btrfs_free_path(path);
}
+/*
+ * very simple check to peek ahead in the leaf looking for xattrs. If we
+ * don't find any xattrs, we know there can't be any acls.
+ *
+ * slot is the slot the inode is in, objectid is the objectid of the inode
+ */
+static noinline int acls_after_inode_item(struct extent_buffer *leaf,
+ int slot, u64 objectid)
+{
+ u32 nritems = btrfs_header_nritems(leaf);
+ struct btrfs_key found_key;
+ int scanned = 0;
+
+ slot++;
+ while (slot < nritems) {
+ btrfs_item_key_to_cpu(leaf, &found_key, slot);
+
+ /* we found a different objectid, there must not be acls */
+ if (found_key.objectid != objectid)
+ return 0;
+
+ /* we found an xattr, assume we've got an acl */
+ if (found_key.type == BTRFS_XATTR_ITEM_KEY)
+ return 1;
+
+ /*
+ * we found a key greater than an xattr key, there can't
+ * be any acls later on
+ */
+ if (found_key.type > BTRFS_XATTR_ITEM_KEY)
+ return 0;
+
+ slot++;
+ scanned++;
+
+ /*
+ * it goes inode, inode backrefs, xattrs, extents,
+ * so if there are a ton of hard links to an inode there can
+ * be a lot of backrefs. Don't waste time searching too hard,
+ * this is just an optimization
+ */
+ if (scanned >= 8)
+ break;
+ }
+ /* we hit the end of the leaf before we found an xattr or
+ * something larger than an xattr. We have to assume the inode
+ * has acls
+ */
+ return 1;
+}
+
/*
* read an inode from the btree into the in-memory inode
*/
struct btrfs_timespec *tspec;
struct btrfs_root *root = BTRFS_I(inode)->root;
struct btrfs_key location;
+ int maybe_acls;
u64 alloc_group_block;
u32 rdev;
int ret;
alloc_group_block = btrfs_inode_block_group(leaf, inode_item);
+ /*
+ * try to precache a NULL acl entry for files that don't have
+ * any xattrs or acls
+ */
+ maybe_acls = acls_after_inode_item(leaf, path->slots[0], inode->i_ino);
+ if (!maybe_acls) {
+ BTRFS_I(inode)->i_acl = NULL;
+ BTRFS_I(inode)->i_default_acl = NULL;
+ }
+
BTRFS_I(inode)->block_group = btrfs_find_block_group(root, 0,
alloc_group_block, 0);
btrfs_free_path(path);