BufferOverflowException while converting int to byte

Question

I have an counter which counts from 0 to 32767

At each step, I want to convert the counter (int) to an 2 byte array.

I've tried this, but I got a BufferOverflowException exception:

byte[] bytearray = ByteBuffer.allocate(2).putInt(counter).array();

Answer 1

Yes, this is because an int takes 4 bytes in a buffer, regardless of the value.

ByteBuffer.putInt is clear about both this and the exception:

Writes four bytes containing the given int value, in the current byte order, into this buffer at the current position, and then increments the position by four.

...

Throws:
BufferOverflowException - If there are fewer than four bytes remaining in this buffer

To write two bytes, use putShort instead... and ideally change your counter variable to be a short as well, to make it clear what the range is expected to be.

Answer 2

First, you seem to assume that the int is big endian. Well, this is Java so it will certainly be the case.

Second, your error is expected: an int is 4 bytes.

Since you want the two last bytes, you can do that without having to go through a byte buffer:

public static byte[] toBytes(final int counter)
{
    final byte[] ret = new byte[2];
    ret[0] = (byte) ((counter & 0xff00) >> 8);
    ret[1] = (byte) (counter & 0xff);
    return ret;
}

You could also use a ByteBuffer, of course:

public static byte[] toBytes(final int counter)
{
    // Integer.BYTES is there since Java 8
    final ByteBuffer buf = ByteBuffer.allocate(Integer.BYTES);
    buf.put(counter);
    final byte[] ret = new byte[2];
    // Skip the first two bytes, then put into the array
    buf.position(2);
    buf.put(ret);
    return ret;
}