On Thu, 2007-23-08 at 13:04 -0600, Charlie Savage wrote: > > > Instead of trying to ensure that we free the ruby > > object when the xmlNode is freed, we have to ensure that such nodes > > referenced by ruby objects are not freed at all by xmlFree. > > Boy - that's tough in a tree. What if you have a reference to a leaf > node, but no longer to the root. How are you supposed to free the whole > tree when you are done with the leaf?
I'm not sure why this seems tough. If we have a reference to a leaf
when we want to garbage collect the rest of the tree, I am suggesting
removing that leaf from the tree prior to freeing it.
> > I don't think it requires any more mapping than is in place now, with
> > each ruby object knowing the xmlNode it represents. Beyond that the
> > reference counting should be enough.
>
> Not if libxml frees the memory from under you. So your point of view is
> that Ruby entirely controls the memory allocation/deallocation. But see
> my point above for why that could be tough.
Ruby controls which libxml trees are freed, but has no say about how.
If I understand things correctly, calling xmlFree on a node, frees that
node and all of its descendants, so removing any referenced node from
the tree before freeing it should be enough.
> > Right now, each xmlNode knows how many ruby objects it is referenced
> > from (assuming the reference counting works). When a ruby object is
> > freed, it decrements the reference count and if it is the last object
> > referencing the xmlNode, and that xmlNode is not referenced by parent
> > nodes, then the xmlNode is freed using xmlFree. The problem is that
> > xmlFree will free child nodes without checking whether they are
> > referenced from ruby. That's why I think we need to walk the tree and
> > remove such nodes prior to letting xmlFree do its stuff.
>
> Walk what tree? The Ruby tree (thereby creating a bunch of new ruby
> objects and unneeded references)? The libxml c tree? If so, you are
> back to having to know which Ruby objects point to which libxml object.
I meant walk the libxml c tree. I don't think we need to care *which*
ruby objects reference an xmlNode, just whether *any* ruby object does.
If any ruby object references a node, that node must not be be freed by
xmlFree. This, of course, assumes that the ruby libxml reference
counting approach works which may not be true.
> > I don't see how this would be possible. My target documents are built
> > one node at a time from ruby objects. I think ruby needs to own those
> > objects.
>
> When you add the Ruby objects to a tree, the Ruby objects can give up
> ownership (that's simple, set the free method to nil).
That would be quite a change to the API. As it is I can do either of
these things and both seem reasonable:
doc = XML::Document.new
root = XML::node.new('node')
doc.root = root
root['wibble'] = 'wibble'
or
doc = XML::Document.new
root = XML::node.new('node')
root['wibble'] = 'wibble'
doc.root = root
I suspect my app does both, so I wouldn't like to see that change.
__
Marc
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