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vdev_draid_min_asize() ignores reserved space
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 accomodate 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. Closes openzfs#11459 Signed-off-by: Matthew Ahrens <[email protected]>
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