Btrfs: use bit operation for ->fs_state

There is no lock to protect fs_info->fs_state, it will introduce
some problems, such as the value may be covered by the other task
when several tasks modify it. For example:
	Task0 - CPU0		Task1 - CPU1
	mov %fs_state rax
	or $0x1 rax
				mov %fs_state rax
				or $0x2 rax
	mov rax %fs_state
				mov rax %fs_state
The expected value is 3, but in fact, it is 2.

Though this problem doesn't happen now (because there is only one
flag currently), the code is error prone, if we add other flags,
the above problem will happen to a certainty.

Now we use bit operation for it to fix the above problem.
In this way, we can make the code more robust and be easy to
add new flags.

Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Josef Bacik <jbacik@fusionio.com>

1 files changed, 1 insertions, 1 deletions

diff --git a/fs/btrfs/scrub.c b/fs/btrfs/scrub.c
index 67783e0..c78b2a3 100644
--- a/fs/btrfs/scrub.c
+++ b/fs/btrfs/scrub.c
@@ -2708,7 +2708,7 @@ static noinline_for_stack int scrub_supers(struct scrub_ctx *sctx,
 	int	ret;
 	struct btrfs_root *root = sctx->dev_root;
 
-	if (root->fs_info->fs_state & BTRFS_SUPER_FLAG_ERROR)
+	if (test_bit(BTRFS_FS_STATE_ERROR, &root->fs_info->fs_state))
 		return -EIO;
 
 	gen = root->fs_info->last_trans_committed;