On Jan 30 2007 09:45, Robert P. J. Day wrote:
>On Tue, 30 Jan 2007, Jan Engelhardt wrote:
>>
>> >> Why the qualifier? Zero *is* not a power of 2, is it?
>>
>> No, it is not:
>>
>> In[1]:= Solve[2^n == 0, n]
>>
>> Out[1]= {}
>>
>> So says Mathematica5.
>
>oooookay, that's kind of like taking a sandblaster to a soup cracker.
Hehe. Well, there is a non-representable solution:
In[2]:= 2^-Infinity
Out[2]= 0
>seriously, though, there is the potential of breaking something with
>this change since you can see how there is some inconsistency in how
>it's done *now* just for powerpc which, in some places, defines its
>own versions of this:
>
>./arch/ppc/mm/pgtable.c:
> #define is_power_of_2(x) ((x) != 0 && (((x) & ((x) - 1)) == 0))
>./arch/ppc/syslib/ppc85xx_rio.c:
> #define is_power_of_2(x) (((x) & ((x) - 1)) == 0)
>./arch/powerpc/mm/pgtable_32.c:
> #define is_power_of_2(x) ((x) != 0 && (((x) & ((x) - 1)) == 0))
>
>note how the first and third macros *won't* consider zero a power of
>two, while the second one *will*. hence the need for a single
>standard for all of this, just to play it safe.
Hmpf. Perhaps a second macro "is_intdivisible_by_power_of_2" or so could
catch the "am I zero or 2^n" question.
Jan
--
ft: http://freshmeat.net/p/chaostables/
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