How to convert a uint16 stored as a little endian byte array to an int?

Question

I'm receiving some data that is a little endian byte array that represents a uint16 but I want to store this in an int. I have wrapped this data into a ByteBuffer so I can call ByteBuffer.get() to get the next byte(s), but I am not sure how to convert these 2 bytes from uint16 to int.

    byte[] data = //From UDP socket
    ByteBuffer bb = ByteBuffer.wrap(data);
    bb = bb.order(ByteOrder.LITTLE_ENDIAN);
    while(bb.hasRemaining()){
        int n = //What goes here?
    }

Thanks for any help :)

Answer 1

The main complexity is in my opinion how to represent a uint16 in Java. I'd suggest to use Java int or Java long, because the least 16 bits have the same meaning as in uint16 (but not Java short: it has 16 bits, but is signed).

1) If you receive uint16 only, no chars, no other types, then the easiest would be following. Read 16 bits into a Java short, then convert it to int:

byte[] data = //From UDP socket
ByteBuffer bb = ByteBuffer.wrap(data);
bb = bb.order(ByteOrder.LITTLE_ENDIAN);

while(bb.hasRemaining()){
    short s = bb.getShort();
    int n = 0xFFFF & s;
    ...
}

2) If you receive not only uint16, but some other types like chars, then you can construct the value from 2 bytes:

byte[] data = //From UDP socket
ByteBuffer bb = ByteBuffer.wrap(data);

while(bb.hasRemaining()) {
    if (consider next bytes as uint16) {
        byte b1 = bb.get();
        byte b2 = bb.get();
        int i1 = 0xFF & b1; // Consider b1 as int, not the same as "(int) b1"
        int i2 = 0xFF & b2; // Consider b2 as int, not the same as "(int) b2"
        int n = b2 * 256 + b1;
        ...
    } else {
        // Read next bytes as char or something other
    }
}