Processing: Creative Coding and Computational Art

I’m not going to provide examples for the other two operators, but I’ll tell you about them.
^ is the XOR, or exclusive OR, operator. It works by comparing parallel bits, as with the first
two operators. However, XOR is looking for differences. If the parallel bits are different,
then the result is a one; otherwise the result is a zero.
~ is the bitwise complement and is a unary operator, meaning it works like a binary minus
sign on a single operand, inverting zeros to ones and vice versa.
Putting it all together
Finally returning to the original bitwise color expression, float r = myColor >> 16 &
0xFF;, you should now have some sense of what is happening in the expression. The last
mystery to solve is the bit mask 0xFF. A bit mask is just a pattern of bits used to perform
some type of operation on another binary value, using one of the bitwise operators you
just looked at. The bit mask value itself (0xFF) hopefully looks familiar to you, as it’s just a
number expressed in hexadecimal notation (base 16). Remember, colors are commonly
represented as hexadecimal values. The 0x part of 0xFF tells the compiler that the characters
FF should be evaluated in base 16. Since F is the highest value in base 16 (equal to 15
in decimal), FF evaluates to 15 * 16 1 + 15 * 16 0 = 255, which you’ll remember is the same
thing as 11111111 (in base 2).
For those of you who actually like this bit-mathy stuff, break the binary value
11111111 into two even groups of four bits and then evaluate each separately
in base 2. Compare the results to the hexadecimal value 0xFF. Do you
see the relationship between the binary and hexadecimal values?
Finishing up this long discussion, you now have all the ingredients to understand the bitwise
color expression myColor >> 16 & 0xFF. Next is a description of how the expression
actually returns the red component of a color.
I’ll begin by creating a new color:
color myColor = color(200, 190, 75, 255);
This color looks like this in binary (please note that I divided the continuous 32-bit string
into byte-sized components, for readability):
11111111 11001000 10111110 01001011
Performing the 16-bit right shift on the number yields the following:
11111111111111111111111111001000
Next I’ll use the bitwise AND operator and bit mask 0xFF (11111111 in binary). Remember,
the & operator works by comparing the parallel bits. If the bits are both 1, then the result
for that place is a one. All other results evaluate out to zero.
MATH REFERENCE
769
B