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Subject: Re: [RFC v1 00/19] Modify zonelist to nodelist v1
To:     "lixinhai.lxh@gmail.com" <lixinhai.lxh@gmail.com>,
        Pengfei Li <fly@kernel.page>, akpm <akpm@linux-foundation.org>
Cc:     mgorman <mgorman@techsingularity.net>,
        Michal Hocko <mhocko@kernel.org>,
        Vlastimil Babka <vbabka@suse.cz>, cl <cl@linux.com>,
        "iamjoonsoo.kim" <iamjoonsoo.kim@lge.com>, guro <guro@fb.com>,
        "linux-kernel@vger.kernel.org" <linux-kernel@vger.kernel.org>,
        "linux-mm@kvack.org" <linux-mm@kvack.org>
References: <20191121151811.49742-1-fly@kernel.page>
 <2019112215245905276118@gmail.com>
From:   David Hildenbrand <david@redhat.com>
Organization: Red Hat GmbH
Message-ID: <7dbec505-ce53-e1f0-6ed4-8cb0328dfc79@redhat.com>
Date:   Fri, 22 Nov 2019 11:14:15 +0100
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On 22.11.19 08:25, lixinhai.lxh@gmail.com wrote:
> On 2019-11-21=C2=A0at 23:17=C2=A0Pengfei Li=C2=A0wrote:
>> Motivation
>> ----------
>> Currently if we want to iterate through all the nodes we have to
>> traverse all the zones from the zonelist.
>>
>> So in order to reduce the number of loops required to traverse node,
>> this series of patches modified the zonelist to nodelist.
>>
>> Two new macros have been introduced:
>> 1) for_each_node_nlist
>> 2) for_each_node_nlist_nodemask
>>
>>
>> Benefit
>> -------
>> 1. For a NUMA system with N nodes, each node has M zones, the number
>>  =C2=A0=C2=A0 of loops is reduced from N*M times to N times when travers=
ing node.
>>
>=20
> It looks to me that we don't really have system which has N nodes and
> each node with=C2=A0M zones in its address range.
> We may have systems which has several nodes, but only the first node has
> all zone types, other nodes only have NORMAL zone. (Evenly distribute the
> !NORMAL zones on all nodes is not reasonable, as those zones have limited
> size)
> So iterate over zones to reach nodes should at N level, not M*N level.

I guess NORMAL/MOVABLE/DEVICE would be common for most nodes, while I do=20
agree that usually we will only have 1 or 2 zones per node (when we have=20
many nodes). So it would be something like c*N, whereby c is most=20
probably on average 2.

--=20

Thanks,

David / dhildenb