Re: [RFC PATCH 5/7] Btrfs: add btrfs_compare_trees function

[Alex Lyakas <alex.bolshoy.btrfs@gmail.com>]

Alexander,

>>> +               if (advance_left && !left_end_reached) {
>>> +                       ret = tree_advance(left_root, left_path, &left_level,
>>> +                                       left_root_level,
>>> +                                       advance_left != ADVANCE_ONLY_NEXT,
>>> +                                       &left_key);
>>> +                       if (ret < 0)
>>> +                               left_end_reached = ADVANCE;
>>> +                       advance_left = 0;
>>> +               }
>>> +               if (advance_right && !right_end_reached) {
>>> +                       ret = tree_advance(right_root, right_path, &right_level,
>>> +                                       right_root_level,
>>> +                                       advance_right != ADVANCE_ONLY_NEXT,
>>> +                                       &right_key);
>>> +                       if (ret < 0)
>>> +                               right_end_reached = ADVANCE;
>>> +                       advance_right = 0;
>>> +               }
>> Do you think it's worth it to put a check/warning/smth before that,
>> that either advance_right or advance_left is non-zero, or we have
>> reached ends in both trees?
>>
>>
> Having the left or right end reached before the other sides end is
> reached is something that is completely normal and expected.
What I meant was that when we start the loop, either advance_left!=0
or advance_right!=0. So I thought it would be good to notice that.
However, on the very first loop iteration, both of them are zero, so I
was wrong.


>>> +               } else if (left_level == right_level) {
>> ...
>>> +               } else if (left_level < right_level) {
>>> +                       advance_right = ADVANCE;
>>> +               } else {
>>> +                       advance_left = ADVANCE;
>>> +               }
>> Can you pls explain why it is correct?
>> Why if we are on lower level in the "newer" tree than we are in the
>> "older" tree, we need to advance the "older" tree? I.e., why this
>> implies that we are on the lower key in the "older" tree? (And
>> vice-versa). I.e., how difference in levels indicates relation between
>> keys?
> Difference in levels has no relation to the keys. These advances
> basically try to keep the two trees positions "in-sync". The compare
> always tries to get both trees to a point where they are at the same
> level, as only then we can compare keys. Also, the two trees may have
> different root levels, this code also handles that case.

Can you pls tell me if I understand your algorithm correctly:
Basically, we need to get to the leaf levels and compare the items in
the leafs. Only when we are on the leaf level, we can safely signal
deletions and additions of items, not on upper levels.
There is only one optimization: we want to find nodes that are shared,
and such nodes can be only on the same level. To make this
optimization happen, we try to always match the levels of the tree.
This is the purpose of:
		} else if (left_level < right_level) {
			advance_right = ADVANCE;
		} else {
			advance_left = ADVANCE;
		}

Note: I think that instead of comparing levels, we could always
compare keys and ADVANCE the lower key. (Because on ADVANCing we never
loose information, we just get closer to leafs, so we don't skip
anything.) But then there is less chance of optimization. Does this
make sense? So what you said that we can compare keys only on the same
level...we can always compare them, correct?

I will now study the rest of your patchset.

Thanks!
Alex.