When talking about bitwise operators, what does the symbol "0xf00f" mean?

Question

Copied out the Java reference manual:

For example, the result of the expression: 0xff00 & 0xf0f0 is: 0xf000

The result of the expression: 0xff00 ^ 0xf0f0 is: 0x0ff0

The result of the expression: 0xff00 | 0xf0f0 is: 0xfff0

What do these letters and numbers put together mean?

Answer 1

Numbers that start in 0x are in hexadecimal (base 16) notation. They use digits 0-9 for hex digits 0-9, and letters A-F for hex digits 10 through 15.

Hexadecimal notation is more convenient than decimal to show bit operations, because the base 16 is 2^4, so a digit corresponds to four bits; the results of bitwise operations are confined to a single digit; they do not "bleed" into adjacent digits.

Conversion table of binary to hex is as follows:

0000 - 0
0001 - 1
0010 - 2
0011 - 3
0100 - 4
0101 - 5
0110 - 6
0111 - 7
1000 - 8
1001 - 9
1010 - A
1011 - B
1100 - C
1101 - D
1110 - E
1111 - F

You can convert a hex number to binary by direct replacement of digits using this table. For example, 0xF0F0 becomes 1111000011110000 in binary.

With these substitution rules in hand, you can follow the text from the documentation in binary:

0xff00 & 0xf0f0 is: 0xf000

becomes

  1111 1111 0000 0000
& 1111 0000 1111 0000
---------------------
  1111 0000 0000 0000

In binary it makes more sense: the result of a bitwise & is 1 only when both operands have 1 in the corresponding bit; otherwise, the result is 0.

You should be able to follow the remaining operations ^ and | more readily as well.