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From: "Vasiliy Molostov" <molostoff@gmail.com>
Newsgroups: comp.lang.ada
Subject: Re: Little people supporting Ada, possibly through AdaCore?
Date: Thu, 19 Jul 2012 00:57:42 +0400
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Date: 2012-07-19T00:57:42+04:00
List-Id: <comp.lang.ada>

Dmitry A. Kazakov <mailbox@dmitry-kazakov.de> =D0=BF=D0=B8=D1=81=D0=B0=D0=
=BB(=D0=B0) =D0=B2 =D1=81=D0=B2=D0=BE=D1=91=D0=BC =D0=BF=D0=B8=D1=81=D1=8C=
=D0=BC=D0=B5 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 expe=
ct
>> any kind  of intrinsic or 'order of insertion' ordering of elements i=
n
>> 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 elemen=
ts  =

> 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 =3D  =

unordered'access)


-- =

ps. perhaps this pseudo code is not ada, but adaptable well, i think.