btrfs-progs picks the latest_dev based on first probed
greatest trans-id. However below test case proofs that
approach is wrong.
$ mkfs.btrfs -d raid1 -m raid1 /dev/sde /dev/sdf
$ modprobe -r btrfs && modprobe btrfs
$ mount -o degraded /dev/sde /btrfs
$ touch /btrfs/testfile && btrfs fi sync /btrfs
The above steps will make /dev/sdf not part of the btrfs.
and as below when you use /dev/sdf the btrfs dev stat
and dev scrub picks up wrong disk
$ btrfs dev stat /dev/sdf
[/dev/sde].write_io_errs 0
[/dev/sde].read_io_errs 0
[/dev/sde].flush_io_errs 0
[/dev/sde].corruption_errs 0
[/dev/sde].generation_errs 0
$ btrfs scrub start -B /dev/sdf
scrub done for 2e99c881-6abd-4f8a-8290-e2f8d0acc575
scrub started at Mon Feb 24 14:45:06 2014 and finished after 0 seconds
total bytes scrubbed: 256.00KiB with 0 errors
Signed-off-by: Anand Jain <Anand.Jain@xxxxxxxxxx>
---
utils.c | 24 ++++++++++++++++++++++--
1 file changed, 22 insertions(+), 2 deletions(-)
diff --git a/utils.c b/utils.c
index e0c750d..231031e 100644
--- a/utils.c
+++ b/utils.c
@@ -1646,6 +1646,11 @@ int get_fs_info(char *path, struct btrfs_ioctl_fs_info_args *fi_args,
memset(fi_args, 0, sizeof(*fi_args));
if (is_block_device(path)) {
+
+ struct btrfs_super_block *disk_super;
+ char *buf;
+ u64 devid;
+
/* Ensure it's mounted, then set path to the mountpoint */
fd = open(path, O_RDONLY);
if (fd < 0) {
@@ -1665,8 +1670,23 @@ int get_fs_info(char *path, struct btrfs_ioctl_fs_info_args *fi_args,
path = mp;
/* Only fill in this one device */
fi_args->num_devices = 1;
- fi_args->max_id = fs_devices_mnt->latest_devid;
- i = fs_devices_mnt->latest_devid;
+
+ buf = malloc(4096);
+ if (!buf) {
+ ret = -ENOMEM;
+ goto out;
+ }
+ disk_super = (struct btrfs_super_block *)buf;
+ ret = btrfs_read_dev_super(fd, disk_super, BTRFS_SUPER_INFO_OFFSET);
+ if (ret < 0) {
+ ret = -EIO;
+ goto out;
+ }
+ devid = btrfs_stack_device_id(&disk_super->dev_item);
+
+ fi_args->max_id = devid;
+ i = devid;
+
memcpy(fi_args->fsid, fs_devices_mnt->fsid, BTRFS_FSID_SIZE);
close(fd);
}
--
1.8.4.2
--
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