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Date: Wed, 5 Aug 2015 10:01:02 +1000
From: David Gibson <david@gibson.dropbear.id.au>
Message-ID: <20150805000102.GB7739@voom.redhat.com>
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Subject: Re: [Qemu-devel] [PATCH][TRIVIAL] i6300esb: fix timer overflow
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To: Laurent Vivier <lvivier@redhat.com>
Cc: qemu-trivial@nongnu.org, Paolo Bonzini <pbonzini@redhat.com>, qemu-devel@nongnu.org, "Richard W.M. Jones" <rjones@redhat.com>


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On Tue, Aug 04, 2015 at 10:27:31AM +0200, Laurent Vivier wrote:
> We use muldiv64() to compute the time to wait:
>=20
>     timeout =3D muldiv64(get_ticks_per_sec(), timeout, 33000000);
>=20
> but get_ticks_per_sec() is 10^9 (30 bit value) and timeout
> is a 35 bit value.
>=20
> Whereas muldiv64 is:
>=20
>     uint64_t muldiv64(uint64_t a, uint32_t b, uint32_t c)
>=20
> So we loose 3 bits of timeout.
>=20
> Swapping get_ticks_per_sec() and timeout fixes it.
>=20
> We can also replace it by a multiplication by 30 ns,
> but this changes PCI clock frequency from 33MHz to 33.333333MHz
> and we need to do this on all the QEMU PCI devices (later...)
>=20
> Signed-off-by: Laurent Vivier <lvivier@redhat.com>

Hah.  32-bit second argument.  Totally missed that when I put the
muldiv64() in there.  So I didn't eliminate the overflow, just pushed
it out some bits.

Thanks for finding this.

Reviewed-by: David Gibson <david@gibson.dropbear.id.au>

--=20
David Gibson			| I'll have my music baroque, and my code
david AT gibson.dropbear.id.au	| minimalist, thank you.  NOT _the_ _other_
				| _way_ _around_!
http://www.ozlabs.org/~dgibson

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