Hi, I am involved in doing a ZFS implementation for a research OS developed at DTU, Denmark. In the process, a question has come up: Does ZFS really provide a 128 bit storage system? Jeff Bonwick seems to think so, as he writes on his blog: "A fully-populated 128-bit [zfs?] storage pool would contain 2128 blocks = 2137bytes = 2140 bits;" I just can''t make out the math for this. The structure used for addressing data in a ZFS pool, the DVA uses 32 bits to address the vdev and 63 bits to address the sector offset on this vdev where a given block resides. The DVA is 128 bits, but 8 of these bits is used for Raid-Z info, 24 bits is used to hold the size of the block, and 1 bit is used as a gang flag, and these does not contribute to the amount of addressable blocks. When people say that ZFS is a 128 bit storage system, they refer to the number of addressable blocks in a storage pool, right? It seems to me that one is only able to address 2^32 * 2^63 = 2^95 blocks in each pool. If anybody can put some light on this matter, it will be greatly appreciated. Regards, Andreas Hindborg -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://mail.opensolaris.org/pipermail/zfs-code/attachments/20100524/0a661dd7/attachment.html>