On Thu, Jan 14, 2016 at 2:52 PM, Alex Williamson <alex.williamson at redhat.com> wrote: > On Thu, 2016-01-14 at 14:03 +0800, Jike Song wrote: >> On Wed, Dec 16, 2015 at 12:38 PM, Alex Williamson >> <alex.williamson at redhat.com> wrote: >> > >> > So it works. Is it acceptable? Useful? Sufficiently complete? Does >> > it imply deprecating the uio interface? I believe the feature that >> > started this discussion was support for MSI/X interrupts so that VFs >> > can support some kind of interrupt (uio only supports INTx since it >> > doesn't allow DMA). Implementing that would be the ultimate test of >> > whether this provides dpdk with not only a more consistent interface, >> > but the feature dpdk wants that's missing in uio. Thanks, >> > >> Hi Alex, >> >> Sorry for jumping in. Just being curious, how does VFIO No-IOMMU mode >> support DMA from userspace drivers? If I understand correctly, due to >> the absence of IOMMU, pcidev has to use physaddr to start a DMA >> transaction, but how it is supposed to get physaddr from userspace >> drivers, /proc//pagemap or something else? > > Hi Jike, > > vfio no-iommu does nothing to facilitate DMA mappings, UIO didn't > either and DPDK managed to work with that. vfio no-iommu is meant to > be an enabler and provide a consistent vfio device model (with MSI/X > interrupts), but fundamentally the idea of a non-iommu protected, user > owned device capable of DMA is unsupportable. This is why vfio no- > iommu taints the kernel. With that in mind, one of the design goals is > to introduce as little code as possible for vfio no-iommu. A new vfio > iommu backend with pinning and virt-to-bus translation goes against > that design goal. I don't know the details of how DPDK did this with > UIO, but the same lack of DMA mapping facilities is present with vfio > no-iommu. It really just brings the vfio device model, nothing more. > Thanks, > > Alex
Thanks! - that addressed my question :) By the way, my previous assumption(consulting /proc/<pid>/pagemap) apparently doesn't work: one cannot assume a usespace buffer is physically continuous. -- Thanks, Jike
