[MERGE] Tree.revision_tree

[Aaron Bentley]

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Robert Collins wrote:
>>Now that we've got a clean separation between repositories, branches,
>>and trees, I think we should be hesitant to go and blur those boundaries
>>again.
> 
> 
> I agree. Thats why the api is strict about only returning a tree if it
> actually has the data, rather than having it ever go and fetch the
> inventory from the repository.

No, I don't think we actually do agree here.  Because I'm thinking about
this in terms of responsibilities, and you're thinking about it in terms
of data layout.

I think the responsibility for providing revision_trees should lie with
Repository.  That matches all our existing code's expectations.  What's
the expected usage of this api?  It seems like it would be this:

try:
    base_tree = tree.revision_tree(base_revision_id)
except NoSuchRevision:
    base_tree = branch.repository.revision_tree(base_revision_id)

Is there any case in which we would do something else?  If not, I think
we should try to fix the API so that we don't have to repeat chunks like
this everywhere.


>>And since only part of a tree is ever local to the WorkingTree (the
>>inventory), it's hard to say that the basis tree is "locally present",
>>either.
> 
> 
> Well like I say above, its about exposing what dirstate offers cleanly.
> 
> Other suggestions appreciated, but I need some way of getting at it that
> is of broad scope, to change other code to use it.

My suggestion is to register working trees with their repositories, so
that repository.revision_tree() will use a cached inventory if one is
available in any live working tree that uses the repository.

A solution like this automatically upgrades all our code.  It also gets
away from special-casing basis_tree.

What do you think?

Aaron
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