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From: bobduff@world.std.com (Robert A Duff)
Subject: Re: QUIZ: To be or not to be (able to post)
Date: 1997/10/03
Message-ID: <EHHp4E.AC4@world.std.com>#1/1
X-Deja-AN: 277874151
References: <610a7h$a7n@news3.his.com> <EHG0r1.K47.0.-s@inmet.camb.inmet.com>
 <6114h5$pmk@news3.his.com>
Organization: The World Public Access UNIX, Brookline, MA
Newsgroups: comp.lang.ada
Date: 1997-10-03T00:00:00+00:00
List-Id: <comp.lang.ada>


In article <6114h5$pmk@news3.his.com>,
Joel VanLaven  <jvl@ocsystems.com> wrote:
>...  I am just a little
>surprised that the "obvious" big difference (between float and float'base) 
>wasn't really a difference at all.

If you say:

    type T1 is range -10..10;
    type T2 is digits 5 range -10.0..10.0;

then T1 and T2 are constrained.  On the other hand, if you say:

    type T3 is digits 5; -- no range given here

then T3 is unconstrained.  So T3 is the same thing as T3'Base (both are
unconstrained).  But T1 and T1'Base are different, and T2 and T2'Base
are different (T1 and T2 are constrained; T1'Base and T2'Base are
unconstrained).

The predefined floating point types, such as Float, are like T3 in this
regard.  That is, Float is unconstrained, so Float is the same thing as
Float'Base.  This is because the declaration of Float has no "something
.. something else".  See A.1(20).

The predefined integer types, such as Integer, are constrained.

This all makes sense to me.  The "surprise" probably comes from the fact
that this is different from Ada 83, where the first subtype was always
constrained in the floating-point case.

- Bob