Re: extended membership tests

[Dan <dan@irvine.com>]

On Mar 31, 12:55 am, Dan <d...@irvine.com> wrote:
> On Mar 31, 12:34 am, AdaMagica <christ-usch.gr...@t-online.de> wrote:
>
>
>
>
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> > On 31 Mrz., 09:04, Dan <d...@irvine.com> wrote:
>
> > > Here's a fun quiz to test your Ada2012 knowledge, based on an
> > > observation by Yannick Moy.
>
> > > Consider this Ada 2012 procedure:
>
> > > with ada.containers.ordered_sets;
> > > procedure membership is
> > >     package int_sets is new ada.containers.ordered_sets(integer);
> > >     function "="(x,y: integer) return boolean is begin return true;
> > > end;
> > >     x: integer := 6;
> > >     y: integer := 12;
> > >     b1: boolean := x in y;
> > >     b2: boolean := x in int_sets.to_set(new_item => y);
> > > begin
> > >     null;
> > > end;
>
> > My first reaction was WTF!
>
> > But RM 4.5.2 seems to make x in y legal, but false here.
>
> > x in y | 1..7
>
> > would be true.
>
> > b2 is illegal, the simple expression must be of an elementary type.
>
> > Hope this is correct.- Hide quoted text -
>
> > - Show quoted text -
>
> I think if the choice(s) resolve to the same type they don't have to
> be elementary,
> according to 4.5.2(3/3).  (other than that, good answer!)- Hide quoted text -
>
> - Show quoted text -

but you are right, they don't the lhs doesn't resolve to the same type
as the rhs,
so in this case the rhs does need to be elementary.