将int转换为byte时的BufferOverflowException

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;
}

Answer 3

This should work

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

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