> -----Original Message----- > From: Mark Cranness [mailto:[email protected]] > Sent: Wednesday, November 07, 2012 4:09 AM > To: Community mailing list of VirtualBox users > Subject: Re: [VBox-users] removing redundant snapshots > > [comments inline, below] > > On 6 November 2012 14:27, John A. Wallace <[email protected]> wrote: > > As I am setting up a new VM and making several and frequent changes, > I > > end up making a number of snapshots along the way, especially when it > > is a server to which I am adding features incrementally and assessing > > effects, etc.. Eventually, I get to a stable state, one which is not > > likely to be changed much, if any, indefinitely, and I make a > snapshot > > of the current condition. At this point it would be good to get rid > of > > some of the interim snapshots in order to recover disk space. What is > > the right way (best > > practices) to eliminate these middle snapshots or, say, all of the > > preceding snapshots except possibly the very first one and very last > > one in the sequence? > > If you intend to eventually delete all snapshots, then deleting from > the top tends to minimise total disk IO. > If you intend to leave some snapshots (as you do), then delete from the > bottom up towards those you will keep. > > Deleting the topmost snapshot (so I understand) is a 'backwards merge' > : the topmost differencing disk is merged into the base hard disk. > Deleting a snapshot other than the topmost is a 'forwards merge' : the > differencing disk is merged into the following differencing disk (which > might be for the "Current State"). > > A post here: https://forums.virtualbox.org/viewtopic.php?p=29269, has > some information. > The description there is still accurate, but, terminology has changed a > bit (now "Delete" rather than the older "Discard"). > > An example, imagine 4 snapshots. > There will be 5 VDI files:, the main base VDI, which is frozen > (readonly) and represents snapshot T (top). Then 3 differencing VDIs > for snapshots M1, M2 and M3 (middle). Then a final differencing VDI > for the "Current State". > You intend to delete the two middle snapshots, leaving the last (M3) > and the first (T). > Suppose Then 3 differencing VDIs for snapshots M1, M2 and M3 are each > 1GB in size. > If snapshot M1 is deleted first, its 1GB differencing disk is forward > merged into that for snapshot M2, giving 1GB of data processed, and > snapshot M2 is now 2GB. > Repeat by deleting snapshot M2, and its 2GB is merged forward into > snapshot M3. > Total data merged forward 3GB. Each merge forward has to re-handle the > data carried from the previous snapshots. In a chain of snapshots, the > data from the (other than very very top-most) top snapshots is > repeatedly re-copied forward, duplicating effort. > > If instead snapshot M2 is deleted first, its 1GB is merged into M3, > then M1 is merged forward into M3. > Total data merged forwards 2GB. No double handling. > > (At least that is the theory, if I have it right.) > > At all times make sure you have free space on the host drive. > If you run out of space during a snapshot delete, likely bad things > will happen to your VM! > I recommend as much free space as the total of the base VDI and the > differencing VDIs at all times (check before each snapshot delete). > > > Moreover, I would just as soon remove as many as I can at one time, > if > > possible, using the "vboxmanage.exe" tool. > > They can only be dealt with one at a time, AFAIK. >
Frankly, the only way I can make sense of their order is to reference them chronologically as in earliest or most recent, etc., not physically as in top to bottom, etc. Additionally, two questions now occur to me. First of all, I feel inclined to clarify the merging outcome in the following manner. Secondly, if I understand you right, it sounds as though deleting snapshots cannot result in my loosing the state I would be in as of the most recent snapshot, regardless of which point in time I start with. I mean, if I delete any of them, all I am doing is merging them in this sort of way: 1+1+1+1=4 Now if these represent snapshots taken in time from left to right, and if I delete the first one, it would appear to me necessarily that it becomes merged with the second one, like so: (1+1)+1+1=4 Or I can represent it like this: 2+1+1=4 So now the next question becomes clearer, what happens if I were to delete the middle one of these. I mean, would doing so cause it to become like this: (2+1)+1=4 Or would it become like this: 2+(1+1)=4 It seems clearly that one could easily test this if in fact deleting a snapshot is nothing more than merging two points into one and removing their differences. It would necessarily end up either with a substantially larger first disk image in comparison to the last one, all other things being equal, or with two disk images comparable in size (i.e., roughly equivalent). Regardless of how it ended up happening, what would it matter one way or the other? Why should I care? At the same time, going back to my second point above, if it happens one way or another, it would still always be true, it seems to me, that I needn't worry ever about losing my current state by deleting a snapshot because deleting does not discard changes, it merely merges two adjacent states. Is that not so? Or am I missing something else in this picture? Thanks. John ------------------------------------------------------------------------------ LogMeIn Central: Instant, anywhere, Remote PC access and management. Stay in control, update software, and manage PCs from one command center Diagnose problems and improve visibility into emerging IT issues Automate, monitor and manage. Do more in less time with Central http://p.sf.net/sfu/logmein12331_d2d _______________________________________________ VBox-users-community mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/vbox-users-community _______________________________________________ Unsubscribe: mailto:[email protected]?subject=unsubscribe
