How to work with the bits in a byte

Question

I have a single byte which contains two values. Here\'s the documentation:

The authority byte is split into two fields. The three least significant bits

Answer 1

To get a value of the five most significant bits in a byte as an integer, shift the byte to the right by 3 (i.e. by 8-5), and set the three upper bits to zero using bitwise AND operation, like this:

byte orig = ...
int rejThreshold = (orig >> 3) & 0x1F;

• >> is the "shift right" operator. It moves bits 7..3 into positions 4..0, dropping the three lower bits.
• 0x1F is the binary number 00011111, which has the upper three bits set to zero, and the lower five bits set to one. AND-ing with this number zeroes out three upper bits.

This technique can be generalized to get other bit patterns and other integral data types. You shift the bits that you want into the least-significant position, and apply a mask that "cuts out" the number of bits that you want. In some cases, shifting would not be necessary (e.g. when you get the least significant group of bits). In other cases, such as above, the masking would not be necessary, because you get the most significant group of bits in an unsigned type (if the type is signed, ANDing would be required).