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Subject: Re: [geometry] [index] r-tree concerns
From: feverzsj (feverzsj_at_[hidden])
Date: 2012-06-23 02:13:44

hi, Adam

> 1. Should MAX and MIN be compile-time? rtree.v2.01 took run-time MAX and
> MIN. It was just a little slower but took more space than the current
> one because static size nodes were unable to use (vectors takes more
> space). But what are real-life use cases?

If there is only slight impact on performance, I think a more dynamic way would be prefered for lib usage.
The template <Max Min> can be still present for convenience, e.g.:
template<..., int Max = defaultMax, int Min, defaultMin>
struct rtree
    rtree(..., int max_ = Max, int min_ = Min );

> 2. Both versions' queries returns values by copy. This means that if you
> want to modify your values, you must store them somewhere else. When
> e.g. The difference is that rtree.v2 may take various value types. If
> e.g. Point is passed to the rtree.v2, the container behaves as std::set,
> it returns copies of Point. If std::pair<Point, Data> is passed it
> behaves as std::map. But the difference between the rtree and std
> container is that they returns iterators not copies. What is more from
> known reasons they make Keys const and returns iterator<const Point> and
> iterator<pair<Point, Data>> respectively.
> The main difference between std containers and rtree is that rtree's
> iterators probably won't be preserved after the modification of the
> rtree because internally values are stored in arrays/vectors. So, if the
> interface is too close to the std containers users may assume that it
> works exactly the same way.
> My questions are:
> Should rtree.v2 return some iterator<const Point> or
> iterator<std::pair<const Point, Data>> respectively?
> Maby there should be two query interfaces one returning copies, one
> returning iterators?
> What should be the interface of the rtree? Maby there shouldn't be only
> one rtree<Value> but rtree_set<KeyValue> and rtree_map<Kay, Value>?

In old style, rtree (or query result) is traversed by visiter pattern.
While in GP style, concerning the characteristic of rtree, the single pass range concept could be a good choice. But the implementation of this range and iterator may be complex.
So, my suggestion is using the visiter pattern as the basic traverse concept, you or users may develope more complex usage based on this.
For convenience, there may be a interface that return query results as a sequence of values(like rtree.v1 dose);

> 3. The reason why rtree takes rtree<Value> and dispatches it's type is
> that it allows to use user-defined value types. It uses a Translator
> object to get a Key from Value. The default one handles Points, Boxes,
> std::pairs, pointers and iterators. The user may provide whatever type
> he wants along with the translator. I now realized that this may be not
> safe because he may modify the Key outside the rtree after the
> insertion. Should this stay this way? or allow only predefined types
> e.g. Point, Box, pair<Point, T>, pair<Box, T> and tuples?
> Still, if someone wants to brake something he will do it. Even with std
> containers one may use some e.g. wraped pointer as a Key and mess up
> with it.

Actually, my most concern is, besides in-memory rtree, if the rtree.v2 is able to handle a external-memory rtree. All in all, that’s what R-tree for.
So there should be more indirect concepts to be involved. Except this translator to exact bbox, there may be concept for access child nodes.

> 4. It is related to the previous point
> Default translator uses SFINAE, which I believe won't work on all
> compilers. Should there be:
> - hard-coded types instead (less default functionality but someone may
> still write his translator and to be honest those additional types
> probably won't be used),
> - some #ifdef BOOST_SFINAE_SUPPORTED (different interface for some
> compilers),
> - as is (don't compilable on some compilers)?

Generally, traits/policy class would be prefered. You could put the default behaviour in to a default transltor. All these default handled types should be handled there.


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