Re: Ada.Strings.Bounded

[Eric Hughes <eric.eh9@gmail.com>]

On Apr 21, 6:49 pm, Adam Beneschan <a...@irvine.com> wrote:
> See 4.9(33).

Oh.  OK.  For those insufficiently motivated to look at this section,
it says that static expressions are evaluated exactly, without
performing overflow checks.  To me, that word "exactly" means that
static expressions are evaluated as universal_integer, since that's
the only way to get them defined exactly.

Previously, Bob Duff had said:
> An integer literal has type universal_integer, as you said, but
> universal_integer has no "+" operator.  There is a "+" for root_integer,
> and also for user-defined integer types.  Universal_integers can be
> implicitly converted to various integer types.

Something's got to give here.  I see the following possibilities:
1) universal_integer actually does have a "+" operator.  It would only
be available at compile time, since there's no way to declare a
variable of type universal_integer.
2) The addition chosen is the context-dependent one, as for dynamic
expressions, but when overflow happens, the result is magically
converted to universal_integer.
3) The addition of the ordinary type is tried with the base type.  If
it overflows, a bigger type is tried, and so on up to
universal_integer, which will always succeed.
4) The evaluation happens purely mathematically, without regard for
the Ada type system.
5) There are different versions of the type and its addition operation
for static and run-time use.  The static ones have arbitrarily large
capacity and exact arithmetic.

I'm sure my imagination is not large enough to guess at all the
possibilities.  What's going on here, really?

Eric