Re: highest bit, statically determined

[Georg Bauhaus <rm.dash-bauhaus@futureapps.de>]

On 29.09.12 20:11, Pascal Obry wrote:

>>     Highest_Bit_In_Octet : constant :=
>>       Natural'Max
>>       (8 * Boolean'Pos (N / 2**7 > 0),
...
>>            (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))))))));
>
> if N > 128 then
>     return 8;
> elsif N > 64
>     return 7;
> elsif ...
>
> elsif N > 0 then
>     return 1;
> end if;
>


The N > 2**X part is good, thanks for the answer that removed
a fair bit of fog. But the "return X" indicates that the solution
cannot be static, or can it?

There might be an expression in Ada 2012 that does it. Alas,
I cannot use Ada 2012 yet---which I should have mentioned!

    Highest_Bit_In_Octet_2012 : constant :=
      (if N > 2**7 then 8
      elsif N > 2**6 then 7
      elsif N > 2**5 then 6
      elsif N > 2**4 then 5
      elsif N > 2**3 then 4
      elsif N > 2**2 then 3
      elsif N > 2**1 then 2
      elsif N > 2**0 then 1
      else 0);