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I have a RAID5 array with 3 devices as follows:

~$ cat /proc/mdstat 
Personalities : [raid1] [linear] [multipath] [raid0] [raid6] [raid5] [raid4] [raid10] 
md127 : active raid5 sdb[4] sdc1[2] sda1[3]
      955537408 blocks super 1.2 level 5, 64k chunk, algorithm 2 [3/3] [UUU]
      bitmap: 2/4 pages [8KB], 65536KB chunk

As you can see, sdb isn't partitioned while the other devices are.

The sector count for each is:

Device     Sectors 
/dev/sda1  955799596 
/dev/sdc1  955799596 
/dev/sdb   976773168

If one of the drives fails, will a non-partitioned member make it harder to replace the failed drive?

If so, should I remove sdb now and give it the same partition table as the others before adding it back (at the cost of slow I/O while the array rebuilds)?

TommyPeanuts
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1 Answers1

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Actually non-partitioned drives are easier to replace: you don't need to create a partition of the right size, just to use a big enough replacement drive. You also can't misalign partitions with hardware blocks (most drives are physical 4k nowadays). If you're using only one partition on each drive, you can as well simply use the full drives without any partitioning.

wazoox
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  • Oh OK, so in effect (and as long as any replacement drive is the same size or bigger than the smallest drive) a single partition is treated as the same as no partition? – TommyPeanuts May 03 '22 at 16:08
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    @TommyPeanuts yes, a partition table with a single partition simply adds a small gap at the beginning of the disk and nothing else. You don't need a partition table unless you need to install a bootloader (or obviously, wish to create several partitions). Windows can't use hard drives without partitioning them first OTOH. – wazoox May 03 '22 at 16:48