Re: [Patch] RBTree iteration interface improvement

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От Aleksander Alekseev
Тема Re: [Patch] RBTree iteration interface improvement
Дата
Msg-id 20160826130743.GA9505@e733
обсуждение исходный текст
Ответ на Re: [Patch] RBTree iteration interface improvement  (Heikki Linnakangas <hlinnaka@iki.fi>)
Ответы Re: [Patch] RBTree iteration interface improvement  (Heikki Linnakangas <hlinnaka@iki.fi>)
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Hello, Heikki.

Thank you for your attention to this patch!

> This also seems to change the API so that instead of a single 
> rb_begin_iterate()+rb_iterate() pair, there is a separate begin and 
> iterate function for each traversal order. That seems like an unrelated 
> change. Was there a particular reason for that? I think the old 
> rb_begin_iterate()+rb_iterate() functions were fine, we just need to 
> have a RBTreeIterator struct that's different from the tree itself.

I'm trying to avoid calling procedures by a pointer, an old habit. When
I started to work on this patch I just needed a RB-tree implementation
for a pet project. I took one from PostgreSQL code. Then I found this
flaw of iteration interface and decided to fix it. The idea to merge
this fix back to PostgreSQL code appeared later so I just wrote code the
way I like.

These days code performance depends on many factors like whether code
fits into cache, i.e not only on whether you call a procedure directly
or using a pointer. Until someone finds a real bottleneck here I think
current rb_begin_iterate()+rb_iterate() interface should do just fine.
> Another unrelated change in this patch is the addition of 
> rb_rightmost(). It's not used for anything, so I'm not sure what the 
> point is. Then again, there don't seem to be any callers of 
> rb_leftmost() either.

It's just something I needed in tests to reach a good percent of code
coverage. Implementation of rb_rightmost is trivial so we probably can do
without it.

> I think we should something like in the attached patch. It seems to pass 
> your test suite, but I haven't done any other testing on this. Does it
> look OK to you?

Looks good to me.

-- 
Best regards,
Aleksander Alekseev



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