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From: Ludovic Brenta <ludovic@ludovic-brenta.org>
Newsgroups: comp.lang.ada
Subject: Re: Ada-Singleton-Why does it work like this?
Date: Wed, 25 Mar 2009 04:46:23 -0700 (PDT)
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On Mar 25, 11:57=A0am, "patrick.gu...@googlemail.com"
<patrick.gu...@googlemail.com> wrote:
> Yet another question, if my Singleton_Type is unconstrained, why does
> Ada allow creating Access-Types to unconstrained types allowing to
> access there record-properties, while not allowing to create an
> instance of Singleton_Type directly? I thought that Ada demands a
> declaration of an instance before creating an Access-Type to access
> it? Maybe I=B4m mixing in C++, but if I declare a pointer in C++ I
> either have to create a new instance using "new" in heap-memory where
> the pointer points to or I have to use references and then let the
> pointer point to a reference. In both cases I first have to create an
> object and then let the pointer point to it. When I use access types
> in Ada in my example it=B4s possible to create a pointer to a class
> which can=B4t be instantiated, because it is unconstrained. Though it
> works to access the record-properties using the access-type. Could
> anyone tell me what I=B4m getting wrong concerning Ada?
>
> Thank you!

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.

--
Ludovic Brenta.