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] with the high byte and data[1] with the low byte.

That is binary value of 400 = 1 1001 0000 so data[0] should contain 0000 0001 and data[1] should contain 1001 0000

score 95 · Accepted Answer · edited May 23 '17 at 12:34

95

Using simple bitwise operations:

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

How it works:

& 0xFF masks all but the lowest eight bits.
>> 8 discards the lowest 8 bits by moving all bits 8 places to the right.
The cast to byte is necessary because these bitwise operations work on an int and return an int, which is a bigger data type than byte. The case is safe, since all non-zero bits will fit in the byte. For more information, see Conversions and Promotions.

Edit: Taylor L correctly remarks that though >> works in this case, it may yield incorrect results if you generalize this code to four bytes (since in Java an int is 32 bits). In that case, it's better to use >>> instead of >>. For more information, see the Java tutorial on Bitwise and Bit Shift Operators.

edited May 23 '17 at 12:34

Community

1
1

answered Nov 14 '09 at 22:37

Stephan202

59,965
13
127
133

4

>> might not be what you want. >>> doesn't carry the sign bit when you shift so that is probably better to use in this case if you want to capture all 4 bytes. – Taylor Leese Nov 14 '09 at 22:41
if I do "new byte[]{width & 0xFF, (width >> 8) & 0xFF}" the compiler says "possible loss of precision required byte found int" – Kevin Boyd Nov 14 '09 at 22:54
@Kevin: I see I forgot to cast to `byte`. Will update the answer. – Stephan202 Nov 14 '09 at 23:01
@Taylor: Will check it out, but why is Stephan's code not working? – Kevin Boyd Nov 14 '09 at 23:02
@Kevin: I added an explanation for that fact as well. – Stephan202 Nov 14 '09 at 23:04
@Stephan: Thanks, its doing fine now – Kevin Boyd Nov 14 '09 at 23:34
1

You don't need the mask, the cast will cut off the excess bits. It doesn't matter whether you use >> or >>>, the ones due to sign extension will be cut off. Though I'd consider >>> better style. – starblue Nov 15 '09 at 10:17

score 13 · Answer 2 · edited May 23 '17 at 12:34

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.

score 4 · Answer 3 · answered Nov 14 '09 at 22:57

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;
}

score 1 · Answer 4 · answered Nov 14 '09 at 22:37

1

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.

answered Nov 14 '09 at 22:37

Dmitry

3,740
15
17

score 0 · Answer 5 · answered Nov 14 '09 at 22:37

0

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;

answered Nov 14 '09 at 22:37

R Hyde

10,301
1
32
28

Apologies - was trying to illustrate the principle, but you're quite right! – R Hyde Nov 15 '09 at 10:18

score 0 · Answer 6 · answered Nov 14 '09 at 22:47

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);
    }
}

score 0 · Answer 7 · answered Nov 15 '09 at 10:39

0

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 ;)

answered Nov 15 '09 at 10:39

Peter Lawrey

525,659
79
751
1,130

How do I split an integer into 2 byte binary?

7 Answers7

Linked