On Mer, 2003-04-30 at 22:59, Falk Hueffner wrote:
> > ffs(x^(x-1)) == fls(x)
>
> I don't think so. Maybe you are thinking of ffs(x) == fls(x & -x). So
> you can calculate ffs with fls, but not easily the other way round.
Well I asked my CPU to double check. If I got the code right it thinks that
ffs(i^(i-1))==fls(i) is true for 1->2^32-1
If you think about it it seems to make sense
x-1 turns the lowest 100..0 sequence into 01......1
The xor removes unchanged higher bits
so
10100
-1 10011
XOR 00111
-
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This archive was generated by hypermail 2b29 : Wed Apr 30 2003 - 22:00:37 EST