Re: Operator visibility: why are modular types different??!

[bobduff@world.std.com (Robert A Duff)]

In article <4mbcuu$4ne@fozzie.sun3.iaf.nl>,
Geert Bosch <geert@fozzie.sun3.iaf.nl> wrote:
>procedure UseTest is
>
>   package A is
>--      type I is range 0..2**32 - 1;   -- Program compiles with this def.
>      type I is mod 2**32;		-- This gives visibility error
>   end A;
>
>   package B is
>      subtype J is A.I;
>      use type J;
>   end B;
>
>   use B;
>
>   X   : J := 1;
>
>begin
>   if X /= 1 then

This is illegal.  The "use type J" makes the operators directly visible
inside B.  But use clauses are not transitive.  The "use B" makes things
declared in B directly visible.  It does not grab everything directly
visible in B.

If you moved the "use type J" outside B, then it would work.

There is no difference between signed integers and modular integers in
this respect.  So it sounds like it may be a compiler bug, that the
compiler didn't detect the error in the signed case.

- Bob