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
