Re: highest bit, statically determined

["Yannick Duchêne (Hibou57)" <yannick_duchene@yahoo.fr>]

Le Sat, 29 Sep 2012 21:16:01 +0200, Bill Findlay  
<yaldnif.w@blueyonder.co.uk> a écrit:

>
> 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 = 1 in a static numeric constant?
>>>
>>> If N is such a constant, e.g. Some_Type'Last where
>>> Some_Type'Size = 8 and its bound are static, then
>>>
>>>     Highest_Bit_In_Octet : constant :=
>>>       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 less
>> efficient than one would hope, and more obvious code works faster.
>>
>> Something like the following can be readily extended to greater operand
>> widths:
>>
>>    function first_1_bit (y : octet)
>>    return Natural is
>>       x : octet;
>>       r : Natural;
>>    begin
>>       if y = 0 then return 0; end if;
>>
>>       if (y / 16) /= 0 then
>>          r := 4; x := y / 16;
>>       else
>>          r := 0; x := y;
>>       end if;
>>
>>       if (x / 4) /= 0 then
>>          r := r + 2; x := x / 4;
>>       end if;
>>
>>       if (x / 2) /= 0 then
>>          r := 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 fold,
> 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?


-- 
“Syntactic sugar causes cancer of the semi-colons.” [1]
“Structured Programming supports the law of the excluded muddle.” [1]
[1]: Epigrams on Programming — Alan J. — P. Yale University