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Newsgroups: comp.lang.ada
Date: Mon, 3 Jul 2017 10:24:25 -0700 (PDT)
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Subject: Re: derived formal types and known discriminants
From: Jere <jhb.chat@gmail.com>
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Xref: news.eternal-september.org comp.lang.ada:47267
Date: 2017-07-03T10:24:25-07:00
List-Id: <comp.lang.ada>

On Thursday, June 22, 2017 at 8:45:22 PM UTC-4, sbelm...@gmail.com wrote:
> Can any language lawyer explain the reasoning behind precluding known dis=
criminants for general formal derived types?  For example, I can write this=
:
>=20
> generic
>   type T is new P with private; -- constrained (i.e. none)
>  =20
> and I can write this:
>=20
> generic
>   type T (<>) is new P with private;  -- unconstrained (i.e. any)
>  =20
> But I can't write
>=20
> generic
>   type T (foo : Integer) is new P with private;  -- (just one integer)
>  =20
> presumably because 12.5.1~11/3 says "The declaration of a formal derived =
type shall not have a known_discriminant_part", to wit GNAT complains "disc=
riminants not allowed for this formal type".  It would seems as though if u=
nknown discriminants are legal, known would be an easier subset, and my int=
erest has been piqued to the reason why.
>=20
> Thank you for any insight
>=20
> -sb

I've often wondered this myself.  The best I came up with was to use an int=
ermediate type:

type NP(Foo : Integer) is new P with null record;

and then do the generic like this:

generic
   type T is new NP with private;
package G is
   subtype Test_Type is T;
end G;

A bit more cumbersome, but that's how I managed it.