Java mutable byte array data structure

Question

I'm trying to find an easy way to create a mutable byte array that can automatically append any primitive Java data type. I've been searching but could not find anything useful.

I'm looking for something like this

ByteAppender byteStructure = new ByteAppender();
byteStructure.appendInt(5);
byteStructure.appendDouble(10.0);

byte[] bytes = byteStructure.toByteArray();

There is ByteByffer which is great, but you have to know the size of the buffer before you start, which won't work in my case. There is a similar thing (StringBuilder) for creating Strings, but I cannot find one for Bytes.

I thought this would be obvious in Java.

Why can't you use an ArrayList<Byte> and use .toArray() afterwards? — Jeroen Vannevel
– Jeroen Vannevel, Commented Jan 17, 2014 at 1:58
@JoshM, speed mostly. I have a collection of ordered elements, and for primitive types, I'll just use their bytes. But for strings, I'll need to calculate their length and convert them to a byte array. If I have to calculate the byte array size first, then I have to iterate the list twice, calculating the size on the first pass and then translating on the second. — Jason
– Jason, Commented Jan 17, 2014 at 2:01
@JeroenVannevel, ArrayList<Byte> doesn't accept primitive types as input, so I would have to constantly do something like this: ByteBuffer.allocate(4).putInt(yourInt).array(); and then for(Byte b : bytes) arrayBytes.add(b);. It is doable but very messy. — Jason
– Jason, Commented Jan 17, 2014 at 2:03

BlackJoker · Accepted Answer · 2014-01-17 02:03:05Z

16

I guess you are looking for java.io.DataOutputStream

ByteArrayOutputStream out = new ByteArrayOutputStream();
DataOutputStream dout = new DataOutputStream(out);
dout.writeInt(1234);
dout.writeLong(123L);
dout.writeFloat(1.2f);
byte[] storingData = out.toByteArray();

How to use storingData?

//how to use storingData?
ByteArrayInputStream in = new ByteArrayInputStream(storingData);
DataInputStream din = new DataInputStream(in);
int v1 = din.readInt();//1234
long v2 = din.readLong();//123L
float v3 = din.readFloat();//1.2f

answered Jan 17, 2014 at 2:03

BlackJoker

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2 Comments

Jason Over a year ago

Perfect. Exactly what I was looking for.

Jason Over a year ago

PS don't forget to close

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