diff options
author | Ned Bass <[email protected]> | 2012-02-29 10:08:20 -0800 |
---|---|---|
committer | Brian Behlendorf <[email protected]> | 2012-03-05 09:49:50 -0800 |
commit | 613d88eda89945bb0011ddc05d3bd064bf1a8e1b (patch) | |
tree | 15d24445dc675b198b50e47800e49c803264598d /lib | |
parent | ec2626ad3f695a2ced3946c4197ef64cbcac4959 (diff) |
Align parition end on 1 MiB boundary
Some devices have exhibited sensitivity to the ending alignment of
partitions. In particular, even if the first partition begins at 1
MiB, we have seen many sd driver task abort errors with certain SSDs
if the first partition doesn't end on a 1 MiB boundary. This occurs
when the vdev label is read during pool creation or importation and
causes a delay of about 30 seconds per device. It can also be
simulated with dd when the pool isn't imported:
dd if=/dev/sda1 of=/dev/null bs=262144 count=1
For the record, this problem was observed with SMARTMOD
SG9XCA2E200GE01 200GB SSDs. Unfortunately I don't have a good
explanation for this behavior. It seems to have something to do with
highly fragmented single-sector requests being issued to the device,
which it may not support. With end-aligned partitions at least
page-sized requests were queued and issued to the driver according
to blktrace. In any case, aligning the partition end is a fairly
innocuous work-around, wasting at most 1 MiB of space.
Signed-off-by: Brian Behlendorf <[email protected]>
Closes #574
Diffstat (limited to 'lib')
-rw-r--r-- | lib/libzfs/libzfs_pool.c | 9 |
1 files changed, 6 insertions, 3 deletions
diff --git a/lib/libzfs/libzfs_pool.c b/lib/libzfs/libzfs_pool.c index 4b4d81f7b..311def6b6 100644 --- a/lib/libzfs/libzfs_pool.c +++ b/lib/libzfs/libzfs_pool.c @@ -672,10 +672,12 @@ zpool_expand_proplist(zpool_handle_t *zhp, zprop_list_t **plp) * Don't start the slice at the default block of 34; many storage * devices will use a stripe width of 128k, other vendors prefer a 1m * alignment. It is best to play it safe and ensure a 1m alignment - * give 512b blocks. When the block size is larger by a power of 2 - * we will still be 1m aligned. + * given 512B blocks. When the block size is larger by a power of 2 + * we will still be 1m aligned. Some devices are sensitive to the + * partition ending alignment as well. */ -#define NEW_START_BLOCK 2048 +#define NEW_START_BLOCK 2048 +#define PARTITION_END_ALIGNMENT 2048 /* * Validate the given pool name, optionally putting an extended error message in @@ -3786,6 +3788,7 @@ zpool_label_disk(libzfs_handle_t *hdl, zpool_handle_t *zhp, char *name) if (start_block == MAXOFFSET_T) start_block = NEW_START_BLOCK; slice_size -= start_block; + slice_size = P2ALIGN(slice_size, PARTITION_END_ALIGNMENT); vtoc->efi_parts[0].p_start = start_block; vtoc->efi_parts[0].p_size = slice_size; |