I\'m messing with assembly language programming and I\'m curious how I could tell if a number is a multiple of 4 using the logic operator AND?
I know how to do it us
A number is a multiple of 4 if its lower 2 bits are 0, so you can simply shift the number right twice and check shifted bits for 0.
Well, to detect if a number is a multiple of another, you simply need to do x MOD y
. If the result is 0
, then it is an even multiple.
It is also true that for every y
that is a power of 2
, (x MOD y)
is equivalent to (x AND (y - 1))
.
Therefore:
IF (x AND 3) == 0 THEN
/* multiple of 4 */
EDIT:
ok, you want to know why (x MOD y) == (x AND (y - 1))
when y
is a power of 2. I'll do my best to explain.
Basically, if a number is a power of 2, then it has a single bit set (since binary is base 2). This means that all of the lower bits are unset. So for example: 16 == 10000b, 8 == 1000b
, etc.
If you subtract 1 from any of these values. You end up with the bit that was set being unset and all bits below it being set.
15 = 01111b, 7 = 0111b
, etc. So basically it is creates a mask which can be used to test if the any of the lower bits were set. I hope that was clear.
EDIT: Bastien Léonard's comment covers it well too:
if you divide (unsigned) by 4, you shift two bits to the right. Thus the remainder is those two bits, which get lost when you divide. 4 - 1 = 11b, that is, a mask that yields the two rightmost bits when you AND it with a value.
EDIT: see this page for possibly clearer explanations: http://en.wikipedia.org/wiki/Power_of_two#Fast_algorithm_to_check_if_a_positive_number_is_a_power_of_two.
It covers detecting powers of 2 and using AND as a fast modulo operation if it is a power of 2.
In x86 assembly:
test eax, 3
jnz not_multiple_of_4
; action to be taken if EAX is a multiple of 4
not_multiple_of_4:
; ...
(x & 3) == 0
W.r.t. assembly language, use TST if available, otherwise AND, and check the zero flag.