On 13.04.2018 03:04, John Snow wrote:
>
>
> 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
>
I prefer not talking of active checkpoint. It is a kind of controversial.
Better just keep in mind that last bitmap is active one. So for checkpoints 1 2 3 4
we have bitmaps:
1 1->2 2->3 3->4
Note the first bitmap name. When it was created name 2 was unknown so we'd better
have this name for the first bitmap.
Checkpoint 4 can not be used without checkpoint 5 by design to it is not a problem
that 3->4 is active.
Nikolay
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