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From: =?utf-8?Q?Yannick_Duch=C3=AAne_=28Hibou57?=
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Newsgroups: comp.lang.ada
Subject: Re: highest bit, statically determined
Date: Sun, 04 Nov 2012 21:45:28 +0100
Organization: Ada @ Home
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Date: 2012-11-04T21:45:28+01:00
List-Id: <comp.lang.ada>

Le Sat, 29 Sep 2012 21:16:01 +0200, Bill Findlay  =

<yaldnif.w@blueyonder.co.uk> a =C3=A9crit:

>
> On 29/09/2012 19:57, in article  =

> CC8D0314.1E2CC%yaldnif.w@blueyonder.co.uk,
> "Bill Findlay" <yaldnif.w@blueyonder.co.uk> wrote:
>
>> On 29/09/2012 18:34, in article
>> 50673111$0$9505$9b4e6d93@newsspool1.arcor-online.net, "Georg Bauhaus"=

>> <rm.dash-bauhaus@futureapps.de> wrote:
>>
>>> Is there a shorter/better way of having the compiler
>>> find the highest bit =3D 1 in a static numeric constant?
>>>
>>> If N is such a constant, e.g. Some_Type'Last where
>>> Some_Type'Size =3D 8 and its bound are static, then
>>>
>>>     Highest_Bit_In_Octet : constant :=3D
>>>       Natural'Max
>>>       (8 * Boolean'Pos (N / 2**7 > 0),
>>>        Natural'Max
>>>        (7 * Boolean'Pos (N / 2**6 > 0),
>>>         Natural'Max
>>>         (6 * Boolean'Pos (N / 2**5 > 0),
>>>          Natural'Max
>>>          (5 * Boolean'Pos (N / 2**4 > 0),
>>>           Natural'Max
>>>           (4 * Boolean'Pos (N / 2**3 > 0),
>>>            Natural'Max
>>>            (3 * Boolean'Pos (N / 2**2 > 0),
>>>             Natural'Max
>>>             (2 * Boolean'Pos (N / 2**1 > 0),
>>>              Natural'Max
>>>              (1 * Boolean'Pos (N / 2**0 > 0), 0))))))));
>>
>> In my experience that sort of code, applied in non-static cases, is l=
ess
>> efficient than one would hope, and more obvious code works faster.
>>
>> Something like the following can be readily extended to greater opera=
nd
>> widths:
>>
>>    function first_1_bit (y : octet)
>>    return Natural is
>>       x : octet;
>>       r : Natural;
>>    begin
>>       if y =3D 0 then return 0; end if;
>>
>>       if (y / 16) /=3D 0 then
>>          r :=3D 4; x :=3D y / 16;
>>       else
>>          r :=3D 0; x :=3D y;
>>       end if;
>>
>>       if (x / 4) /=3D 0 then
>>          r :=3D r + 2; x :=3D x / 4;
>>       end if;
>>
>>       if (x / 2) /=3D 0 then
>>          r :=3D r + 1;
>>       end if;
>>
>>       return r + 1;
>>    end first_1_bit;
>>
>> It looks fairly inline-able, and foldable for a static value of y.
>
> I can now confirm that with GNAT GPL 2012 at -O3 it does inline and fo=
ld,
> but I now see that you want the result to be static as well as the  =

> operand,
> and this does not achieve that.
>

What means folding in this context?


-- =

=E2=80=9CSyntactic sugar causes cancer of the semi-colons.=E2=80=9D [1]
=E2=80=9CStructured Programming supports the law of the excluded muddle.=
=E2=80=9D [1]
[1]: Epigrams on Programming =E2=80=94 Alan J. =E2=80=94 P. Yale Univers=
ity