Re: Little people supporting Ada, possibly through AdaCore?

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

On Thu, 19 Jul 2012 00:57:42 +0400, Vasiliy Molostov wrote:

> Dmitry A. Kazakov <mailbox@dmitry-kazakov.de> яПНяПНяПНяПНяПН(яПН) яПН яПНяПНяПНяПНяПН яПНяПНяПНяПНяПНяПН Wed,  
> 18 Jul 2012 21:14:35 +0400:
> 
>> On Wed, 18 Jul 2012 06:33:34 -0700 (PDT), Marc C wrote:
>>
>>> An "unordered" set is an *unordered* set. It is inappropriate to expect
>>> any kind  of intrinsic or 'order of insertion' ordering of elements in
>>> such a set.
>>
>> True. However a set of truly unordered elements cannot be implemented
>> otherwise than on top of an ordered set. E.g. to be able to implement
>> membership test. Either the container imposes some order on the elements  
>> or else the elements have some arbitrary order.
> 
> how about the following:
> 
> type Unordered_Set is
> record
> 	null;
> end;
>
> type Unordered_Element (Belongs_to: access Unordered_Set)is
> record
> 	null; -- data
> end;
> 
> elements has no order between each other, except membership itself, and a  
> set has no ordering of elements also.
> 
> The membership test is very simple here: (element.belongs_to =  
> unordered'access)

How would you copy such sets?

P.S. If you put book keeping in the elements, that is no more a set of
original element. Yes it is equivalent to it in some sense. But so an
ordered set would too.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de