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Re: question regarding QEMU adding overlapping memory regions to VFIO


From: Alex Williamson
Subject: Re: question regarding QEMU adding overlapping memory regions to VFIO
Date: Fri, 7 May 2021 09:42:05 -0600

On Fri, 7 May 2021 13:51:52 +0000
Thanos Makatos <thanos.makatos@nutanix.com> wrote:

> I've noticed that QEMU adds overlapping memory regions to VFIO, e.g.:
> 
> vfio_listener_region_add_ram region_add [ram] 0xc0000 - 0xc0fff 
> [0x7f6702c00000]
> vfio_listener_region_del region_del 0xc4000 - 0xdffff
> vfio_listener_region_add_ram region_add [ram] 0xc1000 - 0xc3fff 
> [0x7f66406c1000]
> vfio_listener_region_del region_del 0xe0000 - 0xfffff
> vfio_listener_region_add_ram region_add [ram] 0xc4000 - 0xdffff 
> [0x7f6702c04000]
> vfio_listener_region_add_ram region_add [ram] 0xc0000 - 0xc0fff 
> [0x7f66406c0000]
> 2021-05-05T09:38:16.158864Z qemu-system-x86_64: vfio_dma_map(0x557b8fd281b0, 
> 0xc0000, 0x1000, 0x7f66406c0000) = -22 (Resource temporarily unavailable)
> 
> Region 0xc0000 - 0xc0fff is added first and then region 0xc0000 -
> 0xc0fff is added again? Is this legitimate? What is the implication
> of this? Is the previous region replaced by the more recent one?

This might be where the hack we have in hw/vfio/common.c:vfio_dma_map()
comes from:

    /*
     * Try the mapping, if it fails with EBUSY, unmap the region and try
     * again.  This shouldn't be necessary, but we sometimes see it in
     * the VGA ROM space.
     */
    if (ioctl(container->fd, VFIO_IOMMU_MAP_DMA, &map) == 0 ||
        (errno == EBUSY && vfio_dma_unmap(container, iova, size, NULL) == 0 &&
         ioctl(container->fd, VFIO_IOMMU_MAP_DMA, &map) == 0)) {
        return 0;
    }

Clearly that's only triggered with -EBUSY and you're getting -EINVAL,
did we unintentionally change the errno for this?  What's the host
kernel version?

It's my expectation that this really shouldn't happen, the above is a
lazy workaround, but a listener being told to map two different things
at the same address range without an unmap in between seems like it
should violate the MemoryListener protocol.  Thanks,

Alex




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