Subject: Re: BOOKE KVM calling load_up_fpu from C?

From: Scott Wood <[email-masked]>
Date: Tue, 12 Feb 2013 19:23:14 -0600

On 02/12/2013 07:18:14 PM, Bhushan Bharat-R65777 wrote:
>=20
>=20
> > -----Original Message-----
> > From: Wood Scott-B07421
> > Sent: Wednesday, February 13, 2013 12:03 AM
> > To: Bhushan Bharat-R65777
> > Cc: Michael Neuling; Wood Scott-B07421; =20
> [email-masked]
> > Subject: Re: BOOKE KVM calling load_up_fpu from C?
> >
> > On 02/12/2013 03:01:07 AM, Bhushan Bharat-R65777 wrote:
> > > To me this looks wrong. And this seems to works because the
> > > thread->reg->msr is not actually used to write SRR1 (and =20
> eventually
> > > the thread MSR) when doing rfi to enter guest. Infact
> > > Guest(shadow_msr) MSR is used as SRR1 and which will have proper =20
> MSR
> > > (including FP set).
> > >
> > > But Yes, Scott is right person to comment, So let us wait for him
> > > comment.
> >
> > I don't think it's actually a problem on 32-bit, since r9 is =20
> modified but never
> > actually used for anything.
>=20
> Is not the epilog loads srr1 in r9 and load_up_fpu() changes r9 and =20
> then r9 is written back in srr1 ?

What epilog?  We're talking about the case where it's called from C =20
code.

When it's called from an exception handler, then r9 is used, but in =20
that case it's also initialized before calling load_up_fpu, by the =20
prolog.

> >  On 64-bit, though, there's a store to the caller's
> > stack frame (yuck) which the kvm/booke.h caller is not prepared for.
>=20
> So if caller is using r12 then it can lead to come corruption, right ?

No, r12 is a volatile register in the ABI, as is r9.  The issue is that =20
the stack can be corrupted.

-Scott=