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From: bobduff@world.std.com (Robert A Duff)
Subject: Re: Java vs Ada 95 (Was Re: Once again,
 Ada absent from DoD SBIR solicitation)
Date: 1996/10/17
Message-ID: <DzFBKE.HCA@world.std.com>#1/1
X-Deja-AN: 190069638
references: <325D7F9B.2A8B@gte.net> <ROGOFF.96Oct14095012@sccm.Stanford.EDU>
 <DzBrFw.IAs@world.std.com> <3266DC85.D45@dynamite.com.au>
organization: The World Public Access UNIX, Brookline, MA
newsgroups: comp.lang.ada
Date: 1996-10-17T00:00:00+00:00
List-Id: <comp.lang.ada>


In article <3266DC85.D45@dynamite.com.au>,
Alan Brain  <aebrain@dynamite.com.au> wrote:
>If you really have to have 32 bits, just use
>
>type INTEGER_32 is new INTEGER;
>for INTEGER_32'size use 32;

No, that doesn't work.  What you want is "type Integer_32 is range
-2**31..2**31-1;".  A Size clause does not affect the semantics of
arithmetic.  So your Integer_32 might use 16 bits or 64 bits for
intermediate results.

> But unless interfacing with an IO device or some such, what the heck
>does it matter whether it's 8,16,32,48,12,19 or whatever bits, providing
>it's enough?

The portability issue arises for intermediate results within expressions
(as in (X + Y)/2).  The compiler is not required to do the arithmetic in
"enough" bits to hold the intermediate result, so some compilers can
overflow where others do not.  Java doesn't use "enough" bits either,
but at least it's always 32 or 64 bits (depending on the expression, not
on the compiler).

- Bob