On Sun, Apr 25, 2010 at 5:43 PM, Jonathan Ellis <jbel...@gmail.com> wrote: > On Sun, Apr 25, 2010 at 5:40 PM, Tatu Saloranta <tsalora...@gmail.com> wrote: >>> Now with TimeUUIDType, if two UUID have the same timestamps, they are >>> ordered >>> by bytes order. >> >> Naively for the whole UUID? That would not be good, given that >> timestamp within UUID is not stored in expected lexical order, but >> with sort of little-endian mess (first bytes are least-significant >> bytes of timestamp). > > I think the code here is clearer than explaining in English. :) > > comparing timeuuids o1 and o2: > > long t1 = LexicalUUIDType.getUUID(o1).timestamp(); > long t2 = LexicalUUIDType.getUUID(o2).timestamp(); > return t1 < t2 ? -1 : (t1 > t2 ? 1 : > FBUtilities.compareByteArrays(o1, o2));
:-) Yes, this makes sense, so it is a two-part sort, not just latter part. -+ Tatu +- ps. Not sure if this matters, but I am finally working on Java Uuid Generator v3, which might help with time-location based UUIDs. Will announce it on the list when it's ready (in couple of weeks)
