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
"Bit" is short for "binary digit". And yes, it's a 0 or 1. There are almost always 8 in a byte, and they're written kinda like decimal numbers are -- with the most significant digit on the left, and the least significant on the right.
In your example, w1 & 3 masks everything but the two least significant (rightmost) digits because 3, in binary, is 00000011. (2 + 1) The AND operation returns 0 if either bit being ANDed is 0, so everything but the last two bits are automatically 0.
w1 = ????...??ab
3 = 0000...0011
--------------------
& = 0000...00ab
0 & any bit N = 0
1 & any bit N = N
So, anything bitwise anded with 3 has all their bits except the last two set to 0. The last two bits, a and b in this case, are preserved.
@cHao & all: No! Bits are not numbers. They’re not zero or one!
Well, 0 and 1 are possible and valid interpretations. Zero and one is the typical interpretation.
But a bit is only a thing, representing a simple alternative. It says “it is” or “it is not”. It doesn’t say anything about the thing, the „it“, itself. It doesn’t tell, what thing it is.
In most cases this won’t bother you. You can take them for numbers (or parts, digits, of numbers) as you (or the combination of programming languages, cpu and other hardware, you know as being “typical”) usaly do – and maybe you’ll never have trouble with them.
But there is no principal problem if you switch the meaning of “0“ and “1”. Ok, if doing this while programming assembler, you’ll find it a bit problematic as some mnemonics will do other logic then they tell you with their names, numbers will be negated and such things.
Have a look at http://webdocs.cs.ualberta.ca/~amaral/courses/329/webslides/Topic2-DeMorganLaws/sld017.htm if you want.
Greetings