Re: bitmap migration bug with -drive while block mirror runs

[Kevin Wolf]

Am 01.10.2019 um 17:57 hat Vladimir Sementsov-Ogievskiy geschrieben:
> 01.10.2019 17:10, John Snow wrote:
> > 
> > 
> > On 10/1/19 10:00 AM, Vladimir Sementsov-Ogievskiy wrote:
> >>> Otherwise: I have a lot of cloudy ideas on how to solve this, but
> >>> ultimately what we want is to be able to find the "addressable" name for
> >>> the node the bitmap is attached to, which would be the name of the first
> >>> ancestor node that isn't a filter. (OR, the name of the block-backend
> >>> above that node.)
> >> Not the name of ancestor node, it will break mapping: it must be name of 
> >> the
> >> node itself or name of parent (may be through several filters) 
> >> block-backend
> >>
> > 
> > Ah, you are right of course -- because block-backends are the only
> > "nodes" for which we actually descend the graph and add the bitmap to
> > its child.
> > 
> > So the real back-resolution mechanism is:
> > 
> > - Find the first non-filter ancestor, A
> > - if A is not a block-backend, we must use our node-local name.
> > - if A's name is empty, we must use our node-local name.
> > - If the name we have chosen is not id_wellformed, we have no
> > migration-stable addressable name for this bitmap and the migration must
> > fail!
> > 
> > 
> > For resolving bitmap addresses via QMP (node, name) pairs; the
> > resolution method would be this:
> > 
> > - if the node-name N is a block-backend, descend the tree until we find
> > the first non-filter node V.
> > - if the node-name N is a BlockDriverState, use this node directly.
> > 
> 
> Looks good for me.
> 
> I also think if on destination we have both block-backend with name N and
> block-node with name N and the latter is not (filtered) child of the former,
> we should fail migration of at least that bitmap. (Hope, nobody reuse
> block-backend names as node-names in practice.. (should we restrict it
> explicitly ?))

You can't have a node and a BlockBackend of the same name, they share a
single namespace. If you try to do so, you get an error.

Kevin