Instantiating private types with discriminants?

[rick H <rik_nntp@dsl.pipex.com>]

Hello,

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?

Thanks very much,
Rick