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From: Shark8 <onewingedshark@gmail.com>
Newsgroups: comp.lang.ada
Subject: Re: How do I write directly to a memory address?
Date: Wed, 9 Feb 2011 12:25:22 -0800 (PST)
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Date: 2011-02-09T12:25:22-08:00
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On Feb 9, 1:00=A0pm, Hyman Rosen <hyro...@mail.com> wrote:
> On 2/9/2011 2:41 PM, Shark8 wrote:
>
> > Yes. It takes a bit of doing but check out the proofs of recursive
> > & =A0iterative implementations of functions being identical. The
> > methods used in the implementations are *NOT* identical, but
> > the methods themselves are identical in that for ANY input
> > given they produce the EXACT SAME output.
>
> Can I play too? I'll write my routine in C, but the translation
> to Ada should be readily apparent:
>
> unsigned Exponent(unsigned base, unsigned exp)
> {
> =A0 =A0 =A0if (exp =3D=3D 0) return 1;
> =A0 =A0 =A0if (base =3D=3D 0) return 0;
> =A0 =A0 =A0while (true)
> =A0 =A0 =A0{
> =A0 =A0 =A0 =A0 =A0unsigned guess =3D random();
> =A0 =A0 =A0 =A0 =A0unsigned g =3D guess;
> =A0 =A0 =A0 =A0 =A0unsigned i =3D 0;
> =A0 =A0 =A0 =A0 =A0while (g % base =3D=3D 0 && i < exp)
> =A0 =A0 =A0 =A0 =A0{
> =A0 =A0 =A0 =A0 =A0 =A0 =A0g =3D g / base;
> =A0 =A0 =A0 =A0 =A0 =A0 =A0i =3D i + 1;
> =A0 =A0 =A0 =A0 =A0}
> =A0 =A0 =A0 =A0 =A0if (g =3D=3D 1 && i =3D=3D exp) return guess;
> =A0 =A0 =A0}
>
> }
>
> Is it OK if I replace your Exponent method with my implementation?

No, yours rejects signed values for the bases; mine does not...
thus it violates the equal-inputs requirement.