vdev_draid_min_asize() returns the minimum size of a child vdev. This
is used when determining if a disk is big enough to replace a child.
It's also used by zdb to determine how big of a child to make to test
replacement.
vdev_draid_min_asize() says that the child’s asize has to be at least
1/Nth of the entire draid’s asize, which is the same logic as raidz.
However, this contradicts the code in vdev_draid_open(), which
calculates the draid’s asize based on a reduced child size:
An additional 32MB of scratch space is reserved at the end of each
child for use by the dRAID expansion feature
So the problem is that you can replace a draid disk with one that’s
vdev_draid_min_asize(), but it actually needs to be larger to accommodate
the additional 32MB. The replacement is allowed and everything works at
first (since the reserved space is at the end, and we don’t try to use
it yet), but when you try to close and reopen the pool,
vdev_draid_open() calculates a smaller asize for the draid, because of
the smaller leaf, which is not allowed.
I think the confusion is that vdev_draid_min_asize() is correctly
returning the amount of required *allocatable* space in a leaf, but the
actual *size* of the leaf needs to be at least 32MB more than that.
ztest_vdev_attach_detach() assumes that it can attach that size of
device, and it actually can (the kernel/libzpool accepts it), but it
then later causes zdb to not be able to open the pool.
This commit changes vdev_draid_min_asize() to return the required size
of the leaf, not the size that draid will make available to the metaslab
allocator.
Reviewed-by: Brian Behlendorf <behlendorf1@llnl.gov>
Reviewed-by: Mark Maybee <mark.maybee@delphix.com>
Signed-off-by: Matthew Ahrens <mahrens@delphix.com>
Closes#11459Closes#12221
7e2212990b