java Serialize Object to ByteArray

Question

Hy guys! I've been searching for some old posts about Serialization in Java. I want to convert an Object into a ByteArray. So far I´ve done this:

public class test 
{

    public static void main(String[] args) throws IOException 
    {
        // 00111111 00111111 00111111 11110000  - in bytes: 63, 63, 63, 240
        int x = 1061109744; 
        byte[] bytes = null;

        ByteArrayOutputStream bos = new ByteArrayOutputStream();
        ObjectOutput out = null;
        try {
            out = new ObjectOutputStream(bos);   
            out.writeObject(x);
            bytes = bos.toByteArray();

        } 
        finally 
        {
            out.close();
            bos.close();
        }

        System.out.println("Lenght: " + bytes.length);

        for(int i=0; i < bytes.length; i++)
        {
            System.out.println(bytes[i]);
        }
    }
}

Apparently it works good but it introduces a lot of "junk" into the ByteArray. The values that really have interest to me are the last 4 bytes that correspond to my "int x = 1061109744;".

Why does this happen?

Is it possible to avoid the "junk" values?

Is it possible to surpass the "signed" values? (I have to write byte values bigger than 128).

You can check here http://stackoverflow.com/questions/2836646/java-serializable-object-to-byte-array — Ozan Deniz, May 04 '13 at 21:55

score 1 · Accepted Answer · answered May 04 '13 at 21:58

1

Is it possible to avoid the "junk" values?

Yes. Write out an int, not an Object. By passing an int to writeObject, you're promoting it (autoboxing it) to Integer, which means that the serialized information contains header information saying that it's an Integer object, not an int. Your ByteArrayOutputStream has a write(int) method that writes a byte (the low-order 8 bits of the int) to the output stream. Whether those are signed or not is purely a matter of interpretation, not the bits in the stream.

answered May 04 '13 at 21:58

T.J. Crowder

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In the example I shown, the variable "x" is just for test purpose. In reality I want to write a complete object called "EventParser" that contains int and long types. And Im still having the "signed" problem. – eduardo May 04 '13 at 22:13
2

@eduardo: Well, it makes no sense to show an example using a primitive if you're going to use an object in your "real" code. If you don't want "junk," write your own serialization format. The Java serialization format includes header information about the types of the objects being written to the stream; that's kind of the point. And again: Sign is a matter of *interpretation*, not bits in the stream. – T.J. Crowder May 04 '13 at 22:14
Sorry for my mistake. Now I got it about the "signed"! If I try to print it of course it will show signed but as you said, in the stream it is only bits. – eduardo May 04 '13 at 22:21

score 1 · Answer 2 · answered May 04 '13 at 23:15

Use DataOutputStream and field by field serialization:

ByteOutputStream baos = new ByteOuputStream();
DataOutputStream dos = new DataOutputStream(baos);

dos.writeInt(x);

dos.close();
bytes = baos.toByteArray();

In the byte array all elements are signed (naturally). But if you want to print them unsigned:

for(int i=0; i < bytes.length; i++)
{
    System.out.println(0xFF & bytes[i]);
}

java Serialize Object to ByteArray

2 Answers2