xfs: compare btree block keys to parent block's keys during scrub

When we're done checking all the records/keys in a btree block, compute
the low and high key of the block and compare them to the associated key
in the parent btree block.

Signed-off-by: Darrick J. Wong <darrick.wong@oracle.com>
Reviewed-by: Brian Foster <bfoster@redhat.com>
This commit is contained in:
Darrick J. Wong 2017-10-25 15:03:46 -07:00
Родитель 8210f4dda2
Коммит 2fdbec5cbe
3 изменённых файлов: 52 добавлений и 2 удалений

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@ -2027,7 +2027,7 @@ error0:
}
/* Find the high key storage area from a regular key. */
STATIC union xfs_btree_key *
union xfs_btree_key *
xfs_btree_high_key_from_key(
struct xfs_btree_cur *cur,
union xfs_btree_key *key)
@ -2101,7 +2101,7 @@ xfs_btree_get_node_keys(
}
/* Derive the keys for any btree block. */
STATIC void
void
xfs_btree_get_keys(
struct xfs_btree_cur *cur,
struct xfs_btree_block *block,

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@ -541,5 +541,9 @@ int64_t xfs_btree_diff_two_ptrs(struct xfs_btree_cur *cur,
void xfs_btree_get_sibling(struct xfs_btree_cur *cur,
struct xfs_btree_block *block,
union xfs_btree_ptr *ptr, int lr);
void xfs_btree_get_keys(struct xfs_btree_cur *cur,
struct xfs_btree_block *block, union xfs_btree_key *key);
union xfs_btree_key *xfs_btree_high_key_from_key(struct xfs_btree_cur *cur,
union xfs_btree_key *key);
#endif /* __XFS_BTREE_H__ */

Просмотреть файл

@ -357,6 +357,50 @@ xfs_scrub_btree_get_block(
return xfs_scrub_btree_block_check_siblings(bs, *pblock);
}
/*
* Check that the low and high keys of this block match the keys stored
* in the parent block.
*/
STATIC void
xfs_scrub_btree_block_keys(
struct xfs_scrub_btree *bs,
int level,
struct xfs_btree_block *block)
{
union xfs_btree_key block_keys;
struct xfs_btree_cur *cur = bs->cur;
union xfs_btree_key *high_bk;
union xfs_btree_key *parent_keys;
union xfs_btree_key *high_pk;
struct xfs_btree_block *parent_block;
struct xfs_buf *bp;
if (level >= cur->bc_nlevels - 1)
return;
/* Calculate the keys for this block. */
xfs_btree_get_keys(cur, block, &block_keys);
/* Obtain the parent's copy of the keys for this block. */
parent_block = xfs_btree_get_block(cur, level + 1, &bp);
parent_keys = xfs_btree_key_addr(cur, cur->bc_ptrs[level + 1],
parent_block);
if (cur->bc_ops->diff_two_keys(cur, &block_keys, parent_keys) != 0)
xfs_scrub_btree_set_corrupt(bs->sc, cur, 1);
if (!(cur->bc_flags & XFS_BTREE_OVERLAPPING))
return;
/* Get high keys */
high_bk = xfs_btree_high_key_from_key(cur, &block_keys);
high_pk = xfs_btree_high_key_addr(cur, cur->bc_ptrs[level + 1],
parent_block);
if (cur->bc_ops->diff_two_keys(cur, high_bk, high_pk) != 0)
xfs_scrub_btree_set_corrupt(bs->sc, cur, 1);
}
/*
* Visit all nodes and leaves of a btree. Check that all pointers and
* records are in order, that the keys reflect the records, and use a callback
@ -418,6 +462,7 @@ xfs_scrub_btree(
/* End of leaf, pop back towards the root. */
if (cur->bc_ptrs[level] >
be16_to_cpu(block->bb_numrecs)) {
xfs_scrub_btree_block_keys(&bs, level, block);
if (level < cur->bc_nlevels - 1)
cur->bc_ptrs[level + 1]++;
level++;
@ -442,6 +487,7 @@ xfs_scrub_btree(
/* End of node, pop back towards the root. */
if (cur->bc_ptrs[level] > be16_to_cpu(block->bb_numrecs)) {
xfs_scrub_btree_block_keys(&bs, level, block);
if (level < cur->bc_nlevels - 1)
cur->bc_ptrs[level + 1]++;
level++;