Situation: A six disk RAID5/6 array with a completely failed disk. The failed disk is removed and an identical replacement drive is plugged in.
Here I have two options for replacing the disk, assuming the old drive is device 6 in the superblock and the replacement disk is /dev/sda. 'btrfs replace start 6 /dev/sda /mnt' This will start a rebuild of the array using the new drive, copying data that would have been on device 6 to the new drive from the parity data. btrfs add /dev/sda /mnt && btrfs device delete missing /mnt This adds a new device (the replacement disk) to the array and dev delete missing appears to trigger a rebalance before deleting the missing disk from the array. The end result appears to be identical to option 1. A few weeks back I recovered an array with a failed drive using 'delete missing' because 'replace' caused a kernel panic. I later discovered that this was not (just) a failed drive but some other failed hardware that I've yet to start diagnosing. Either motherboard or HBA. The drives are in a new server now and I am currently rebuilding the array with 'replace', which is believe is the "more correct" way to replace a bad drive in an array. Both work, but 'replace' seems to be slower so I'm curious what the functional differences are between the two. I thought the replace would be faster as I assumed it would need to read fewer blocks since instead of a complete rebalance it's just rebuilding a drive from parity data. What are the differences between the two under the hood? The only obvious difference I could see is that when I ran `replace` the space on the replacement drive was instantly allocated under 'filesystem show' while when I used 'device delete' the drive usage slowly crept up through the course of the rebalance. -- To unsubscribe from this list: send the line "unsubscribe linux-btrfs" in the body of a message to [email protected] More majordomo info at http://vger.kernel.org/majordomo-info.html
