linux-kernel - Re: [PATCH] x86: Hyper-V: register clocksource only if its advertised

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Date:	Tue, 29 Jan 2013 10:26:52 +0100
From:	Olaf Hering <olaf@...fle.de>
To:	Jan Beulich <JBeulich@...e.com>
Cc:	Stefano Stabellini <stefano.stabellini@...citrix.com>,
	KY Srinivasan <kys@...rosoft.com>,
	Greg KH <gregkh@...uxfoundation.org>,
	"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>
Subject: Re: [PATCH] x86: Hyper-V: register clocksource only if its advertised

On Tue, Jan 29, Jan Beulich wrote:

> >>> On 28.01.13 at 18:44, Stefano Stabellini <stefano.stabellini@...citrix.com> wrote:
> > I think that Olaf made his point very clear: a feature A should only be
> > enabled if the corresponding flag A is set.
> > In fact it seems to me that this patch is correct on its own merits,
> > regardless of Xen does or does not.
> > 
> > The Xen tools might or might not know whether a guest is going to be
> > Linux, Windows, FreeBSD or whatever else people use nowadays.  Setting
> > viridian=1 is the safe choice, given that it shouldn't create any
> > issues: after all guests are supposed to check for feature flags before
> > using them.
> > 
> > If Xen is going to implement "Partition Reference Counter", it is also
> > going to set the corresponding flag, so the guest OS (Windows, Linux,
> > my pet OS) can check whether the feature is available and decide whether
> > it wants to use it.
> 
> While I agree in general, the specific case of the callback vector
> seems a little more difficult: As KY says, there's no feature flag
> for this (or perhaps more precisely for it being deliverable across
> all CPUs), and hence there's both the problem of detection and
> the problem of disambiguation (as otherwise both the Hyper-V
> code and the Xen code in Linux could be trying to use the same
> vector).

This is true, but belongs to that other thread about the interrupt
vector. I agree that the detection in ms_hyperv_platform() should be
extended, with a DMI check for example.

The patch which started this thread is still valid because it enables
feature B only if the featurebit for B is enabled.

Olaf
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