It is possible for a single SGL to span an aligned boundary, eg if the SGL
is
61440 -> 90112
Then the length is 28672, which currently limits the block size to
32k. With a 32k page size the two covering blocks will be:
32768->65536 and 65536->98304
However, the correct answer is a 128K block size which will span the whole
28672 bytes in a single block.
Instead of limiting based on length figure out which high IOVA bits don't
change between the start and end addresses. That is the highest useful
page size.
Fixes: 4a35339958f1 ("RDMA/umem: Add API to find best driver supported page size in an MR")
Link: https://lore.kernel.org/r/1-v2-270386b7e60b+28f4-umem_1_jgg@nvidia.com
Reviewed-by: Leon Romanovsky <leonro@nvidia.com>
Reviewed-by: Shiraz Saleem <shiraz.saleem@intel.com>
Signed-off-by: Jason Gunthorpe <jgg@nvidia.com>
return 0;
va = virt;
- /* max page size not to exceed MR length */
- mask = roundup_pow_of_two(umem->length);
+ /* The best result is the smallest page size that results in the minimum
+ * number of required pages. Compute the largest page size that could
+ * work based on VA address bits that don't change.
+ */
+ mask = pgsz_bitmap &
+ GENMASK(BITS_PER_LONG - 1,
+ bits_per((umem->length - 1 + virt) ^ virt));
/* offset into first SGL */
pgoff = umem->address & ~PAGE_MASK;