On Sat, Mar 21, 2015 at 10:46 AM, shylesh kumar <shylesh.mo...@gmail.com> wrote: > Hi , > > I was going through this simplified crush algorithm given in ceph website. > > def crush(pg): > all_osds = ['osd.0', 'osd.1', 'osd.2', ...] > result = [] > # size is the number of copies; primary+replicas > while len(result) < size: > --> r = hash(pg) > chosen = all_osds[ r % len(all_osds) ] > if chosen in result: > # OSD can be picked only once > continue > result.append(chosen) > return result > > 10:24 PM (51 minutes ago) > In the line where r = hash(pg) , will it gives the same hash value in every > iteration ? > if that is the case we always endup choosing the same osd from the list > or will the pg number be used as seed for the hashing so that r value > changes in the next iteration. > > Am I missing something really basic ?? > Can somebody please provide me some pointers ?
I'm not sure where this bit of documentation came from, but the selection process includes the "attempt" number as one of the inputs. Where the attempt starts at 0 (or 1, I dunno) and increments each time we try to map a new OSD to the PG. -Greg > > > > -- > Thanks, > Shylesh Kumar M > > > _______________________________________________ > ceph-users mailing list > ceph-users@lists.ceph.com > http://lists.ceph.com/listinfo.cgi/ceph-users-ceph.com > _______________________________________________ ceph-users mailing list ceph-users@lists.ceph.com http://lists.ceph.com/listinfo.cgi/ceph-users-ceph.com
