Re: disjoint ranges ?

[Robert A Duff <bobduff@world.std.com>]

Lao Xiao Hai <laoxhai@ix.netcom.com> writes:

> So, making a feature more powerful might be something such as being
> able to overload the "in" membership test.

It might be in the abstract, but in the context of Ada, probably not.

>...   For some reason, this
> operator is not really a first class operator in Ada.   I have often
> wondered about the rationale for that decision.

What is the type of the right-hand side of "in" here:

    if X in Integer range 1..10 ...

?  Ada has no way of describing that type.  In order to make "in" a
normal operator, we would need to invent a way to pass types as
parameters, and talk about the type of a type (and subtypes).  This is
not entirely unreasonable in the abstract, but it would be an enormous
change to Ada.

Consider:

    type T is range 1..100;
    function "in"(X: T; Y: type-of-T?) return Boolean;

Or maybe it should be T'in?  Anyway, it raises all kinds of questions.
What *is* the range of T (given that "in" changes it)?

- Bob