Nick Hibma scribbled this message on Aug 21:
> Does anyone know an inexpensive algorithm (O(1)) to go from an number to
> the next (lower or higher) power of two.
>
> 1 -> 1
> 2,3 -> 2
> 4,5,6,7 -> 4
> 8,9,10,11,12,13,14,15 -> 8
> etc.
>
> So %1101 should become either %10000 or %1000.
>
> The only solution I have so far is a table. That is a possibility as the
> the highest number will be 32 I think.
hmmm.... how about:
int
fls(int n)
{
/* courtesy of fortune, of course this is O(lgn) */
/* swap the bits around */
n = ((n >> 1) & 0x55555555) | ((n << 1) & 0xaaaaaaaa);
n = ((n >> 2) & 0x33333333) | ((n << 2) & 0xcccccccc);
n = ((n >> 4) & 0x0f0f0f0f) | ((n << 4) & 0xf0f0f0f0);
n = ((n >> 8) & 0x00ff00ff) | ((n << 8) & 0xff00ff00);
n = ((n >> 16) & 0x0000ffff) | ((n << 16) & 0xffff0000);
/* mask all but the lowest bit off */
n = n & -n;
/* swap them back to where they were originally */
n = ((n >> 1) & 0x55555555) | ((n << 1) & 0xaaaaaaaa);
n = ((n >> 2) & 0x33333333) | ((n << 2) & 0xcccccccc);
n = ((n >> 4) & 0x0f0f0f0f) | ((n << 4) & 0xf0f0f0f0);
n = ((n >> 8) & 0x00ff00ff) | ((n << 8) & 0xff00ff00);
n = ((n >> 16) & 0x0000ffff) | ((n << 16) & 0xffff0000);
return n;
}
now this is O(lgn) (where n is number of bits in an int), but I'm not
sure how this will perform wrt the other posting.. it probably won't be
that fast because of all the arithmetic that happens, and a binary search
of the number would probably be faster (which I have implemented a few
times)...
--
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