When I strated out, I was really confused about bit shift and hexadecimal number and binary. This is a good example how confusing it could get sometimes. Occasionally you see this in the code, for exmaple in C++:
byte a = ( b >> 8) & 0xff; // b is an integer
Okay, I know what “>> 8” means, we are shifting variable “b” to the right by 8 bits. But, what the heck is “& 0xff” doing there?
Well, 0xff is the hexadecimal number FF which has a integer value of 255. And the binary representation of FF is 00000000000000000000000011111111 (under the 32-bit integer).
The & operator performs a bitwise AND operation. a & b will give you an integer with a bit pattern that has a 0 in all positions where b has a 0, while in all positions where b has a 1, the corresponding bit value froma is used (this also goes the other way around). For example, the bitwise AND of 10110111 and00001101 is 00000101.
In a nutshell, “& 0xff” effectively masks the variable so it leaves only the value in the last 8 bits, and ignores all the rest of the bits.
It’s seen most in cases like when trying to transform color values from a special format to standard RGB values (which is 8 bits long).