Re: extended membership tests

[AdaMagica <christ-usch.grein@t-online.de>]

On 31 Mrz., 16:33, Robert A Duff <bobd...@shell01.TheWorld.com> wrote:
> For extra credit, what happens if we replace b2 with:
>
>     b3: boolean := int_sets.to_set(new_item => x)
>                 in int_sets.to_set(new_item => x)
>                 or int_sets.to_set(new_item => y);

OK, I'll try. We should write this as

   b3: boolean :=
   int_sets.to_set(new_item => x) in
      int_sets.to_set(new_item => x) or int_sets.to_set(new_item =>
y);

So the right-hand side is a choice_expression of the form
simple_expression or simple_expression.
So if there isn't an or operator for sets (I don't know, I didn't
check the set container), it's illegal.
If there is one, it's a set (presumably the union) and it's the
tested_set.
The lhs is a simple_expression which is tagged (I presume sets are
tagged) which must resolve to a type convertible to the tested type,
which it is.
Now we have an individual membership test lhs in rhs.
The simple expression (lhs) is not equal to the value of the
membership_choice (rhs). So the membership test returns False.

   int_sets.to_set(new_item => x) in
      int_sets.to_set(new_item => x) | int_sets.to_set(new_item => y);

would return True.

This doesn't seem very intuitive to me. (Provided that my reading of
the RM is correct.)