Re: [PATCH v12 Kernel 4/7] vfio iommu: Implementation of ioctl to for dirty pages tracking.

Alex Williamson
2020-02-17T20:55:31Z

On Tue, 18 Feb 2020 00:43:48 +0530
Kirti Wankhede <kwankhede@nvidia.com> wrote:

> On 2/14/2020 4:50 AM, Alex Williamson wrote:
> > On Fri, 14 Feb 2020 01:41:35 +0530
> > Kirti Wankhede <kwankhede@nvidia.com> wrote:
> >
> >> <snip>
> >>
> >>>>>>
> >>>>>> +static int vfio_iova_dirty_bitmap(struct vfio_iommu *iommu, dma_addr_t iova,
> >>>>>> + size_t size, uint64_t pgsize,
> >>>>>> + unsigned char __user *bitmap)
> >>>>>> +{
> >>>>>> + struct vfio_dma *dma;
> >>>>>> + dma_addr_t i = iova, iova_limit;
> >>>>>> + unsigned int bsize, nbits = 0, l = 0;
> >>>>>> + unsigned long pgshift = __ffs(pgsize);
> >>>>>> +
> >>>>>> + while ((dma = vfio_find_dma(iommu, i, pgsize))) {
> >>>>>> + int ret, j;
> >>>>>> + unsigned int npages = 0, shift = 0;
> >>>>>> + unsigned char temp = 0;
> >>>>>> +
> >>>>>> + /* mark all pages dirty if all pages are pinned and mapped. */
> >>>>>> + if (dma->iommu_mapped) {
> >>>>>> + iova_limit = min(dma->iova + dma->size, iova + size);
> >>>>>> + npages = iova_limit/pgsize;
> >>>>>> + bitmap_set(dma->bitmap, 0, npages);
> >>>>>
> >>>>> npages is derived from iova_limit, which is the number of bits to set
> >>>>> dirty relative to the first requested iova, not iova zero, ie. the set
> >>>>> of dirty bits is offset from those requested unless iova == dma->iova.
> >>>>>
> >>>>
> >>>> Right, fixing.
> >>>>
> >>>>> Also I hope dma->bitmap was actually allocated. Not only does the
> >>>>> START error path potentially leave dirty tracking enabled without all
> >>>>> the bitmap allocated, when does the bitmap get allocated for a new
> >>>>> vfio_dma when dirty tracking is enabled? Seems it only occurs if a
> >>>>> vpfn gets marked dirty.
> >>>>>
> >>>>
> >>>> Right.
> >>>>
> >>>> Fixing error paths.
> >>>>
> >>>>
> >>>>>> + } else if (dma->bitmap) {
> >>>>>> + struct rb_node *n = rb_first(&dma->pfn_list);
> >>>>>> + bool found = false;
> >>>>>> +
> >>>>>> + for (; n; n = rb_next(n)) {
> >>>>>> + struct vfio_pfn *vpfn = rb_entry(n,
> >>>>>> + struct vfio_pfn, node);
> >>>>>> + if (vpfn->iova >= i) {
> >>>>>> + found = true;
> >>>>>> + break;
> >>>>>> + }
> >>>>>> + }
> >>>>>> +
> >>>>>> + if (!found) {
> >>>>>> + i += dma->size;
> >>>>>> + continue;
> >>>>>> + }
> >>>>>> +
> >>>>>> + for (; n; n = rb_next(n)) {
> >>>>>> + unsigned int s;
> >>>>>> + struct vfio_pfn *vpfn = rb_entry(n,
> >>>>>> + struct vfio_pfn, node);
> >>>>>> +
> >>>>>> + if (vpfn->iova >= iova + size)
> >>>>>> + break;
> >>>>>> +
> >>>>>> + s = (vpfn->iova - dma->iova) >> pgshift;
> >>>>>> + bitmap_set(dma->bitmap, s, 1);
> >>>>>> +
> >>>>>> + iova_limit = vpfn->iova + pgsize;
> >>>>>> + }
> >>>>>> + npages = iova_limit/pgsize;
> >>>>>
> >>>>> Isn't iova_limit potentially uninitialized here? For example, if our
> >>>>> vfio_dma covers {0,8192} and we ask for the bitmap of {0,4096} and
> >>>>> there's a vpfn at {4096,8192}. I think that means vpfn->iova >= i
> >>>>> (4096 >= 0), so we break with found = true, then we test 4096 >= 0 +
> >>>>> 4096 and break, and npages = ????/pgsize.
> >>>>>
> >>>>
> >>>> Right, Fixing it.
> >>>>
> >>>>>> + }
> >>>>>> +
> >>>>>> + bsize = dirty_bitmap_bytes(npages);
> >>>>>> + shift = nbits % BITS_PER_BYTE;
> >>>>>> +
> >>>>>> + if (npages && shift) {
> >>>>>> + l--;
> >>>>>> + if (!access_ok((void __user *)bitmap + l,
> >>>>>> + sizeof(unsigned char)))
> >>>>>> + return -EINVAL;
> >>>>>> +
> >>>>>> + ret = __get_user(temp, bitmap + l);
> >>>>>
> >>>>> I don't understand why we care to get the user's bitmap, are we trying
> >>>>> to leave whatever garbage they might have set in it and only also set
> >>>>> the dirty bits? That seems unnecessary.
> >>>>>
> >>>>
> >>>> Suppose dma mapped ranges are {start, size}:
> >>>> {0, 0xa000}, {0xa000, 0x10000}
> >>>>
> >>>> Bitmap asked from 0 - 0x10000. Say suppose all pages are dirty.
> >>>> Then in first iteration for dma {0,0xa000} there are 10 pages, so 10
> >>>> bits are set, put_user() happens for 2 bytes, (00000011 11111111b).
> >>>> In second iteration for dma {0xa000, 0x10000} there are 6 pages and
> >>>> these bits should be appended to previous byte. So get_user() that byte,
> >>>> then shift-OR rest of the bitmap, result should be: (11111111 11111111b)
> >>>>
> >>>> Without get_user() and shift-OR, resulting bitmap would be
> >>>> 111111 00000011 11111111b which would be wrong.
> >>>
> >>> Seems like if we use a put_user() approach then we should look for
> >>> adjacent vfio_dmas within the same byte/word/dword before we push it to
> >>> the user to avoid this sort of inefficiency.
> >>>
> >>
> >> Won't that add more complication to logic?
> >
> > I'm tempted to think it might be less complicated.
> >
> >>>>> Also why do we need these access_ok() checks when we already checked
> >>>>> the range at the start of the ioctl?
> >>>>
> >>>> Since pointer is updated runtime here, better to check that pointer
> >>>> before using that pointer.
> >>>
> >>> Sorry, I still don't understand this, we check access_ok() with a
> >>> pointer and a length, therefore as long as we're incrementing the
> >>> pointer within that length, why do we need to retest?
> >>>
> >>
> >> Ideally caller for put_user() and get_user() must check the pointer with
> >> access_ok() which is used as argument to these functions before calling
> >> this function. That makes sure that pointer is correct after pointer
> >> arithematic. May be lets remove previous check of pointer and length,
> >> but keep these checks.
> >
> > So we don't trust that we can increment a pointer within a range that
> > we've already tested with access_ok() and expect it to still be ok? I
> > think the point of having access_ok() and __put_user() is that we can
> > batch many __put_user() calls under a single access_ok() check. I
> > don't see any justification here why if we already tested
> > access_ok(ptr, 2) that we still need to test access_ok(ptr + 0, 1) and
> > access_ok(ptr + 1, 1), and removing the initial test is clearly the
> > wrong optimization if we agree there is redundancy here.
> >
>
> access_ok(ptr + x, 1), where x is variable, then x shouldn't be out of
> range. If we go with initial test, then there should be check for x,
> such that x is within range.

That logic should already exist though, we shouldn't be trying to fill
a bitmap beyond what the user requested and therefore what we've
already tested that it's sized for and we have access to.

> >>>>>> + if (ret)
> >>>>>> + return ret;
> >>>>>> + }
> >>>>>> +
> >>>>>> + for (j = 0; j < bsize; j++, l++) {
> >>>>>> + temp = temp |
> >>>>>> + (*((unsigned char *)dma->bitmap + j) << shift);
> >>>>>
> >>>>> |=
> >>>>>
> >>>>>> + if (!access_ok((void __user *)bitmap + l,
> >>>>>> + sizeof(unsigned char)))
> >>>>>> + return -EINVAL;
> >>>>>> +
> >>>>>> + ret = __put_user(temp, bitmap + l);
> >>>>>> + if (ret)
> >>>>>> + return ret;
> >>>>>> + if (shift) {
> >>>>>> + temp = *((unsigned char *)dma->bitmap + j) >>
> >>>>>> + (BITS_PER_BYTE - shift);
> >>>>>> + }
> >>>>>
> >>>>> When shift == 0, temp just seems to accumulate bits that never get
> >>>>> cleared.
> >>>>>
> >>>>
> >>>> Hope example above explains the shift logic.
> >>>
> >>> But that example is when shift is non-zero. When shift is zero, each
> >>> iteration of the loop just ORs in new bits to temp without ever
> >>> clearing the bits for the previous iteration.
> >>>
> >>>
> >>
> >> Oh right, fixing it.
> >>
> >>>>>> + }
> >>>>>> +
> >>>>>> + nbits += npages;
> >>>>>> +
> >>>>>> + i = min(dma->iova + dma->size, iova + size);
> >>>>>> + if (i >= iova + size)
> >>>>>> + break;
> >>>>>
> >>>>> So whether we error or succeed, we leave cruft in dma->bitmap for the
> >>>>> next pass. It doesn't seem to make any sense why we pre-allocated the
> >>>>> bitmap, we might as well just allocate it on demand here. Actually, if
> >>>>> we're not going to do a copy_to_user() for some range of the bitmap,
> >>>>> I'm not sure what it's purpose is at all. I think the big advantages
> >>>>> of the bitmap are that we can't amortize the cost across every pinned
> >>>>> page or DMA mapping, we don't need the overhead of tracking unmapped
> >>>>> vpfns, and we can use copy_to_user() to push the bitmap out. We're not
> >>>>> getting any of those advantages here.
> >>>>>
> >>>>
> >>>> That would still not work if dma range size is not multiples of 8 pages.
> >>>> See example above.
> >>>
> >>> I don't understand this comment, what about the example above justifies
> >>> the bitmap?
> >>
> >> copy_to_user() could be used if dma range size is not multiple of 8 pages.
> >
> > s/is not/is/ ?
> >
>
> My bad, you're right.
>
> > And we expect that to be a far more common case, right? I don't think
> > there are too many ranges for a guest that are only mapped in sub-32KB
> > chucks.
> >
> >>> As I understand the above algorithm, we find a vfio_dma
> >>> overlapping the request and populate the bitmap for that range. Then
> >>> we go back and put_user() for each byte that we touched. We could
> >>> instead simply work on a one byte buffer as we enumerate the requested
> >>> range and do a put_user() ever time we reach the end of it and have bits
> >>> set. That would greatly simplify the above example. But I would expect
> >>> that we're a) more likely to get asked for ranges covering a single
> >>> vfio_dma
> >>
> >> QEMU ask for single vfio_dma during each iteration.
> >>
> >> If we restrict this ABI to cover single vfio_dma only, then it
> >> simplifies the logic here. That was my original suggestion. Should we
> >> think about that again?
> >
> > But we currently allow unmaps that overlap multiple vfio_dmas as long
> > as no vfio_dma is bisected, so I think that implies that an unmap while
> > asking for the dirty bitmap has even further restricted semantics. I'm
> > also reluctant to design an ABI around what happens to be the current
> > QEMU implementation.
> >
> > If we take your example above, ranges {0x0000,0xa000} and
> > {0xa000,0x10000} ({start,end}), I think you're working with the
> > following two bitmaps in this implementation:
> >
> > 00000011 11111111b
> > 00111111b
> >
> > And we need to combine those into:
> >
> > 11111111 11111111b
> >
> > Right?
> >
> > But it seems like that would be easier if the second bitmap was instead:
> >
> > 11111100b
> >
> > Then we wouldn't need to worry about the entire bitmap being shifted by
> > the bit offset within the byte, which limits our fixes to the boundary
> > byte and allows us to use copy_to_user() directly for the bulk of the
> > copy. So how do we get there?
> >
> > I think we start with allocating the vfio_dma bitmap to account for
> > this initial offset, so we calculate bitmap_base_iova as:
> > (iova & ~((PAGE_SIZE << 3) - 1))
> > We then use bitmap_base_iova in calculating which bits to set.
> >
> > The user needs to follow the same rules, and maybe this adds some value
> > to the user providing the bitmap size rather than the kernel
> > calculating it. For example, if the user wanted the dirty bitmap for
> > the range {0xa000,0x10000} above, they'd provide at least a 1 byte
> > bitmap, but we'd return bit #2 set to indicate 0xa000 is dirty.
> >
> > Effectively the user can ask for any iova range, but the buffer will be
> > filled relative to the zeroth bit of the bitmap following the above
> > bitmap_base_iova formula (and replacing PAGE_SIZE with the user
> > requested pgsize). I'm tempted to make this explicit in the user
> > interface (ie. only allow bitmaps starting on aligned pages), but a
> > user is able to map and unmap single pages and we need to support
> > returning a dirty bitmap with an unmap, so I don't think we can do that.
> >
>
> Sigh, finding adjacent vfio_dmas within the same byte seems simpler than
> this.

How does KVM do this? My intent was that if all of our bitmaps share
the same alignment then we can merge the intersection and continue to
use copy_to_user() on either side. However, if QEMU doesn't do the
same, it doesn't really help us. Is QEMU stuck with an implementation
of only retrieving dirty bits per MemoryRegionSection exactly because
of this issue and therefore we can rely on it in our implementation as
well? Thanks,

Alex