btrfs: simplify leaf traversal after path release in btrfs_next_old_leaf()

After releasing the path in btrfs_next_old_leaf(), we need to re-check
the leaf because a balance operation may have added items or removed the
last item. The original code handled this with two separate conditional
blocks, the second marked with a lengthy comment explaining a "missed
case".

Merge these two blocks into a single logical structure that handles both
scenarios more clearly.

Also update the comment to be more concise and accurate, incorporating the
explanation directly into the main block rather than a separate annotation.

Signed-off-by: Sun YangKai <sunk67188@gmail.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>

Author: SaltyKitkat

fs/btrfs/ctree.c

@@ -4853,34 +4853,23 @@ int btrfs_next_old_leaf(struct btrfs_root *root, struct btrfs_path *path,
 
 	nritems = btrfs_header_nritems(path->nodes[0]);
 	/*
-	 * by releasing the path above we dropped all our locks.  A balance
-	 * could have added more items next to the key that used to be
-	 * at the very end of the block.  So, check again here and
-	 * advance the path if there are now more items available.
-	 */
-	if (nritems > 0 && path->slots[0] < nritems - 1) {
-		if (ret == 0)
-			path->slots[0]++;
-		ret = 0;
-		goto done;
-	}
-	/*
-	 * So the above check misses one case:
-	 * - after releasing the path above, someone has removed the item that
-	 *   used to be at the very end of the block, and balance between leafs
-	 *   gets another one with bigger key.offset to replace it.
+	 * By releasing the path above we dropped all our locks.  A balance
+	 * could have happened and
 	 *
-	 * This one should be returned as well, or we can get leaf corruption
-	 * later(esp. in __btrfs_drop_extents()).
+	 * 1. added more items after the previous last item
+	 * 2. deleted the previous last item
 	 *
-	 * And a bit more explanation about this check,
-	 * with ret > 0, the key isn't found, the path points to the slot
-	 * where it should be inserted, so the path->slots[0] item must be the
-	 * bigger one.
+	 * So, check again here and advance the path if there are now more
+	 * items available.
 	 */
-	if (nritems > 0 && ret > 0 && path->slots[0] == nritems - 1) {
-		ret = 0;
-		goto done;
+	if (nritems > 0 && path->slots[0] <= nritems - 1) {
+		if (ret == 0 && path->slots[0] != nritems - 1) {
+			path->slots[0]++;
+			goto done;
+		} else if (ret > 0) {
+			ret = 0;
+			goto done;
+		}
 	}
 
 	while (level < BTRFS_MAX_LEVEL) {