Re: Ada-Singleton-Why does it work like this?

[Adam Beneschan <adam@irvine.com>]

On Mar 25, 7:40 am, Ludovic Brenta <ludo...@ludovic-brenta.org> wrote:
> On Mar 25, 3:10 pm, "patrick.gu...@googlemail.com"
>
>
>
> <patrick.gu...@googlemail.com> wrote:
> > > 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 := new T'(...);
> > > -- storage pool. Possible only in the package declaring T, since T is
> > > private.
>
> > > Other_Access_Value : Access_T := Access_Value; -- possible anywhere
>
> > > Neither Access_Value nor Other_Access_Value allow you to see the
> > > components of T because T is private.
>
> > I´m 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/Generics
> > 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´t use this construct together with the keyword generic, I
> > simply don´t 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´t limited and private?
>
> > Thank you!
>
> Please re-read the wikibook chapter on generics; there is a difference
> between:
>
> generic
>    type T1 is <>; -- any discrete type

should be

  generic
     type T1 is (<>);

                            -- Adam