[PATCH 1/5] btrfs: Ensure at least 1g is free for balance

[Nikolay Borisov <nborisov@suse.com>]

The first part of balance operation is to shrink every constituting
device to ensure there is free space for chunk allocation. However, the code
has been buggy ever since its introduction since calculating the space to shrink
the device by was bounded by 1 mb. Most likely the original intention was to
have an upper bound of 1g and not 1m, since the largest chunk size is 1g. This 
means the first stage in __btrfs_balance so far has been a null op since it
effectively freed just a single megabyte.

Fix this by setting an upper bound of size_to_free of 1g. 

Signed-off-by: Nikolay Borisov <nborisov@suse.com>
---
 fs/btrfs/volumes.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/fs/btrfs/volumes.c b/fs/btrfs/volumes.c
index f435d397019e..8b0fd7bf3447 100644
--- a/fs/btrfs/volumes.c
+++ b/fs/btrfs/volumes.c
@@ -3467,7 +3467,7 @@ static int __btrfs_balance(struct btrfs_fs_info *fs_info)
 	list_for_each_entry(device, devices, dev_list) {
 		old_size = btrfs_device_get_total_bytes(device);
 		size_to_free = div_factor(old_size, 1);
-		size_to_free = min_t(u64, size_to_free, SZ_1M);
+		size_to_free = min_t(u64, size_to_free, SZ_1G);
 		if (!test_bit(BTRFS_DEV_STATE_WRITEABLE, &device->dev_state) ||
 		    btrfs_device_get_total_bytes(device) -
 		    btrfs_device_get_bytes_used(device) > size_to_free ||
-- 
2.7.4