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xfs: split indlen reservations fairly when under reserved

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Certain workoads that punch holes into speculative preallocation can
cause delalloc indirect reservation splits when the delalloc extent is
split in two. If further splits occur, an already short-handed extent
can be split into two in a manner that leaves zero indirect blocks for
one of the two new extents. This occurs because the shortage is large
enough that the xfs_bmap_split_indlen() algorithm completely drains the
requested indlen of one of the extents before it honors the existing
reservation.

This ultimately results in a warning from xfs_bmap_del_extent(). This
has been observed during file copies of large, sparse files using 'cp
--sparse=always.'

To avoid this problem, update xfs_bmap_split_indlen() to explicitly
apply the reservation shortage fairly between both extents. This smooths
out the overall indlen shortage and defers the situation where we end up
with a delalloc extent with zero indlen reservation to extreme
circumstances.

Reported-by: Patrick Dung <mpatdung@gmail.com>
Signed-off-by: Brian Foster <bfoster@redhat.com>
Reviewed-by: Darrick J. Wong <darrick.wong@oracle.com>
Signed-off-by: Darrick J. Wong <darrick.wong@oracle.com>
This commit is contained in:
Brian Foster 2017-02-13 22:48:30 -08:00 committed by Darrick J. Wong
parent 0e339ef855
commit 75d65361cf
1 changed files with 43 additions and 18 deletions
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61
fs/xfs/libxfs/xfs_bmap.c
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@ -4795,34 +4795,59 @@ xfs_bmap_split_indlen(
xfs_filblks_t len2 = *indlen2; xfs_filblks_t len2 = *indlen2;
xfs_filblks_t nres = len1 + len2; /* new total res. */ xfs_filblks_t nres = len1 + len2; /* new total res. */
xfs_filblks_t stolen = 0; xfs_filblks_t stolen = 0;
xfs_filblks_t resfactor;
/* /*
* Steal as many blocks as we can to try and satisfy the worst case * Steal as many blocks as we can to try and satisfy the worst case
* indlen for both new extents. * indlen for both new extents.
*/ */
while (nres > ores && avail) { if (ores < nres && avail)
nres--; stolen = XFS_FILBLKS_MIN(nres - ores, avail);
avail--; ores += stolen;
stolen++;
} /* nothing else to do if we've satisfied the new reservation */
if (ores >= nres)
return stolen;
/* /*
* The only blocks available are those reserved for the original * We can't meet the total required reservation for the two extents.
* extent and what we can steal from the extent being removed. * Calculate the percent of the overall shortage between both extents
* If this still isn't enough to satisfy the combined * and apply this percentage to each of the requested indlen values.
* requirements for the two new extents, skim blocks off of each * This distributes the shortage fairly and reduces the chances that one
* of the new reservations until they match what is available. * of the two extents is left with nothing when extents are repeatedly
* split.
*/ */
while (nres > ores) { resfactor = (ores * 100);
if (len1) { do_div(resfactor, nres);
len1--; len1 *= resfactor;
nres--; do_div(len1, 100);
len2 *= resfactor;
do_div(len2, 100);
ASSERT(len1 + len2 <= ores);
ASSERT(len1 < *indlen1 && len2 < *indlen2);
/*
* Hand out the remainder to each extent. If one of the two reservations
* is zero, we want to make sure that one gets a block first. The loop
* below starts with len1, so hand len2 a block right off the bat if it
* is zero.
*/
ores -= (len1 + len2);
ASSERT((*indlen1 - len1) + (*indlen2 - len2) >= ores);
if (ores && !len2 && *indlen2) {
len2++;
ores--;
}
while (ores) {
if (len1 < *indlen1) {
len1++;
ores--;
} }
if (nres == ores) if (!ores)
break; break;
if (len2) { if (len2 < *indlen2) {
len2--; len2++;
nres--; ores--;
} }
} }