Re: A simple ADA puzzle (I haven't the answer)

[Georg Bauhaus <sb463ba@l1-hrz.uni-duisburg.de>]

Frank J. Lhota <NOSPAM.lhota.adarose@verizon.net> wrote:
: "Hyman Rosen" <hyrosen@mail.com> wrote in message
:>
:> Because when you write
:>
:>      type STRING_TYPE (LEN : INTEGER := 0) is
:>        record STRING : LEN_TYPE (1 .. LEN); end record;
:>
:> it's perfectly reasonable to expect that the array is
:> allocated just enough room to hold whatever LEN is
:> specified.

I guess this is true all the more there is a default
value for the discriminant. How is one to know that
a default value for a length does not actually
extend to the type's size? 

with Ada.Text_IO;

procedure str is

   type T(size: Natural := 0) is record
      comp: String(1.. size);
   end record;

   package TIO renames Ada.Text_IO;

begin
   TIO.put_line("type's size is" & Natural'base'image(T'size));
end str;

There is not even an explicit declaration of a T object
anywhere in the program, yet it fails miserably. The
compiler warns, but nevertheless it doesn't explain
why (a) object of this type may raise Storage_Error,
and (b), more subtly I guess, the following:

 12.    TIO.put_line("type's size is" & Natural'base'image(T'size));
                                                            |
     >>> warning: Constraint_Error will be raised at run-time

(Seen in the light of a Natural size constraint intended
for the type.)

: The C equivalent of these declarations would have used
: 
:    char STRING[ 0x7FFFFFFF ];

I don't think so, the literal in the Ada code is no greater
than 0, and who knows, maybe people are lead to believe that
a default value of 0 (not?) used for the last index of a string
will give them a rather short string.