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Date:	Wed, 23 Apr 2014 12:22:28 -0700
From:	Zi Shen Lim <zlim@...adcom.com>
To:	Mark Brown <broonie@...nel.org>
CC:	Catalin Marinas <catalin.marinas@....com>,
	Lorenzo Pieralisi <lorenzo.pieralisi@....com>,
	Mark Rutland <mark.rutland@....com>,
	Will Deacon <will.deacon@....com>,
	<linux-arm-kernel@...ts.infradead.org>,
	<linux-kernel@...r.kernel.org>
Subject: Re: [PATCH 2/2] arm64: topology: add MPIDR-based detection

On Wed, Apr 23, 2014 at 07:26:11PM +0100, Mark Brown wrote:
> On Wed, Apr 23, 2014 at 10:27:20AM -0700, Zi Shen Lim wrote:
> 
> It will at least ensure that all clusters get assigned a unique ID and
> we don't end up discarding some of the information and coming out with
> two identically numbered clusters which then have identically numbered
> CPUs inside of them which doesn't seem clever.

I agree with you. Simply ignoring aff3 is not acceptable, whether or not
someone is using it.

> 
> When I was looking at this it wasn't sufficiently clear to me that the
> cluster clustering would be well modelled by sockets as the scheduler
> currently assumes them, nor what to do with additional levels of that
> (the DT binding allows for infinite levels).  Punting and just putting
> all clusters at the same level avoids active bugs and seems fairly
> conservative.
> 

Sounds like you prefer "cluster of clusters" over "socket", correct?

In any case, with only 4 affinity levels defined in the arch, as long as
we also have 4 variables to capture that information, we should be good,
right?

Anything more exotic not expressable by these 4 affinity levels in MPIDR
will require additional information from other sources such as DT or ACPI.

> > Perhaps we should just add a new 'socket_id' and that will accommodate
> > all cases (up to aff3).
> 
> Not in the non-MT case where we've got two levels above the cluster ID
> in affinity level 1 unless we just combine 2 and 3 (which would be
> reasonable enough of course).

Is the following an accurate description of your proposal for non-MT?

	thread_id   = -1
	core_id     = aff0
	cluster_id  = aff1
	clusters_id = combine(aff2,aff3)

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