ByteBuffer byte padding in Java

Question

I'm working on a Java program in which I need to convert a short into two bytes (which will then be packed into a data packet). I'm using a ByteBuffer to perform the conversion, and it seems to work, but I'm seeing some apparent byte padding which I don't quite understand.

Here's a simple example I wrote:

import java.lang.*;
import java.io.*;
import java.nio.ByteOrder;
import java.nio.ByteBuffer;

public class Test {
    public static void main(String args[]) {
        short i = 27015;
        String s = Integer.toHexString(i);

        System.out.println( "i = " + i );
        System.out.println( "s = " + s );
        System.out.println( "---" );

        ByteBuffer b = ByteBuffer.allocate(2);
        b.order(ByteOrder.BIG_ENDIAN);
        b.putShort(i);

        System.out.printf("0x%H\n", b.getShort(0));
        System.out.println( "---" );

        byte[] a = b.array();

        for( int j = 0; j < a.length; j++ )
            System.out.printf("a[" + j + "] = 0x%H\n", a[j]);

        System.exit(0);
    }
}

This program produces the following output:

i = 27015
s = 6987
---
0x6987
---
a[0] = 0x69
a[1] = 0xFFFFFF87

When the ByteBuffer is converted to a byte array, why is the second byte padded with 0xFF? It seems like the second element of the array should be 0x87 instead of 0xFFFFFF87. Am I missing something?

Thanks!

The byte 0x87 is negative, has its high bit set. Converted to an int other one get 0xFFFFFF87, a negative number. — Joop Eggen
– Joop Eggen, Commented Aug 5, 2013 at 15:14

Peter Lawrey · Accepted Answer · 2013-08-05 15:09:13Z

1

When the ByteBuffer is converted to a byte array, why is the second byte padded with 0xFF?

When you print with %H it casts the byte into an int. It id not the byte in the ByteBuffer which has has this many bit, but the way it is being printed. I suggest using Byte.toHexString(a[j]) instead.

answered Aug 5, 2013 at 15:09

Peter Lawrey

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2 Comments

Dave Over a year ago

Thanks for condensing my confusion down to a much more concise question, and thanks for your answer. That makes complete sense. It never occurred to me that the issue was in the printf() conversion.

Peter Lawrey Over a year ago

When I have a test fail, at least 50% of the time it's due to something wrong with the test.

Martijn Courteaux · Accepted Answer · 2013-08-05 15:12:57Z

0

Try this instead:

for (int j = 0; j < a.length; j++)
   System.out.printf("a[" + j + "] = 0x%H\n", ((int) a[j]) & 0xFF);

The & 0xFF operation will cut the highest three bytes, that are created after the conversion to an int.

FYI: ((int) a[j]) & 0xFF can be written as a[j] & 0xFF and will compile exactly the same. But I wrote it initially like that to clarify what is happening.

answered Aug 5, 2013 at 15:12

Martijn Courteaux

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1 Comment

Dave Over a year ago

Thanks for the suggestion. I tried it and that does indeed make the output more clear.

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