Re: Instantiating private types with discriminants?

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

On Tue, 09 May 2006 13:17:36 GMT, rick H wrote:

> I'm slowly learning Ada for my own amusement, and I've ground to a halt
> trying to understand something.  If some kind sole could explain it to
> me, I'd be very grateful.
> 
> I've defined two types, each with a descriminant, and each with its own
> access type.  One of the type's implementations is, however, private:
> 
> package Discrim is
>    type Type_A (Param : Integer := 100) is null record;
>    type Type_A_Ptr is access Type_A;
> 
>    --  same as above, but implementation now private...
>    type Type_B (Param : Integer := 100) is private;
>    type Type_B_Ptr is access Type_B;
> private
>    type Type_B (Param : Integer := 100) is null record;
> end Discrim;
> 
> When I use "new" on two variables declared as Type_A_Ptr and Type_B_Ptr,
> one requires a type conversion for the discriminant, whereas the other
> requires a qualified expression:
> 
> with Discrim; use Discrim;
> procedure Use_Discrim is
>    A : Type_A_Ptr;                   --  public  implementation
>    B : Type_B_Ptr;                   --  private implementation
> begin
>    A := new Type_A'(Param => 100);   --  qualified expression
>    B := new Type_B (Param => 123);   --  type conversion
> end Use_Discrim;
> 
> So, my question to the experts is: Why does "privatising" a type's
> details change the way that you "new" instantiations of it?

They don't:

   A := new Type_A (Param => 100);   --  That's OK

The only difference is that you can't use aggregates with Type_B because
they are private. You don't know members. Note that you can still use
qualified expressions with Type_B if you have some other way of
constructing Type_B, without aggregates. So if Discrim had a function:

   function Get_Me_Some_B return Type_B;

then:

   B := new Type_B'(Get_Me_Some_B);  -- qualified expression

The only purpose of a qualified expression is to specify the expected type.

Further

   B := new Type_B (Param => 123);

is not a type conversion, it is constraining of Type_B. Type_B (Param =>
123) is a subtype specification. It is same as in:

   B_Object : Type_B (Param => 123); -- subtype of Type_B

You can also compare:

   A := new Type_A'(Param => 100);

with

   A_Object : Type_A := (Param => 100);

Type_A is the type of A_Object, (Param => 100) is an expression resulting
in the initial value. It could also well be:

   A_Object :  Type_A (Param => 100) := (Param => 100);

Here Type_A ((Param => 100) is the subtype of Type_A for all values of
A_Object. (Param => 100) appearing after ":=" is an expression in the form
of an aggregate. You can qualify it if you want:

   A_Object :  Type_A (Param => 100) := Type_A'(Param => 100);

You can do all this with Type_B, except for record aggregates, because they
are private.

Note that both:

   A_Object : Type_A;
   A := new Type_A;

are OK, because you have provided a default value for the discriminant.

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