Re: Ada.Strings.Bounded

[Robert A Duff <bobduff@shell01.TheWorld.com>]

Eric Hughes <eric.eh9@gmail.com> writes:

> On Apr 22, 9:41 am, Robert A Duff <bobd...@shell01.TheWorld.com>
> wrote:
>> The above-quoted paragraph 3.5.4(8) is normative.  It comes right out
>> and says in plain language exactly what you were trying to prove in a
>> circuitous manner -- that the the set of values is infinite.
>
> Here's the relevant sentence in 3.5.4(8) again:
>> The set of values for a signed integer type is the (infinite) set
>> of mathematical integers, though only values of the base range of
>> the type are fully supported for run-time operations.
>
> So now wait.  When I declare
>     type X is range 0 .. 99 ;
> does that mean that the value 100 is a value of type X?

Yes, if you're using "type X" as I explained in my other message to mean
"the type whose first subtype is X".

Subtype X has the range you expect (0..99).  But it has an underlying
type that has all the integer values, including 100.

Note that you could say:

    subtype S is X'Base range 0..100;

>...Because how
> that definition reads in any sort of facially-reasonable way is that
> 100 is not a value of type X.

100 is not a value of SUBtype X.

>> Note that it does not say anything special about root_int or univ_int.
>> It applies to ALL signed integer types, including user-defined ones.
>
> Apparently so.
>
> And so now I have developed a Strong Opinion.  This is a travesty.

I agree it's confusing.  But "travesty" is too strong.  In practise,
none of this matters much.

Don't you want to be able to say "(Blah + 1) mod 99" without overflow
(where Blah is of subtype X)?

- Bob