From: Jeffrey Carter <spam.jrcarter.not@spam.not.acm.org>
Subject: Re: How to declare a generic formal type "covered" by another?
Date: Thu, 01 May 2014 13:59:35 -0700
Date: 2014-05-01T13:59:35-07:00 [thread overview]
Message-ID: <ljucjn$in$1@dont-email.me> (raw)
In-Reply-To: <slrnlm3u5u.i0l.lithiumcat@nat.rebma.instinctive.eu>
On 05/01/2014 12:33 AM, Natasha Kerensikova wrote:
>
> Well maybe the problem here is my limited vocabulary. I used the word
> "constrained" there only to mean "Something := new S" or "Variable : S;"
> are both valid, i.e. objects of that type can be created without
> further information. There must be some way to express that in generics.
As Rosen has pointed out, this is the definite/indefinite idea, expressed in
generic formals as lacking/having unknown discriminants ["(<>)"].
Most definite subtypes are constrained, and most indefinite, unconstrained, but
that's not always the case. Consider
subtype VSI is Integer range 0 .. 255;
type VS (Length : VSI := 0) is record
Value : String (1 .. Length);
end record;
VS is definite, because you can say
S : VS;
but it's also unconstrained, because you can change the discriminant:
S := (Length => 1, Value => "A");
S := (Length => 2, Value => "Hi");
For your uses, perhaps you should consider something like
generic -- GP
type T (<>) is limited private;
package GP is
type T_Access is access all T;
type Ref is private;
generic -- F
with function Allocate return T_Access;
function Allocate return Ref;
...
end GP;
Most smart-pointer packages expose the access type; see, for example,
http://www.oopweb.com/Ada/Documents/AdaLinux/Volume/18.html
http://www.adacore.com/adaanswers/gems/gem-97-reference-counting-in-ada-part-1/
HTH.
--
Jeff Carter
"C++ is like jamming a helicopter inside a Miata
and expecting some sort of improvement."
Drew Olbrich
51
next prev parent reply other threads:[~2014-05-01 20:59 UTC|newest]
Thread overview: 16+ messages / expand[flat|nested] mbox.gz Atom feed top
2014-05-01 6:54 How to declare a generic formal type "covered" by another? Natasha Kerensikova
2014-05-01 7:09 ` J-P. Rosen
2014-05-01 7:33 ` Natasha Kerensikova
2014-05-01 13:35 ` J-P. Rosen
2014-05-01 17:56 ` Natasha Kerensikova
2014-05-01 20:59 ` Jeffrey Carter [this message]
2014-05-02 7:58 ` AdaMagica
2014-05-02 8:17 ` Natasha Kerensikova
2014-05-02 15:12 ` Jeffrey Carter
2014-05-02 15:33 ` Dmitry A. Kazakov
2014-05-02 16:00 ` AdaMagica
2014-05-01 9:30 ` Georg Bauhaus
2014-05-01 9:32 ` Georg Bauhaus
2014-05-01 9:33 ` Georg Bauhaus
2014-05-01 16:34 ` Georg Bauhaus
2014-05-01 18:11 ` Natasha Kerensikova
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