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Re: [libvirt] [PATCH] qemu: qxl devices don't support multifunction yet



* Laine Stump (laine laine org) wrote:
> On 09/29/2011 10:30 AM, Laine Stump wrote:
> >On 09/27/2011 05:13 AM, Daniel P. Berrange wrote:
> >>When we originally enabled multifunction for all PCI devices, we did so
> >>in the belief that this was effectively a no-op for guest OSes. ie it
> >>should not have changed guest OS behaviour at all, unless
> >>multiple devices
> >>were actually inserted in the slot.
> >>
> >>Evidentally this turns out to be incorrect. In other words we introduced
> >>a guest ABI change. This is very bad, because our stated goal is to have
> >>a stable guest ABI across libvirt&  QEMU releases.
> >>
> >>Thus IMHO we need to disable *all* setting of the 'multifunction=on'
> >>parameter for all guests by default, to unbreak the ABI compatibility.

Yes, I think that's correct.  It should only be on for multifunction
devices.

> >>Then we should introduce a new parameter 'multifunction='on' in the
> >><address type=pci>  element to allow it to be optionally enabled per
> >>device.
> >
> >
> >A couple of questions:
> >
> >1) Is it possible/useful to have function > 0 without
> >multifunction=on in the qemu commandline?

qemu should blcok function > 0 when function 0 does not have
multifunction=on.

> >2) If the answer to (1) is "No", if the XML has function > 0 but
> >no multifunction=on, should we implicitly add multifunction=on, or
> >should we log an error?

You have to have a function 0 _and_ and function > 0 to be
multifunction.  YOu have to specify multifucntion=on in function 0 (at a
minimum, spec is unclear whether it's required or implied for > 0).

> Also:
> 
> 3) if you explicitly/(implicitly?) set multifunction on for one
> function of a slot, does qemu automatically do so for other
> functions on the slot (ie, if you have a device that uses slot 1
> function 1 (and turns on multifunction), then will slot 1 function 0
> automatically have multifunction turned on too?)

It will not automagically set multifunction=on on func 0.
It will generate an error.

thanks,
-chris


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