How to Convert Int to Unsigned Byte and Back

Question

I need to convert a number into an unsigned byte. The number is always less than or equal to 255, and so it will fit in one byte.

I also need to convert that byte ba

Answer 1

If you just need to convert an expected 8-bit value from a signed int to an unsigned value, you can use simple bit shifting:

int signed = -119;  // 11111111 11111111 11111111 10001001

/**
 * Use unsigned right shift operator to drop unset bits in positions 8-31
 */
int psuedoUnsigned = (signed << 24) >>> 24;  // 00000000 00000000 00000000 10001001 -> 137 base 10

/** 
 * Convert back to signed by using the sign-extension properties of the right shift operator
 */
int backToSigned = (psuedoUnsigned << 24) >> 24; // back to original bit pattern

http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

If using something other than int as the base type, you'll obviously need to adjust the shift amount: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

Also, bear in mind that you can't use byte type, doing so will result in a signed value as mentioned by other answerers. The smallest primitive type you could use to represent an 8-bit unsigned value would be a short.

Answer 2

The solution works fine (thanks!), but if you want to avoid casting and leave the low level work to the JDK, you can use a DataOutputStream to write your int's and a DataInputStream to read them back in. They are automatically treated as unsigned bytes then:

For converting int's to binary bytes;

ByteArrayOutputStream bos = new ByteArrayOutputStream();
DataOutputStream dos = new DataOutputStream(bos);
int val = 250;
dos.write(byteVal);
...
dos.flush();

Reading them back in:

// important to use a (non-Unicode!) encoding like US_ASCII or ISO-8859-1,
// i.e., one that uses one byte per character
ByteArrayInputStream bis = new ByteArrayInputStream(
   bos.toString("ISO-8859-1").getBytes("ISO-8859-1"));
DataInputStream dis = new DataInputStream(bis);
int byteVal = dis.readUnsignedByte();

Esp. useful for handling binary data formats (e.g. flat message formats, etc.)

Answer 3

Handling bytes and unsigned integers with BigInteger:

byte[] b = ...                    // your integer in big-endian
BigInteger ui = new BigInteger(b) // let BigInteger do the work
int i = ui.intValue()             // unsigned value assigned to i

Answer 4

Java 8 provides Byte.toUnsignedInt to convert byte to int by unsigned conversion. In Oracle's JDK this is simply implemented as return ((int) x) & 0xff; because HotSpot already understands how to optimize this pattern, but it could be intrinsified on other VMs. More importantly, no prior knowledge is needed to understand what a call to toUnsignedInt(foo) does.

In total, Java 8 provides methods to convert byte and short to unsigned int and long, and int to unsigned long. A method to convert byte to unsigned short was deliberately omitted because the JVM only provides arithmetic on int and long anyway.

To convert an int back to a byte, just use a cast: (byte)someInt. The resulting narrowing primitive conversion will discard all but the last 8 bits.

Answer 5

If you want to use the primitive wrapper classes, this will work, but all java types are signed by default.

public static void main(String[] args) {
    Integer i=5;
    Byte b = Byte.valueOf(i+""); //converts i to String and calls Byte.valueOf()
    System.out.println(b);
    System.out.println(Integer.valueOf(b));
}

Answer 6

in java 7

public class Main {
    public static void main(String[] args) {
        byte b =  -2;
        int i = 0 ;
        i = ( b & 0b1111_1111 ) ;
        System.err.println(i);
    }
}

result : 254