Re: OO in Ada

["Matthew Heaney" <mheaney@on2.com>]

"Warren W. Gay VE3WWG" <ve3wwg@cogeco.ca> wrote in message
news:3DA31301.80506@cogeco.ca...
>
> I believe that the problem referred to here is simply one
> of a visibility. Either a 'use' or a fully qualified call is
> required.

No.  Primitive operations for a type are inherited during a derivation (it
doesn't matter what kind of type the parent is), and are therefore
"implicitly declared" immediately following the declaration of the derived
type.

For example:

package P is
   type T is range 0 .. 10;
   procedure Op (O : in T);
end:

with P;
package Q is
   type NT is new P.T;
end;

There is an operation called Op implicitly declared for Q.NT.  To call it
you do NOT have to with P.  For example:

declare
   use Q;
   O : NT;
begin
   Op (O);
end;

Op does NOT need any qualification, because it's already declared in Q, and
the operations in Q (even implicitly declared operations) are directly
visible (here, because use Q).


> If I understand correctly, you're saying you have
> some parent package "Pen", and a child package "Thick_Pen",
> and you use the package as in:
>
>    use Pen.Thick_Pen;
>    ...
>    Draw(...); -- Compile error because Draw is defined in Pen

The Draw operation that takes a thick pen type is (implicitly) declared in
package Thick_Pen.  There is NO compile error.  It doesn't even matter
whether package Think_Pen is a child of Pen.

Primitive operations of a type are inherited during a derivation.  That's
what it means for an operation to be "primitive."

> This is solved either by doing something like:
>
>    use Pen, Pen.Thick_Pen;
>    ...
>    Draw(...);  -- OK now, because Pen.Draw is visible.

This will NOT compile, because Pen.Draw takes Pen_Type, not Thick_Pen_Type.

> or qualifying by your call as in:
>
>    use Pen.Thick_Pen;
>    ...
>    Pen.Draw(...);

No.  This won't compile either.  You need to say:

Pen.Thick_Pen.Draw(...);