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btrfs: unexport btrfs_prev_leaf()

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btrfs_prev_leaf() is not used outside ctree.c, so there's no need to
export it at ctree.h - just make it static at ctree.c and move its
definition above btrfs_search_slot_for_read(), since that function
calls it.

Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
This commit is contained in:
Filipe Manana 2023-04-12 11:33:10 +01:00 committed by David Sterba
parent 45a3e24f65
commit f469c8bd90
2 changed files with 81 additions and 81 deletions
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161
fs/btrfs/ctree.c
View file

@ -2378,6 +2378,87 @@ int btrfs_search_old_slot(struct btrfs_root *root, const struct btrfs_key *key,
return ret;
}
/*
* Search the tree again to find a leaf with smaller keys.
* Returns 0 if it found something.
* Returns 1 if there are no smaller keys.
* Returns < 0 on error.
*
* This may release the path, and so you may lose any locks held at the
* time you call it.
*/
static int btrfs_prev_leaf(struct btrfs_root *root, struct btrfs_path *path)
{
struct btrfs_key key;
struct btrfs_key orig_key;
struct btrfs_disk_key found_key;
int ret;
btrfs_item_key_to_cpu(path->nodes[0], &key, 0);
orig_key = key;
if (key.offset > 0) {
key.offset--;
} else if (key.type > 0) {
key.type--;
key.offset = (u64)-1;
} else if (key.objectid > 0) {
key.objectid--;
key.type = (u8)-1;
key.offset = (u64)-1;
} else {
return 1;
}
btrfs_release_path(path);
ret = btrfs_search_slot(NULL, root, &key, path, 0, 0);
if (ret <= 0)
return ret;
/*
* Previous key not found. Even if we were at slot 0 of the leaf we had
* before releasing the path and calling btrfs_search_slot(), we now may
* be in a slot pointing to the same original key - this can happen if
* after we released the path, one of more items were moved from a
* sibling leaf into the front of the leaf we had due to an insertion
* (see push_leaf_right()).
* If we hit this case and our slot is > 0 and just decrement the slot
* so that the caller does not process the same key again, which may or
* may not break the caller, depending on its logic.
*/
if (path->slots[0] < btrfs_header_nritems(path->nodes[0])) {
btrfs_item_key(path->nodes[0], &found_key, path->slots[0]);
ret = comp_keys(&found_key, &orig_key);
if (ret == 0) {
if (path->slots[0] > 0) {
path->slots[0]--;
return 0;
}
/*
* At slot 0, same key as before, it means orig_key is
* the lowest, leftmost, key in the tree. We're done.
*/
return 1;
}
}
btrfs_item_key(path->nodes[0], &found_key, 0);
ret = comp_keys(&found_key, &key);
/*
* We might have had an item with the previous key in the tree right
* before we released our path. And after we released our path, that
* item might have been pushed to the first slot (0) of the leaf we
* were holding due to a tree balance. Alternatively, an item with the
* previous key can exist as the only element of a leaf (big fat item).
* Therefore account for these 2 cases, so that our callers (like
* btrfs_previous_item) don't miss an existing item with a key matching
* the previous key we computed above.
*/
if (ret <= 0)
return 0;
return 1;
}
/*
* helper to use instead of search slot if no exact match is needed but
* instead the next or previous item should be returned.
@ -4473,86 +4554,6 @@ int btrfs_del_items(struct btrfs_trans_handle *trans, struct btrfs_root *root,
return ret;
}
/*
* search the tree again to find a leaf with lesser keys
* returns 0 if it found something or 1 if there are no lesser leaves.
* returns < 0 on io errors.
*
* This may release the path, and so you may lose any locks held at the
* time you call it.
*/
int btrfs_prev_leaf(struct btrfs_root *root, struct btrfs_path *path)
{
struct btrfs_key key;
struct btrfs_key orig_key;
struct btrfs_disk_key found_key;
int ret;
btrfs_item_key_to_cpu(path->nodes[0], &key, 0);
orig_key = key;
if (key.offset > 0) {
key.offset--;
} else if (key.type > 0) {
key.type--;
key.offset = (u64)-1;
} else if (key.objectid > 0) {
key.objectid--;
key.type = (u8)-1;
key.offset = (u64)-1;
} else {
return 1;
}
btrfs_release_path(path);
ret = btrfs_search_slot(NULL, root, &key, path, 0, 0);
if (ret <= 0)
return ret;
/*
* Previous key not found. Even if we were at slot 0 of the leaf we had
* before releasing the path and calling btrfs_search_slot(), we now may
* be in a slot pointing to the same original key - this can happen if
* after we released the path, one of more items were moved from a
* sibling leaf into the front of the leaf we had due to an insertion
* (see push_leaf_right()).
* If we hit this case and our slot is > 0 and just decrement the slot
* so that the caller does not process the same key again, which may or
* may not break the caller, depending on its logic.
*/
if (path->slots[0] < btrfs_header_nritems(path->nodes[0])) {
btrfs_item_key(path->nodes[0], &found_key, path->slots[0]);
ret = comp_keys(&found_key, &orig_key);
if (ret == 0) {
if (path->slots[0] > 0) {
path->slots[0]--;
return 0;
}
/*
* At slot 0, same key as before, it means orig_key is
* the lowest, leftmost, key in the tree. We're done.
*/
return 1;
}
}
btrfs_item_key(path->nodes[0], &found_key, 0);
ret = comp_keys(&found_key, &key);
/*
* We might have had an item with the previous key in the tree right
* before we released our path. And after we released our path, that
* item might have been pushed to the first slot (0) of the leaf we
* were holding due to a tree balance. Alternatively, an item with the
* previous key can exist as the only element of a leaf (big fat item).
* Therefore account for these 2 cases, so that our callers (like
* btrfs_previous_item) don't miss an existing item with a key matching
* the previous key we computed above.
*/
if (ret <= 0)
return 0;
return 1;
}
/*
* A helper function to walk down the tree starting at min_key, and looking
* for nodes or leaves that are have a minimum transaction id.

1
fs/btrfs/ctree.h
View file

@ -633,7 +633,6 @@ static inline int btrfs_insert_empty_item(struct btrfs_trans_handle *trans,
return btrfs_insert_empty_items(trans, root, path, &batch);
}
int btrfs_prev_leaf(struct btrfs_root *root, struct btrfs_path *path);
int btrfs_next_old_leaf(struct btrfs_root *root, struct btrfs_path *path,
u64 time_seq);