get address of object if opCast is overridden

[js.mdnq]

On Saturday, 1 December 2012 at 23:57:27 UTC, Artur Skawina wrote:
> On 12/01/12 20:26, Jonathan M Davis wrote:
>> On Saturday, December 01, 2012 18:43:22 Timon Gehr wrote:
>>> On 12/01/2012 06:23 PM, Jonathan M Davis wrote:
>>>> On Saturday, December 01, 2012 12:05:49 Artur Skawina wrote:
>>>>>> So, unless there's a way to do it without a cast, you're 
>>>>>> stuck. And I
>>>>>> have
>>>>>> no idea how you could possibly do it without a cast.
>>>>>>
>>>>>     *cast(void**)&O // assuming O is a class
>>>>
>>>> Are you sure? I'd be _very_ wary of that, because references 
>>>> aren't the
>>>> same as pointers. I don't believe that there's any guarantee 
>>>> whatsoever
>>>> that that will work, even if it happens to work now (which 
>>>> it may or may
>>>> not).
>>>>
>>>> - Jonathan M Davis
>>>
>>> Certainly works now. Another workaround that would keep 
>>> working if class
>>> references were not pointers is:
>>>
>>> ((Object o)=>cast(void*)o)(O) // assuming O is the class 
>>> object
>> 
>> Ah, good point. Using Object would work just fine and doesn't 
>> rely on the implementation details of references.
>
> References not appearing as if they were pointers is less 
> likely than Object
> growing an opCast...
>
> Another workaround (for both non-empty classes and structs) 
> would be
>
>   cast(void*)&O.tupleof[0]-O.tupleof[0].offsetof
>
> which will let you not worry about that kind of potential 
> changes to the core
> of the language. :)
>
>
> Seriously though, if one only needs to compare the addresses of 
> class objects,
> "is" may be a better solution.
>
> artur

I'm not just comparing them but using them as a unique ID for the 
objects in an algorithm to prevent computing over the same object 
more than once.