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authorDave Chinner <dchinner@redhat.com>2012-11-12 22:09:44 +1100
committerBen Myers <bpm@sgi.com>2012-11-13 14:45:29 -0600
commit07428d7f0ca46087f7f1efa895322bb9dc1ac21d (patch)
treeb3fc138245723f3a82c6844d7a8884c39406da39 /fs/xfs
parent579b62faa5fb16ffeeb88cda5e2c4e95730881af (diff)
xfs: fix attr tree double split corruption
In certain circumstances, a double split of an attribute tree is needed to insert or replace an attribute. In rare situations, this can go wrong, leaving the attribute tree corrupted. In this case, the attr being replaced is the last attr in a leaf node, and the replacement is larger so doesn't fit in the same leaf node. When we have the initial condition of a node format attribute btree with two leaves at index 1 and 2. Call them L1 and L2. The leaf L1 is completely full, there is not a single byte of free space in it. L2 is mostly empty. The attribute being replaced - call it X - is the last attribute in L1. The way an attribute replace is executed is that the replacement attribute - call it Y - is first inserted into the tree, but has an INCOMPLETE flag set on it so that list traversals ignore it. Once this transaction is committed, a second transaction it run to atomically mark Y as COMPLETE and X as INCOMPLETE, so that a traversal will now find Y and skip X. Once that transaction is committed, attribute X is then removed. So, the initial condition is: +--------+ +--------+ | L1 | | L2 | | fwd: 2 |---->| fwd: 0 | | bwd: 0 |<----| bwd: 1 | | fsp: 0 | | fsp: N | |--------| |--------| | attr A | | attr 1 | |--------| |--------| | attr B | | attr 2 | |--------| |--------| .......... .......... |--------| |--------| | attr X | | attr n | +--------+ +--------+ So now we go to replace X, and see that L1:fsp = 0 - it is full so we can't insert Y in the same leaf. So we record the the location of attribute X so we can track it for later use, then we split L1 into L1 and L3 and reblance across the two leafs. We end with: +--------+ +--------+ +--------+ | L1 | | L3 | | L2 | | fwd: 3 |---->| fwd: 2 |---->| fwd: 0 | | bwd: 0 |<----| bwd: 1 |<----| bwd: 3 | | fsp: M | | fsp: J | | fsp: N | |--------| |--------| |--------| | attr A | | attr X | | attr 1 | |--------| +--------+ |--------| | attr B | | attr 2 | |--------| |--------| .......... .......... |--------| |--------| | attr W | | attr n | +--------+ +--------+ And we track that the original attribute is now at L3:0. We then try to insert Y into L1 again, and find that there isn't enough room because the new attribute is larger than the old one. Hence we have to split again to make room for Y. We end up with this: +--------+ +--------+ +--------+ +--------+ | L1 | | L4 | | L3 | | L2 | | fwd: 4 |---->| fwd: 3 |---->| fwd: 2 |---->| fwd: 0 | | bwd: 0 |<----| bwd: 1 |<----| bwd: 4 |<----| bwd: 3 | | fsp: M | | fsp: J | | fsp: J | | fsp: N | |--------| |--------| |--------| |--------| | attr A | | attr Y | | attr X | | attr 1 | |--------| + INCOMP + +--------+ |--------| | attr B | +--------+ | attr 2 | |--------| |--------| .......... .......... |--------| |--------| | attr W | | attr n | +--------+ +--------+ And now we have the new (incomplete) attribute @ L4:0, and the original attribute at L3:0. At this point, the first transaction is committed, and we move to the flipping of the flags. This is where we are supposed to end up with this: +--------+ +--------+ +--------+ +--------+ | L1 | | L4 | | L3 | | L2 | | fwd: 4 |---->| fwd: 3 |---->| fwd: 2 |---->| fwd: 0 | | bwd: 0 |<----| bwd: 1 |<----| bwd: 4 |<----| bwd: 3 | | fsp: M | | fsp: J | | fsp: J | | fsp: N | |--------| |--------| |--------| |--------| | attr A | | attr Y | | attr X | | attr 1 | |--------| +--------+ + INCOMP + |--------| | attr B | +--------+ | attr 2 | |--------| |--------| .......... .......... |--------| |--------| | attr W | | attr n | +--------+ +--------+ But that doesn't happen properly - the attribute tracking indexes are not pointing to the right locations. What we end up with is both the old attribute to be removed pointing at L4:0 and the new attribute at L4:1. On a debug kernel, this assert fails like so: XFS: Assertion failed: args->index2 < be16_to_cpu(leaf2->hdr.count), file: fs/xfs/xfs_attr_leaf.c, line: 2725 because the new attribute location does not exist. On a production kernel, this goes unnoticed and the code proceeds ahead merrily and removes L4 because it thinks that is the block that is no longer needed. This leaves the hash index node pointing to entries L1, L4 and L2, but only blocks L1, L3 and L2 to exist. Further, the leaf level sibling list is L1 <-> L4 <-> L2, but L4 is now free space, and so everything is busted. This corruption is caused by the removal of the old attribute triggering a join - it joins everything correctly but then frees the wrong block. xfs_repair will report something like: bad sibling back pointer for block 4 in attribute fork for inode 131 problem with attribute contents in inode 131 would clear attr fork bad nblocks 8 for inode 131, would reset to 3 bad anextents 4 for inode 131, would reset to 0 The problem lies in the assignment of the old/new blocks for tracking purposes when the double leaf split occurs. The first split tries to place the new attribute inside the current leaf (i.e. "inleaf == true") and moves the old attribute (X) to the new block. This sets up the old block/index to L1:X, and newly allocated block to L3:0. It then moves attr X to the new block and tries to insert attr Y at the old index. That fails, so it splits again. With the second split, the rebalance ends up placing the new attr in the second new block - L4:0 - and this is where the code goes wrong. What is does is it sets both the new and old block index to the second new block. Hence it inserts attr Y at the right place (L4:0) but overwrites the current location of the attr to replace that is held in the new block index (currently L3:0). It over writes it with L4:1 - the index we later assert fail on. Hopefully this table will show this in a foramt that is a bit easier to understand: Split old attr index new attr index vanilla patched vanilla patched before 1st L1:26 L1:26 N/A N/A after 1st L3:0 L3:0 L1:26 L1:26 after 2nd L4:0 L3:0 L4:1 L4:0 ^^^^ ^^^^ wrong wrong The fix is surprisingly simple, for all this analysis - just stop the rebalance on the out-of leaf case from overwriting the new attr index - it's already correct for the double split case. Signed-off-by: Dave Chinner <dchinner@redhat.com> Reviewed-by: Mark Tinguely <tinguely@sgi.com> Signed-off-by: Ben Myers <bpm@sgi.com>
Diffstat (limited to 'fs/xfs')
-rw-r--r--fs/xfs/xfs_attr_leaf.c20
1 files changed, 18 insertions, 2 deletions
diff --git a/fs/xfs/xfs_attr_leaf.c b/fs/xfs/xfs_attr_leaf.c
index d330111ca738..70eec1829776 100644
--- a/fs/xfs/xfs_attr_leaf.c
+++ b/fs/xfs/xfs_attr_leaf.c
@@ -1291,6 +1291,7 @@ xfs_attr_leaf_rebalance(xfs_da_state_t *state, xfs_da_state_blk_t *blk1,
leaf2 = blk2->bp->b_addr;
ASSERT(leaf1->hdr.info.magic == cpu_to_be16(XFS_ATTR_LEAF_MAGIC));
ASSERT(leaf2->hdr.info.magic == cpu_to_be16(XFS_ATTR_LEAF_MAGIC));
+ ASSERT(leaf2->hdr.count == 0);
args = state->args;
trace_xfs_attr_leaf_rebalance(args);
@@ -1361,6 +1362,7 @@ xfs_attr_leaf_rebalance(xfs_da_state_t *state, xfs_da_state_blk_t *blk1,
* I assert that since all callers pass in an empty
* second buffer, this code should never execute.
*/
+ ASSERT(0);
/*
* Figure the total bytes to be added to the destination leaf.
@@ -1422,10 +1424,24 @@ xfs_attr_leaf_rebalance(xfs_da_state_t *state, xfs_da_state_blk_t *blk1,
args->index2 = 0;
args->blkno2 = blk2->blkno;
} else {
+ /*
+ * On a double leaf split, the original attr location
+ * is already stored in blkno2/index2, so don't
+ * overwrite it overwise we corrupt the tree.
+ */
blk2->index = blk1->index
- be16_to_cpu(leaf1->hdr.count);
- args->index = args->index2 = blk2->index;
- args->blkno = args->blkno2 = blk2->blkno;
+ args->index = blk2->index;
+ args->blkno = blk2->blkno;
+ if (!state->extravalid) {
+ /*
+ * set the new attr location to match the old
+ * one and let the higher level split code
+ * decide where in the leaf to place it.
+ */
+ args->index2 = blk2->index;
+ args->blkno2 = blk2->blkno;
+ }
}
} else {
ASSERT(state->inleaf == 1);