andrea Arcangeli wrote:
>> On Mon, Mar 04, 2002 at 10:26:30PM -0300, Rik van Riel wrote:
>> > On Tue, 5 Mar 2002, Andrea Arcangeli wrote:
>> > > On Mon, Mar 04, 2002 at 09:01:31PM -0300, Rik van Riel wrote:
>> > > > This could be expressed as:
>> > > >
>> > > > "node A"  HIGHMEM A -> HIGHMEM B -> NORMAL -> DMA
>> > > > "node B"  HIGHMEM B -> HIGHMEM A -> NORMAL -> DMA
>
>> the example you made doesn't have highmem at all.
>>
>> > has 1 ZONE_NORMAL and 1 ZONE_DMA while it has multiple
>> > HIGHMEM zones...
>>
>> it has multiple zone normal and only one zone dma. I'm not forgetting
>> that.
>
> Your reality doesn't seem to correspond well with NUMA-Q
> reality.

I think the difference is that he has a 64 bit vaddr space,
and I don't ;-) Thus all mem to him is ZONE_NORMAL (not sure
why he still has a ZONE_DMA, unless he reused it for the 4Gb
boundary). Andrea, is my assumtpion correct?

On a 32 bit arch (eg ia32) everything above 896Mb (by default)
is ZONE_HIGHMEM. Thus if I have > 896Mb in the first node,
I will have one ZONE_NORMAL in node 0, and a ZONE_HIGHMEM
in every node. If I have < 896Mb in the first node, then
I have a ZONE_NORMAL in every node up to and including the
896 breakpoint, and a ZONE_HIGHMEM in every node from the
breakpoint up (including the breakpoint node). Thus the number
of zones = number of nodes + 1.

M.

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