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ext4: fix stripe-unaligned allocations 

When a filesystem is created using:

	mkfs.ext4 -b 4096 -E stride=512 <dev>

and we try to allocate 64MB extent, we will end up directly in
ext4_mb_complex_scan_group(). This is because the request is detected
as power-of-two allocation (so we start in ext4_mb_regular_allocator()
with ac_criteria == 0) however the check before
ext4_mb_simple_scan_group() refuses the direct buddy scan because the
allocation request is too large. Since cr == 0, the check whether we
should use ext4_mb_scan_aligned() fails as well and we fall back to
ext4_mb_complex_scan_group().

Fix the problem by checking for upper limit on power-of-two requests
directly when detecting them.

Reported-by: Ross Zwisler <ross.zwisler@linux.intel.com>
Signed-off-by: Jan Kara <jack@suse.cz>
Signed-off-by: Theodore Ts'o <tytso@mit.edu>
This commit is contained in:
Jan Kara 2017-02-10 00:50:56 -05:00 committed by Theodore Ts'o
parent 168316db35
commit d9b22cf9f5
1 changed files with 4 additions and 2 deletions
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6
fs/ext4/mballoc.c
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@ -2146,8 +2146,10 @@ ext4_mb_regular_allocator(struct ext4_allocation_context *ac)
* We search using buddy data only if the order of the request
* is greater than equal to the sbi_s_mb_order2_reqs
* You can tune it via /sys/fs/ext4/<partition>/mb_order2_req
* We also support searching for power-of-two requests only for
* requests upto maximum buddy size we have constructed.
*/
if (i >= sbi->s_mb_order2_reqs) {
if (i >= sbi->s_mb_order2_reqs && i <= sb->s_blocksize_bits + 2) {
/*
* This should tell if fe_len is exactly power of 2
*/
@ -2217,7 +2219,7 @@ repeat:
}
ac->ac_groups_scanned++;
if (cr == 0 && ac->ac_2order < sb->s_blocksize_bits+2)
if (cr == 0)
ext4_mb_simple_scan_group(ac, &e4b);
else if (cr == 1 && sbi->s_stripe &&
!(ac->ac_g_ex.fe_len % sbi->s_stripe))