Question

Say you are having a byte of pattern:

byte b = 0x%1;

How to tell when a byte does have certain values on the “2nd position” – in place of % ?
In this example 1, no matter what the 1st position holds.

Answer 1

Use a mask bits to get the last 8 bits:

int last8bits = b & 0xF;

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Edit: You should read up on bitwise operations.

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Full example:

public static void main(String[] args) {

    byte b = (byte) 0xA1;
    int last8bits = b & 0xF;

    if (last8bits == 0x01)
        System.out.println("\"matches\"");
}