Byte - Integer Conversion

Question

I have four bytes that I want to store directly into an integer and then read back out again. I've spent forever trying bitshifts and division/modulo, but to no avail...with Java not supporting unsigned values, I can't figure it out...what I want is this:
[00000000][00000000][00000000][00001111] (four individual bytes)
STEP ONE: is converted to
[00000000000000000000000000001111] (same sequence of bits in an int)
and then some operation is performed (let's say adding one)
[00000000000000000000000000010000] (addition on the integer)
STEP TWO: and afterwards the integer is converted back
[00000000][00000000][00000000][00010000] (four individual bytes)

Answer 1

use parenthesis a lot until you know operator precedence
and use 0xff & byte to get rid of sign extension
public class ByteInt {
public static void main(String[] args) {
  int i;
  byte b3, b2, b1, b0;  // b3 is highest order byte
  b3 = 0x12;
  b2 = 0x34;
  b1 = 0x56;
  b0 = 0x78;
  i = (0xff & b0) | ( (0xff & b1) << 8  )
                  | ( (0xff & b2) << 16 )
                  | ( (0xff & b3) << 24 );
  System.out.println(Integer.toHexString(i));
  System.out.println("");
  i = 0xf1827456;  // f1 82 74 56
  System.out.println(Integer.toHexString(i));
  System.out.println("");
  b0 = (byte)   (i & 0xff);
  b1 = (byte) ( (i & 0xff00)     >> 8  );
  b2 = (byte) ( (i & 0xff0000)   >> 16 );
  b3 = (byte) ( (i & 0xff000000) >> 24 );
  System.out.println(Integer.toHexString(b3));
  System.out.println(Integer.toHexString(b2));
  System.out.println(Integer.toHexString(b1));
  System.out.println(Integer.toHexString(b0));
  System.out.println("");
  System.out.println(Integer.toHexString(0xff & b3));
  System.out.println(Integer.toHexString(0xff & b2));
  System.out.println(Integer.toHexString(0xff & b1));
  System.out.println(Integer.toHexString(0xff & b0));
  System.out.println("");
} // method
} // class