btrfs: avoid extra memory allocation when copying free space cache

At copy_free_space_cache(), we add a new entry to the block group's ctl
before we free the entry from the temporary ctl. Adding a new entry
requires the allocation of a new struct btrfs_free_space, so we can
avoid a temporary extra allocation by freeing the entry from the
temporary ctl before we add a new entry to the main ctl, which possibly
also reduces the chances for a memory allocation failure in case of very
high memory pressure. So just do that.
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>

fs/btrfs/free-space-cache.c

@@ -923,10 +923,12 @@ static int copy_free_space_cache(struct btrfs_block_group *block_group,
 	while (!ret && (n = rb_first(&ctl->free_space_offset)) != NULL) {
 		info = rb_entry(n, struct btrfs_free_space, offset_index);
 		if (!info->bitmap) {
+			const u64 offset = info->offset;
+			const u64 bytes = info->bytes;
+
 			unlink_free_space(ctl, info, true);
-			ret = btrfs_add_free_space(block_group, info->offset,
-						   info->bytes);
 			kmem_cache_free(btrfs_free_space_cachep, info);
+			ret = btrfs_add_free_space(block_group, offset, bytes);
 		} else {
 			u64 offset = info->offset;
 			u64 bytes = ctl->unit;