enforce()?

[Andrei Alexandrescu]

On 07/19/2010 01:50 PM, Steven Schveighoffer wrote:
> On Mon, 19 Jul 2010 13:47:38 -0400, Andrei Alexandrescu
> <SeeWebsiteForEmail at erdani.org> wrote:
>
>> On 07/19/2010 12:23 PM, Steven Schveighoffer wrote:
>>> On Mon, 19 Jul 2010 12:21:36 -0400, Andrei Alexandrescu
>>> <SeeWebsiteForEmail at erdani.org> wrote:
>>>>
>>>> By the way, I'm still eagerly waiting for your red-black tree
>>>> implementation.
>>>
>>> Sorry for the delay, I've been very busy at work, and I wanted to slip
>>> in a couple druntime fixes for array appending.
>>>
>>> All that is left really is the unit testing, and making the docs more
>>> phobos-ish.
>>>
>>>> I think it would be pure awesomeness if you massaged the red/black bit
>>>> inside one of the pointers. I figured out a way of doing that without
>>>> throwing off the garbage collector:
>>>
>>> Yes, that works (BTW, you don't need the union, I hate unions :), just
>>> substitute _bits for _left everywhere, I think it would even work with a
>>> moving GC).
>>
>> Walter told me that union is instrumental to keeping the compiler in
>> the know about such shenanigans. What does your idea look like? You
>> mean keeping a possibly misaligned RBTreeNode pointer and manipulating
>> that? I think that's a bit worse than unions because it transforms a
>> sure thing into a maybe works thing.
>
> I don't pretend to know what ominous problems Walter knows about
> regarding the compiler's view, but here is what I'm thinking:
>
> If a pointer points to the beginning of a node, and a node has at least
> one pointer in it (which it must, since it's a tree), then pointing one
> byte into the node means you're still pointing at the same block, making
> sure the GC doesn't collect.
>
> Really, the generated code will be exactly the same as your solution,
> but it's less of a misuse of the type system IMO (believe it or not).
> I'd rather use casts when you are trying to use something that's typed
> as one thing as something else. When using unions, I usually expect only
> one member of the union to be valid at any one time.
>
> And wouldn't a union be more egregious with the upcoming mostly-precise
> scanner?

I don't think so (applied to all of the above) for reasons of various 
degree of obviousness, but perhaps it's not worth expanding on such a 
minor issue.

Andrei