Re: Operator visibility question

["Jeff Creem" <jeff@thecreems.com>]

I really should be quoting LRM paragraphs when I say something like this but
it seems to me that  GNAT is wrong here since < is not visible and I can not
think of any reason why it should be visible.


It seems that if both sides of the < are MT.T GNAT does detect it but if
one side is universal integer it does not.


It is times like this that I miss Robert Dewar!

Jeff



"Rod Chapman" <rod@praxis-cs.co.uk> wrote in message
news:ba18d5cb.0106030548.62aa45e2@posting.google.com...
> Can anyone please tell me if the following code is legal?
>
> package MT
> is
>    type T is range 0 .. 4;
> end MT;
>
> with MT;
> --# inherit MT;
> package VT
> is
>    function "+" (Left, Right : in MT.T) return MT.T renames MT."+";
>    A : constant := 2;
>    B : constant MT.T := 3;
>
>    C : constant Boolean := (A + 2) < B; -- Is "<" visible here?
> end VT;
>
> The SPARK Examiner rejects the declaration of C with the message:
>
>    9     C : constant Boolean := (A + 2) < B; -- Is "<" visible here?
>                                          ^
> ***        Semantic Error    :309: Operator not visible for these types.
>
> On the other hand, GNAT 3.13 accepts the code with no errors.  Can anyone
> explain which is right, and (more importantly) why?
>  Cheers,
>   Rod Chapman
>   SPARK Team
>   Praxis Critical Systems