xfs: replace xfs_btree_has_record with a general keyspace scanner

The current implementation of xfs_btree_has_record returns true if it
finds /any/ record within the given range.  Unfortunately, that's not
sufficient for scrub.  We want to be able to tell if a range of keyspace
for a btree is devoid of records, is totally mapped to records, or is
somewhere in between.  By forcing this to be a boolean, we conflated
sparseness and fullness, which caused scrub to return incorrect results.
Fix the API so that we can tell the caller which of those three is the
current state.

Signed-off-by: Darrick J. Wong <djwong@kernel.org>
Reviewed-by: Dave Chinner <dchinner@redhat.com>

1 files changed, 16 insertions, 0 deletions

diff --git a/fs/xfs/libxfs/xfs_rmap_btree.c b/fs/xfs/libxfs/xfs_rmap_btree.c
index 84e2b692f034..66beb87caf1a 100644
--- a/fs/xfs/libxfs/xfs_rmap_btree.c
+++ b/fs/xfs/libxfs/xfs_rmap_btree.c
@@ -444,6 +444,21 @@ xfs_rmapbt_recs_inorder(
 	return 0;
 }
 
+STATIC enum xbtree_key_contig
+xfs_rmapbt_keys_contiguous(
+	struct xfs_btree_cur		*cur,
+	const union xfs_btree_key	*key1,
+	const union xfs_btree_key	*key2)
+{
+	/*
+	 * We only support checking contiguity of the physical space component.
+	 * If any callers ever need more specificity than that, they'll have to
+	 * implement it here.
+	 */
+	return xbtree_key_contig(be32_to_cpu(key1->rmap.rm_startblock),
+				 be32_to_cpu(key2->rmap.rm_startblock));
+}
+
 static const struct xfs_btree_ops xfs_rmapbt_ops = {
 	.rec_len		= sizeof(struct xfs_rmap_rec),
 	.key_len		= 2 * sizeof(struct xfs_rmap_key),
@@ -463,6 +478,7 @@ static const struct xfs_btree_ops xfs_rmapbt_ops = {
 	.diff_two_keys		= xfs_rmapbt_diff_two_keys,
 	.keys_inorder		= xfs_rmapbt_keys_inorder,
 	.recs_inorder		= xfs_rmapbt_recs_inorder,
+	.keys_contiguous	= xfs_rmapbt_keys_contiguous,
 };
 
 static struct xfs_btree_cur *