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From: Stephen Leake <Stephen.Leake@gsfc.nasa.gov>
Subject: Re: Portability of Arithmetic (was: Java vs Ada 95)
Date: 1996/10/17
Message-ID: <326628B8.7724@gsfc.nasa.gov>#1/1
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references: <325D7F9B.2A8B@gte.net> <ROGOFF.96Oct14095012@sccm.Stanford.EDU>
 <DzBrFw.IAs@world.std.com> <1996Oct15.174526.1@eisner>
 <DzDBw3.ILH@world.std.com>
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organization: NASA Goddard Space Flight Center -- Greenbelt, Maryland USA
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Date: 1996-10-17T00:00:00+00:00
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Robert A Duff wrote:
> 
> In article <1996Oct15.174526.1@eisner>,
> Larry Kilgallen <kilgallen@eisner.decus.org> wrote:
> >In article <DzBrFw.IAs@world.std.com>, bobduff@world.std.com (Robert A Duff) writes:
> >> How about portability of arithmetic?  In Java, int is always exactly 32
> >> bits 2's complement.  In Ada, Integer is whatever the machine supports.
> >> And if I say "type T is range 1..100;" I might get 32-bit arithmetic, or
> >> I might get 8-bit arithmetic, or who-knows-what.
> >
> >And why should it matter ?  If your specification is to cover that
> >range, a compiler is meeting that specification even if it converts
> >it to BCD to accomplish it on some _very_ old hardware.
> 
>    type T is range 1..100;
>    X: T := ...;
>    Y: T := ...;
>    Average: T := (X + Y) / 2;
> 
> Because the above code will work fine on some Ada compilers, and blow up
> on others (the addition will overflow using 8-bit signed arithmetic) if
> X and Y happen to be, say, 99.  In Java, the above code is guaranteed to
> work.
> 
> - Bob

Clearly, the Ada type should be defined to cover the expected range of
intermediate values, as well as the final values:

type T is range 1 .. 200;

What if I'm doing some computations in Java that have intermediate
values that need 33 bits? In Ada, the compiler tells me when it can't do
that.

-- 
- Stephe