Brian,
yes this is perfectly what I want.
2: 00000010
3: 00000011
11: 00001011
applying a bit-wise AND 2:
2: 00000010
should yield a 2 in all case, as, it tests, whether in 2, 3, 11, the second but last significant bit is set (which is the case for all of 2, 3, and 11).
Guess, I will...