Re: [libvirt] [RFC v2] external (pull) backup API

On 04/12/2018 10:08 AM, Vladimir Sementsov-Ogievskiy wrote: > > I propose, not to say that bitmap represents a checkpoint. It is simpler > to say (and it reflects the reality) that bitmap is a difference between > two consecutive checkpoints. And we can say, that active state is some > kind of a checkpoint, current point in time. > > So, we have checkpoints (5* is an active state) which are points in time: > > 1 2 3 4 5* > Oh -- the most recent checkpoint there doesn't belong to a ***specific time*** yet. It's a floating checkpoint -- it always represents the most current version. It's not really a checkpoint at all. 1, 2, 3, and 4 however are associated with a specific timestamp though. > And bitmaps, first three are disabled, last is enabled: > > "1->2", "2->3", "3->4", "4->5*" > OK; so 1->2, 2->3 and 3->4 define deltas between two ***defined*** points in time. 4->5* however is only anchored by one specific point in time, and is floating just like the most recent checkpoint is floating. > So, remove first checkpoint: just remove bitmap "A->B". I assume you mean "1->2" here. And... yes, I agree -- if you don't care about your very first checkpoint anymore, you can just delete the first bitmap, too. > Remove any other checkpoint N: create new bitmap "(N-1)->(N+1)" = > merge("(N-1)->N", "N->(N+1)"), drop bitmaps "(N-1)->N" and "N->(N+1)". err, okay, so let's say we want to drop checkpoint 3: create: "2->4" merge: "2->3", "3->4" [and presumably store in "2->4"] drop: 2->3, 3->4 OK, that makes more sense to me. In essence; (1) We could consider this 2->3 absorbing 3->4, or (2) 3->4 absorbing 2->3 and in either case it's the same, really. > If the latter was active, the new one becomes active. And we cant remove > 5* checkpoint, as it is an active state, not an actual checkpoint. OK, crystal. --js