From mboxrd@z Thu Jan  1 00:00:00 1970
From: =?iso-8859-1?Q?Fran=E7ois-Fr=E9d=E9ric_Ozog?= <ff-SpBG2i6TTxU@public.gmane.org>
Subject: Re: PCI device mapping to socket
Date: Wed, 18 Dec 2013 21:42:58 +0100
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To: "'Benson, Bryan'" <bmbenson-vV1OtcyAfmbQT0dZR+AlfA@public.gmane.org>,
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Hi,

It depends on the kernel version. For the latests ones you can use:
cat /sys/class/net/<interface name>/device/numa_node

in all other case, you can use lspci fallback (in case even no driver is =
yet
loaded).

lspci | grep Ethernet
09:00.2 Ethernet controller: Intel Corporation I350 Gigabit Network
Connection (rev 01)
09:00.3 Ethernet controller: Intel Corporation I350 Gigabit Network
Connection (rev 01)

lspci -t
\-[0000:00]-+-00.0
             +-01.0-[01-03]----00.0-[02-03]----08.0-[03]--+-00.0
             |                                            +-00.3
             .
             .
             +-1c.0-[09-0a]--+-00.0
             |               +-00.1
             |               +-00.2
             |               \-00.3

So the PCI bus is 0.

Now transform this to socket number:

Space=3D0x100/<nbsocket>

On a dual socket space=3D0x80, bus(0*space=3D0) is socket 0, =
bus(1*space=3D0x80)
is socket1.

On a quad socket space=3D0x40, bus(0*space=3D0) is socket 0, =
bus(1*space=3D0x40)
is socket1, bus(2*space=3D0x80) is socket2, bus(3*space=3D0xc0) is =
socket3 .

Fran=E7ois-Fr=E9d=E9ric


> -----Message d'origine-----
> De=A0: dev [mailto:dev-bounces-VfR2kkLFssw@public.gmane.org] De la part de Benson, Bryan
> Envoy=E9=A0: mercredi 18 d=E9cembre 2013 21:20
> =C0=A0: dev-VfR2kkLFssw@public.gmane.org
> Objet=A0: [dpdk-dev] PCI device mapping to socket
>=20
> All,
> Does anyone know of a way I can find out which socket a PCI =
device/bridge
> is tied up to?  I have looked into dmidecode and lspci to no avail, =
but I
> may be missing something.  We are looking at putting multiple NICs =
into a
> single dual socket server.
>=20
> This is so that I can tie specific NIC ports to the proper socket to =
take
> advantage of DDIO.
>=20
> Thank you,
> Bryan Benson
> Amazon Web Services