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  1. 22 5月, 2015 9 次提交
  3. 20 5月, 2015 4 次提交
  5. 19 5月, 2015 3 次提交
  7. 06 5月, 2015 4 次提交
  9. 02 5月, 2015 1 次提交
    • I
      rbd: end I/O the entire obj_request on error · 082a75da
      由 Ilya Dryomov 提交于 
      When we end I/O struct request with error, we need to pass
obj_request->length as @nr_bytes so that the entire obj_request worth
of bytes is completed.  Otherwise block layer ends up confused and we
trip on

    rbd_assert(more ^ (which == img_request->obj_request_count));

in rbd_img_obj_callback() due to more being true no matter what.  We
already do it in most cases but we are missing some, in particular
those where we don't even get a chance to submit any obj_requests, due
to an early -ENOMEM for example.

A number of obj_request->xferred assignments seem to be redundant but
I haven't touched any of obj_request->xferred stuff to keep this small
and isolated.

Cc: Alex Elder <elder@linaro.org>
Cc: stable@vger.kernel.org # 3.10+
Reported-by: NShawn Edwards <lesser.evil@gmail.com>
Reviewed-by: NSage Weil <sage@redhat.com>
Signed-off-by: NIlya Dryomov <idryomov@gmail.com>
      082a75da
  11. 22 4月, 2015 1 次提交
  13. 20 4月, 2015 2 次提交
  15. 16 4月, 2015 11 次提交
  17. 15 4月, 2015 1 次提交
  19. 12 4月, 2015 1 次提交
  21. 11 4月, 2015 1 次提交
    • J
      sd, mmc, virtio_blk, string_helpers: fix block size units · b9f28d86
      由 James Bottomley 提交于 
      The current string_get_size() overflows when the device size goes over
2^64 bytes because the string helper routine computes the suffix from
the size in bytes.  However, the entirety of Linux thinks in terms of
blocks, not bytes, so this will artificially induce an overflow on very
large devices.  Fix this by making the function string_get_size() take
blocks and the block size instead of bytes.  This should allow us to
keep working until the current SCSI standard overflows.

Also fix virtio_blk and mmc (both of which were also artificially
multiplying by the block size to pass a byte side to string_get_size()).

The mathematics of this is pretty simple:  we're taking a product of
size in blocks (S) and block size (B) and trying to re-express this in
exponential form: S*B = R*N^E (where N, the exponent is either 1000 or
1024) and R < N.  Mathematically, S = RS*N^ES and B=RB*N^EB, so if RS*RB
< N it's easy to see that S*B = RS*RB*N^(ES+EB).  However, if RS*BS > N,
we can see that this can be re-expressed as RS*BS = R*N (where R =
RS*BS/N < N) so the whole exponent becomes R*N^(ES+EB+1)

[jejb: fix incorrect 32 bit do_div spotted by kbuild test robot <fengguang.wu@intel.com>]
Acked-by: NUlf Hansson <ulf.hansson@linaro.org>
Reviewed-by: NAndrew Morton <akpm@linux-foundation.org>
Signed-off-by: NJames Bottomley <JBottomley@Odin.com>
      b9f28d86
  23. 08 4月, 2015 2 次提交