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From: "patrick.gunia@googlemail.com" <patrick.gunia@googlemail.com>
Newsgroups: comp.lang.ada
Subject: Re: Ada-Singleton-Why does it work like this?
Date: Wed, 25 Mar 2009 07:10:00 -0700 (PDT)
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Date: 2009-03-25T07:10:00-07:00
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> If you want to declare an access type, you do not need an instance;
> you only need a type, which may be unconstrained:
>
> type T (<>) is limited private;
> type Access_T is access T; -- OK
>
> If you want an access *value*, then you do need an object; either one
> from a storage pool, or a pre-existing one:
>
> Access_Value : Access_T :=3D new T'(...);
> -- storage pool. Possible only in the package declaring T, since T is
> private.
>
> Other_Access_Value : Access_T :=3D Access_Value; -- possible anywhere
>
> Neither Access_Value nor Other_Access_Value allow you to see the
> components of T because T is private.

I=B4m currently trying to formulate some rules for using Ada myself,
trying to explain the concepts which are used. I got one last question
concerning the box-operator (<>) in this particular context.
It declares a type to be unconstrained, which means, that it needs to
be initialized. I only found this construction together with generic
types, for example here: http://en.wikibooks.org/wiki/Ada_Programming/Gener=
ics
When I try to define a sort-routine I can use this generic form to
state, that discrete types may be used as input for such a routine.
But as I don=B4t use this construct together with the keyword generic, I
simply don=B4t get the meaning of (<>) in my Singleton-context. Do I say
that the type I declare is a discrete type? And what would I have to
do to use this type if it wasn=B4t limited and private?

Thank you!