Understanding the bitwise AND Operator

Numbers can be expressed in binary like this:

3    = 000011
5    = 000101
10   = 001010

…etc. I’m going to assume you’re familiar with binary.

Bitwise AND means to take two numbers, line them up on top of each other, and create a new number that has a 1 where both numbers have a 1 (everything else is 0).

For example:

    3          =>  00011
  & 5          =>  00101
------           -------
    1              00001

Bitwise OR means to take two numbers, line them up on top of each other, and create a new number that has a 1 where either number has a 1 (everything else is 0).

For example:

    3          =>  00011
  | 5          =>  00101
------           -------
    7              00111

Bitwise XOR (exclusive OR) means to take two numbers, line them up on top of each other, and create a new number that has a 1 where either number has a 1 AND the other number has a 0 (everything else is 0).

For example:

    3          =>  00011
  ^ 5          =>  00101
------           -------
    6              00110

Bitwise NOR (Not OR) means to take the Bitwise OR of two numbers, and then reverse everything (where there was a 0, there’s now a 1, where there was a 1, there’s now a 0).

Bitwise NAND (Not AND) means to take the Bitwise AND of two numbers, and then reverse everything (where there was a 0, there’s now a 1, where there was a 1, there’s now a 0).

Continuing: why does word &= 15 set all but the 4 rightmost bits to 0? You should be able to figure it out now…

     n          =>  abcdefghjikl
  & 15          =>  000000001111
------            --------------
     ?              00000000jikl

(0 AND a = 0, 0 AND b = 0, … j AND 1 = j, i AND 1 = i, …)

How is this useful? In many languages, we use things called “bitmasks”. A bitmask is essentially a number that represents a whole bunch of smaller numbers combined together. We can combine numbers together using OR, and pull them apart using AND. For example:

int MagicMap = 1;
int MagicWand = 2;
int MagicHat = 4;

If I only have the map and the hat, I can express that as myInventoryBitmask = (MagicMap | MagicHat) and the result is my bitmask. If I don’t have anything, then my bitmask is 0. If I want to see if I have my wand, then I can do:

int hasWand = (myInventoryBitmask & MagicWand);
if (hasWand > 0) {
  printf("I have a wand\n");
} else {
  printf("I don't have a wand\n");
}

Get it?

EDIT: more stuff

You’ll also come across the “bitshift” operator: << and >>. This just means “shift everything left n bits” or “shift everything right n bits”.

In other words:

1 << 3 = 0001 << 3 = 0001000 = 8

And:

8 >> 2 = 01000 >> 2 = 010 = 2

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