BinaryHeap is a range so it goes in std.range. Agree?

[Leandro Lucarella]

Andrei Alexandrescu, el  9 de junio a las 14:59 me escribiste:
> On 06/09/2010 02:48 PM, Walter Bright wrote:
> >Andrei Alexandrescu wrote:
> >>On 06/09/2010 07:57 AM, Michel Fortin wrote:
> >>>On 2010-06-08 17:41:22 -0400, "Simen kjaeraas" <simen.kjaras at gmail.com>
> >>>said:
> >>>
> >>>>Now, my favorite way of dealing with this: Where would I look for a
> >>>>binary heap if I wanted one? I would think of it as a container, and
> >>>>thus
> >>>>check std.container. If it was not there, I would use the search
> >>>>function
> >>>>to find it. I can invent reasons, but it's mostly grounded in learned
> >>>>names and categories.
> >>>
> >>>And if you were accustomed to the STL, you'd just look for a binary heap
> >>>header to include instead of trying to philosophize about which category
> >>>of things it fits best. That's why I suggested "std.binaryheap" earlier.
> >>>(Could be "std.binheap" if you want it short.)
> >>
> >>I don't think this will scale. There are quite a few data structures
> >>out there, I'm afraid we'll have too many modules too soon.
> >
> >One solution is to have std.container consist of the following:
> >
> >public import std.containers.binaryheap;
> >public import std.containers.redblacktree;
> 
> I wanted to do that but with the singular:
> 
> public import std.container.binaryheap;
> 
> So if someone imports std.container then they get them all, if
> someone imports std.container.something then they only get
> something.
> 
> Having a module and a package with the same name does not currently
> work, and I don't think there's a good rationale for that
> limitation.

That's why the convetion .all is used. If that worked I guess .all
wasn't necessary in the first place.

-- 
Leandro Lucarella (AKA luca)                     http://llucax.com.ar/
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