Convert short to byte[] in Java

Question

How can I convert a short (2 bytes) to a byte array in Java, e.g.

short x = 233;
byte[] ret = new byte[2];

...

it should be something like this. But not sure.

((0xFF << 8) & x) >> 0;

EDIT:

Also you can use:

java.nio.ByteOrder.nativeOrder();

To discover to get whether the native bit order is big or small. In addition the following code is taken from java.io.Bits which does:

• byte (array/offset) to boolean
• byte array to char
• byte array to short
• byte array to int
• byte array to float
• byte array to long
• byte array to double

And visa versa.

Answer 1

ret[0] = (byte)(x & 0xff);
ret[1] = (byte)((x >> 8) & 0xff);

Answer 2

Figure it out its.

public static byte[] toBytes(short s) {
        return new byte[]{(byte)(s & 0x00FF),(byte)((s & 0xFF00)>>8)};
    }

Answer 3

It depends how you want to represent it:

• big endian or little endian? That will determine which order you put the bytes in.

• Do you want to use 2's complement or some other way of representing a negative number? You should use a scheme that has the same range as the short in java to have a 1-to-1 mapping.

For big endian, the transformation should be along the lines of: ret[0] = x/256; ret[1] = x%256;