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Date: Wed, 23 Feb 2011 17:59:16 +0200
From: Avi Kivity <avi@redhat.com>
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Subject: Re: [Qemu-devel] Re: Strategic decision: COW format
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To: Anthony Liguori <anthony@codemonkey.ws>
Cc: Kevin Wolf <kwolf@redhat.com>, Chunqiang Tang <ctang@us.ibm.com>, qemu-devel@nongnu.org, Markus Armbruster <armbru@redhat.com>, Stefan Hajnoczi <stefanha@linux.vnet.ibm.com>

On 02/23/2011 05:47 PM, Anthony Liguori wrote:
>> I don't see how that works, since the memory image is duplicated for 
>> each snapshot.  So thousands of snapshots = terabytes of storage, and 
>> hours of creating the snapshots.
>
>
> Fabrice wanted to use CoW to as a mechanism to deduplicate the memory 
> contents with the on-disk state specifically to address this problem.  
> For the longest time, there was a comment in the savevm code along 
> these lines.  It might still be there.
>
> I think the lack of on-disk hashes was a critical missing bit to make 
> this feature really work well.

So you have to use dirty logging to see which pages changed, otherwise 
you have to dedup all of them.  Still I think migration/kemari is a 
better fit for this.

>> Can you elaborate?  What's the problem with a linear array of 
>> snapshots (say up to 10,000 snapshots)?
>
> Lots of things.  The array will start to consume quite a bit of 
> contiguous space as it gets larger which means it needs to be relocated.

If you double the space each time, it amortizes out.

A snapshot seems to be around 40 bytes.  So 10K snapshots = 400KB, 
hardly a huge amount (sans pointed-to data which doesn't need to move).

> Deleting a snapshot is a far more expensive operation than it needs to 
> be.
>

Move the last snapshot into the deleted entry?

-- 
error compiling committee.c: too many arguments to function