[PATCH 3.2 034/147] xfs: fix inobt inode allocation search optimization
From: Ben Hutchings
Date: Mon Nov 06 2017 - 18:42:00 EST
3.2.95-rc1 review patch. If anyone has any objections, please let me know.
------------------
From: Omar Sandoval <osandov@xxxxxx>
commit c44245b3d5435f533ca8346ece65918f84c057f9 upstream.
When we try to allocate a free inode by searching the inobt, we try to
find the inode nearest the parent inode by searching chunks both left
and right of the chunk containing the parent. As an optimization, we
cache the leftmost and rightmost records that we previously searched; if
we do another allocation with the same parent inode, we'll pick up the
search where it last left off.
There's a bug in the case where we found a free inode to the left of the
parent's chunk: we need to update the cached left and right records, but
because we already reassigned the right record to point to the left, we
end up assigning the left record to both the cached left and right
records.
This isn't a correctness problem strictly, but it can result in the next
allocation rechecking chunks unnecessarily or allocating inodes further
away from the parent than it needs to. Fix it by swapping the record
pointer after we update the cached left and right records.
Fixes: bd169565993b ("xfs: speed up free inode search")
Signed-off-by: Omar Sandoval <osandov@xxxxxx>
Reviewed-by: Christoph Hellwig <hch@xxxxxx>
Reviewed-by: Darrick J. Wong <darrick.wong@xxxxxxxxxx>
Signed-off-by: Darrick J. Wong <darrick.wong@xxxxxxxxxx>
[bwh: Backported to 3.2: adjust filename]
Signed-off-by: Ben Hutchings <ben@xxxxxxxxxxxxxxx>
---
fs/xfs/xfs_ialloc.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
--- a/fs/xfs/xfs_ialloc.c
+++ b/fs/xfs/xfs_ialloc.c
@@ -894,13 +894,13 @@ nextag:
/* free inodes to the left? */
if (useleft && trec.ir_freecount) {
- rec = trec;
xfs_btree_del_cursor(cur, XFS_BTREE_NOERROR);
cur = tcur;
pag->pagl_leftrec = trec.ir_startino;
pag->pagl_rightrec = rec.ir_startino;
pag->pagl_pagino = pagino;
+ rec = trec;
goto alloc_inode;
}