[tslater2006]

> 10000

> & 1111

> ______

> 1

Wouldn't this result in 0 and not 1?

[knome]

Missing a 'not'.

You would use `bool( nn and not (nn&(nn-1)) )`

Let's try for 15

    start 0b1111
    sans1 0b1110
    anded 0b1110
    not   0b0000
    

Now 8

    start 0b1000
    sans1 0b0111
    anded 0b0000
    not   0b0001

It's basically a claim that, in binary, only powers of two won't have any overlapping bits between the initial number and that number minus one.

Doesn't work for 0 so you have to special case it.

[sumnole]

0 is correct. AND(bit1,bit2) returns 1 if and only if both input bits are 1.

AND'ing each bit of the two numbers yields only 0 bits:

   10000
  &01111
  ______
   00000