Re: Access to function returning class-wide type

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

On Thu, 21 Aug 2008 21:30:28 +0200, Pawe� 'Nivertius' P�azie�ski wrote:

> Dmitry A. Kazakov wrote:
> 
>> So you could replace it with a map condition->type tag and then proceed
>> with Ada.Tags.Generic_Dispatching_Constructor, because the map
>> tag->constructor is for free, here you do not need pointers any more, in
>> Ada 2005. The compiler will force to override the "constructor" for each
>> concrete derived type.
>> -------------
>> Having said that, there still is a difficult problem with construction of
>> the map. I don't know if it is an issue for you, but it is for the
>> persistency layer I designed. The problem is that if the map is
>> open-ended, you need some mechanism of registration of each new type in
>> the map. This is still unresolved in Ada. The pattern I am using is that
>> each package declaring a new type is required to call some class-wide
>> procedure registering the type upon package elaboration. In your case it
>> is the condition associated with the type tag. That cannot be enforced,
>> and it works only at the library level, i.e. for "immortal" types.
> 
> I think that the map condition->type tag is in my case same problem as map
> condition->constructor pointer. I need the map registration anyway, because
> it basically needs to be changed at run-time.

Why then N is a parameter? When each type knows its N you don't need to
pass it back.

> Altrough, all the types I
> need to register are library-level types, so the registration mechanism is
> pretty simple.

Then the pattern is (assuming N is unknown in advance):

package A is 
   type Abstracted is abstract tagged null record;
   procedure Register
             (  N : out Natural; -- Given upon registration
                Constructor : not null access
                   function return Abstracted'Class
             );
end A;
   
package B is
   type Derived is new Abstracted with ...;
end B;

package body B is
   My_N : Natural;

   function Create return Abstracted'Class is
   begin
        return Derived'(...); -- Returns derived
   end Create;

begin
   Register (My_N, Create'Access);
end B;

You don't need to declare Create in A at all. Each type will declare it in
order to call to Register anyway.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de