You would need to have the following occur, to make your scenario plausible:
* you write the object, which places it on a majority of the replica nodes (i.e. 2 out of 3) * replication is slowly churning away, but doesn't quite catch up * both the nodes that have the updated data fail simultaneously, before replication catches up the remaining node. Swift chooses the A & P from CAP. if the swift proxy where to wait till all replicas got updated before it returned a reply, it would be choosing the C but probably dropping both the P and maybe the A (depending on how it would handle a failure). So yes.. you are hitting CAP on the head... On Fri, Jan 20, 2012 at 6:56 PM, Nikolaus Rath <nikol...@rath.org> wrote: > On 01/20/2012 06:35 PM, Pete Zaitcev wrote: > > On Fri, 20 Jan 2012 15:17:32 -0500 > > Nikolaus Rath <nikol...@rath.org> wrote: > > > >> Thanks! So there is no way to reliably get the most-recent version of an > >> object under all conditions. > > > > If you bend the conditions hard enough to hit the CAP theorem, you do. > > From what I have heard so far, it seems to be sufficient if all servers > holding the newest replica are down for me to get old data. I don't > think that this condition is already hitting the CAP theorem, or is it? > > > Best, > > -Nikolaus > > -- > »Time flies like an arrow, fruit flies like a Banana.« > > PGP fingerprint: 5B93 61F8 4EA2 E279 ABF6 02CF A9AD B7F8 AE4E 425C > > _______________________________________________ > Mailing list: https://launchpad.net/~openstack > Post to : openstack@lists.launchpad.net > Unsubscribe : https://launchpad.net/~openstack > More help : https://help.launchpad.net/ListHelp >
_______________________________________________ Mailing list: https://launchpad.net/~openstack Post to : openstack@lists.launchpad.net Unsubscribe : https://launchpad.net/~openstack More help : https://help.launchpad.net/ListHelp