How do I split an integer into 2 byte binary?

Question

Given

private int width = 400;
private byte [] data = new byte [2];  

I want to split the integer \"width\" into two bytes and load data[0]

Answer 1

For converting two bytes the cleanest solution is

data[0] = (byte) width;
data[1] = (byte) (width >>> 8);

For converting an integer to four bytes the code would be

data[0] = (byte) width;
data[1] = (byte) (width >>> 8);
data[2] = (byte) (width >>> 16);
data[3] = (byte) (width >>> 24);

It doesn't matter whether >> or >>> is used for shifting, any one bits created by sign extension will not end up in the resulting bytes.

See also this answer.

Answer 2

I suggest you have a look at the source for HeapByteBuffer. It has the conversion code for all primitive data types. (In fact you could just use a ByteBuffer ;)

Answer 3

To get the high byte, shift right by 8 bits then mask off the top bytes. Similarly, to get the low byte just mask off the top bytes.

data[0] = (width >> 8) & 0xff;
data[1] = width & 0xff;

Answer 4

int width = 400;
byte [] data = new byte [2];

data[0] = (byte) ((width & 0xFF00) >> 8);
data[1] = (byte) (width & 0xFF);

for(int b = 0; b < 2; b++) {
    System.out.println("printing byte " + b);
    for(int i = 7; i >= 0; i--) {
        System.out.println(data[b] & 1);
        data[b] = (byte) (data[b] >> 1);
    }
}

Answer 5

This should do what you want for a 4 byte int. Note, it stores the low byte at offset 0. I'll leave it as an exercise to the reader to order them as needed.

public static byte[] intToBytes(int x) {
    byte[] bytes = new byte[4];

    for (int i = 0; x != 0; i++, x >>>= 8) {
        bytes[i] = (byte) (x & 0xFF);
    }

    return bytes;
}

Answer 6

Integer is 32 bits (=4 bytes) in java, you know?

width & 0xff will give you the first byte, width & 0xff00 >> 8 will give you the second, etc.