The normal fsck code doesn't call key_visible_in_snapshot() with an empty list of snapshot IDs seen (the current snapshot ID will always be on the list), but str_hash_repair_key() -> bch2_get_snapshot_overwrites() can, and that's totally fine as long as we check for it.
Signed-off-by: Kent Overstreet <[email protected]> --- fs/bcachefs/fsck.c | 13 ++----------- 1 file changed, 2 insertions(+), 11 deletions(-) diff --git a/fs/bcachefs/fsck.c b/fs/bcachefs/fsck.c index 57ddc20a5cce..9920f1affc5b 100644 --- a/fs/bcachefs/fsck.c +++ b/fs/bcachefs/fsck.c @@ -728,14 +728,8 @@ static int snapshots_seen_update(struct bch_fs *c, struct snapshots_seen *s, static bool key_visible_in_snapshot(struct bch_fs *c, struct snapshots_seen *seen, u32 id, u32 ancestor) { - ssize_t i; - EBUG_ON(id > ancestor); - /* @ancestor should be the snapshot most recently added to @seen */ - EBUG_ON(ancestor != seen->pos.snapshot); - EBUG_ON(ancestor != darray_last(seen->ids)); - if (id == ancestor) return true; @@ -751,11 +745,8 @@ static bool key_visible_in_snapshot(struct bch_fs *c, struct snapshots_seen *see * numerically, since snapshot ID lists are kept sorted, so if we find * an id that's an ancestor of @id we're done: */ - - for (i = seen->ids.nr - 2; - i >= 0 && seen->ids.data[i] >= id; - --i) - if (bch2_snapshot_is_ancestor(c, id, seen->ids.data[i])) + darray_for_each_reverse(seen->ids, i) + if (*i != ancestor && bch2_snapshot_is_ancestor(c, id, *i)) return false; return true; -- 2.50.0
