On Sun, Jun 12, 2011 at 4:14 PM, Scott Lawson <scott.law...@manukau.ac.nz> wrote: > I have an interesting question that may or may not be answerable from some > internal > ZFS semantics.
This is really standard Unix filesystem semantics. > [...] > > So total storage used is around ~7.5MB due to the hard linking taking place > on each store. > > If hard linking capability had been turned off, this same message would have > used 1500 x 2MB =3GB > worth of storage. > > My question is there any simple ways of determining the space savings on > each of the stores from the usage of hard links? [...] But... you just did! :) It's: number of hard links * (file size + sum(size of link names and/or directory slot size)). For sufficiently large files (say, larger than one disk block) you could approximate that as: number of hard links * file size. The key is the number of hard links, which will typically vary, but for e-mails that go to all users, well, you know the number of links then is the number of users. You could write a script to do this -- just look at the size and hard-link count of every file in the store, apply the above formula, add up the inflated sizes, and you're done. Nico PS: Is it really the case that Exchange still doesn't deduplicate e-mails? Really? It's much simpler to implement dedup in a mail store than in a filesystem... _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
