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Date: Tue, 01 Jul 2014 14:56:49 +0200
From: Jan Kiszka <jan.kiszka@web.de>
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Subject: Re: [Qemu-devel] Why devfn will be -1
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To: Le Tan <tamlokveer@gmail.com>
Cc: qemu-devel <qemu-devel@nongnu.org>

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On 2014-07-01 14:55, Le Tan wrote:
> 2014-07-01 20:52 GMT+08:00 Le Tan <tamlokveer@gmail.com>:
>> Hi Jan,
>>
>> 2014-07-01 15:34 GMT+08:00 Jan Kiszka <jan.kiszka@web.de>:
>>> Hi Le,
>>>
>>> On 2014-07-01 04:34, Le Tan wrote:
>>>> Hi Jan,
>>>> I use pci_setup_iommu() to setup a PCIIOMMUFunc for the q35 pci bus.=

>>>> In the iommu_fn, I print out the devfn parameter and find out that i=
t
>>>> sometimes will be -1. So what does it mean?
>>>> The detail code is here:
>>>>
>>>> In mch_init() function, I write like this:
>>>> PCIBus *pci_bus =3D PCI_BUS(qdev_get_parent_bus(DEVICE(mch)));
>>>> pci_setup_iommu(pci_bus, q35_host_dma_iommu, mch->iommu);
>>>>
>>>> And in q35_host_dma_iommu(PCIBus *bus, void *opaque, int devfn), I
>>>> print out the devfn parameter, sometimes it will be -1.
>>>
>>> Hmm, I have no idea about the reason and would suggest to set a
>>> conditional breakpoint on this function, then print the backtrace to =
see
>>> where this comes from and analyze the device structure from where tha=
t
>>> -1 was most probably taken.
>=20
> I think maybe this is a bug? In the function do_pci_register_device(),
> maybe these two sentence should be reorder?
> dma_as =3D pci_device_iommu_address_space(pci_dev);
> pci_dev->devfn =3D devfn;

Looks like. Give it a try, then possibly send a patch :)

Jan


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