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Missing one byte when transferring an image over Socket Java

I have a problem transferring a file over socket.
I Wrote a simple client / server app and the client takes a screenshot and send it to server.
The problem is the file is not completed whatever i do, It's always missing the first byte from the array which makes the photo damaged.
When I open the photo in any hex editor and compare the original photo with the one that the client sent, I can see the missing byte, as if I add it, the photo opens without the problem. The size of the sent file missing just one byte !
Here is a photo for the problem :
Original photo
sent photo
Here is the code :
Server ( Receiver ) :
byte[] buf;
InputStream inp;
try (BufferedOutputStream out1 = new BufferedOutputStream(new FileOutputStream(new File("final.jpeg")))) {
buf = new byte[s.getReceiveBufferSize()];
inp = new DataInputStream(s.getInputStream());
Thread.sleep(200);
int len = 0;
while ((len = inp.read(buf)) >0){
out1.write(buf,0,len);
}
out1.flush();
inp.close();
out1.close();
}
Client ( Sender ):
BufferedImage screenshot = new Robot().createScreenCapture(new Rectangle(Toolkit.getDefaultToolkit().getScreenSize()));
ByteArrayOutputStream os = new ByteArrayOutputStream();
ImageIO.write(screenshot, "jpeg", os);
ImageIO.write(screenshot, "jpeg", new File("test.jpeg"));
OutputStream out = new BufferedOutputStream( connection.getOutputStream());
out.write(os.toByteArray());
out.close();
I have tried to send the array with the same way I receive it but no lock. I have tried with, and without buffered, I have tried flush in both sides, I tried to turn off Nod antivirus, Tried a sleep when sending length,
I almost tried everything without success .
I have tried on both, My pc and a virtual machine windows 7 !
Any help will be appreciated.
Edit :
First 10 bytes from the original file :
first 10 bytes from the sent file :

The code you posted does not lose data. Somewhere prior to executing the server code you posted, you have executed a single InputStream.read() of one byte, possibly in a misguided attempt to test for end of stream.
The sleep is just literally a waste of time. Remove it. You don't need the DataInput/OutputStreams either.

Please keep in mind that DataInputStream signals end of stream by returning value -1 from read() therefore your server reading loop should look like this:
while ((len = inp.read(buf)) != -1){
out1.write(buf,0,len);
}
Perhaps this helps.

The client code looks fine. Must be the server. You only posted the part when "some" input stream is written to a file. What happens before? Anyone doing a read() on the input stream?
Sorry for writing this in the "answer" section. Apparently, I cannot comment yet.

Ok it was my fault ! I was looking for something wrong in server side but the fault was in client side ! I opened a DataInputStream to read the order coming from server without closing it and that was the problem.

Java: I need to get byte array out of outputstream

In my case the out put stream is basically FileOutputStream. Hence for this code:
ByteArrayOutputStream bos = (ByteArrayOutputStream) streamToEncrypt;
(Where streamToEncrypt is OutputStream) is getting this exception:
java.lang.ClassCastException: java.io.FileOutputStream cannot be cast to java.io.ByteArrayOutputStream
All I need to do is to get the byte array out of this outputstream.
I do NOT have access to the file at this level. All I have is this output stream that I have to encrypt before pushing it to the file

That simply doesn't make any sense.
A ByteArrayOutputStream takes the data you push into it ... and stores them within a byte array that you can retrieve later on.
A FileOutputStream takes data ... and pushes it into a file.
Those are two completely different classes; you simply can't cast one into the other! Thats like you trying to cast a String into an Integer; or turning an Apple into a Banana by saying "you apple, now be a banana". Simply wont work!
So, the real "answer" here: you should step back and clarify what exactly you intend to do with your data; and then you use that stream(s) matching that requirement(s).

You don't get bytes out of an OutputStream. You get them out of an InputStream. You put bytes into an OutputStream.
If bytes have already been written to the OutputStream, it's too late. Opening the file again and reading from it is the only way for you to access those data.
If you want to encrypt an output stream, you should construct the stream and pass it to the code that writes to the stream.
Cipher enc = Cipher.getInstance("...");
enc.init(...);
try (OutputStream fos = Files.newOutputStream(path);
OutputStream os = new CipherOutputStream(fos, enc)) {
writingObject.write(os);
}

Build multipart/form-data using Java and without Apache libraries

I have a java code that builds a multipart/form-data and it all works OK but when the text part is in cyrilic it becomes unreadable on the server side.
The reason this happens is because the DataOutputStream class doesn't handle cyrilic character except if i use its method .writeUTF but then two additional bytes are added the again this makes me some trouble on the server side.
Here is the beggining of the code:
ByteArrayOutputStream bos = new ByteArrayOutputStream();
DataOutputStream dos = new DataOutputStream(bos);
Then the text bytes and image bytes are written to the dos object and finally we have
bos.toByteArray();
Do you know what else i can use instead of DataOutputStream and ByteArrayOutputStream.

DataOutputStream has no proper way to write Strings in specified encoding, but if you can't switch to e.g. BufferedWriter/OutputStreamWriter, you can use the write(byte[] b, int offset, int len) as follows:
byte[] bytes = myString.getBytes("UTF-8"); // or whichever encoding you want
dataout.write(bytes, 0, bytes.length);

Receiving files over socket

I am implementing a Direct Connect client. I am using the NMDC protocol. I can connect to a hub and other connected clients. I am trying to retrieve the file list from each client, I understand that in order to do that one must download the file files.xml.bz2 from the other client. The protocol to download a file is as follows:
-> $ADCGET file <filename> <params>|
<- $ADCSND file <fileName> <params>|
<- (*** binary data is now transfered from client B to client A ***)
I am trying to create a file named files.xml.bz2 using the binary data received. Here's my code:
//filesize is provided through the $ADCSND response from other client
byte[] data = new byte[filesize];
/*
Reading binary data from socket inputstream
*/
int read = 0;
for (int i=0; read<filesize;){
int available = in2.available();
int leftspace = filesize-read;
if (available>0){
in2.read(data, read, available>leftspace? leftspace:available);
++i;
}
read += (available>leftspace? leftspace:available)+1;
}
/*
writing the bytes to an actual file
*/
ByteArrayInputStream f = new ByteArrayInputStream(data);
FileOutputStream file = new FileOutputStream("files.xml.bz2");
file.write(data);
file.close();
The file is created, however, the contents (files.xml) are not readable. Opening it in firefox gives:
XML Parsing Error: not well-formed
Viewing the contents in the terminal only reads binary data. What am i doing wrong?
EDIT
I also tried Decompressing the file using the bz2 libray from Apache Ant.
ByteArrayInputStream f = new ByteArrayInputStream(data);
BZip2CompressorInputStream bzstream = new BZip2CompressorInputStream(f);
FileOutputStream xmlFile = new FileOutputStream("files.xml");
byte[] bytes = new byte[1024];
while((bzstream.read(bytes))!=-1){
xmlFile.write(bytes);
}
xmlFile.close();
bzstream.close();
I get an error, here's the stacktrace:
java.io.IOException: Stream is not in the BZip2 format
at org.apache.commons.compress.compressors.bzip2.BZip2CompressorInputStream.init(BZip2CompressorInputStream.java:240)
at org.apache.commons.compress.compressors.bzip2.BZip2CompressorInputStream.<init>(BZip2CompressorInputStream.java:132)
at org.apache.commons.compress.compressors.bzip2.BZip2CompressorInputStream.<init>(BZip2CompressorInputStream.java:109)
at control.Controller$1.run(Controller.java:196)

Usual, typical misuse of available(). All you need to copy a stream in Java is as follows:
while ((count = in.read(buffer)) >= 0)
{
out.write(buffer, 0, count);
}
Use this with any size buffer greater than zero, but preferably several kilobytes. You don't need a new buffer per iteration, and you don't need to know how much data is available to read without blocking, as you have to block, otherwise you're just smoking the CPU. But you do need to know how much data was actually read per iteration, and this is the first place where your code falls down.

The error java.io.IOException: Stream is not in the BZip2 format is generated by the constructor of class BZip2CompressorInputStream. I decided to scan the bytes, looking for the magic number to make sure that the file was bz2 format, it turns out that Java was right -- it wasnt in bz2 format.
Upon examining the source code of Jucy, I saw that the reason for this was a slight error in the command I sent to the other client, in essence, this error was caused a mistake in my protocol implementation. The solution was:
Replace:
$ADCGET file files.xml.bz2 0 -1 ZL1|
With:
$ADCGET file files.xml.bz2 0 -1|
ZL1 specifies compression of the files being sent (Not necessary).

How to pass an InputStream via RMI

Consider these two functions:
Function A takes inputStream as parameter.
public void processStream(InputStream stream)
{
//Do process routine
}
Function B loads a file content to pass it to Function A as InputStream.
pulic void loadFile()
{
File file =new File("c:\\file.txt");
//Pass file as InputStream
}
How can I pass file from Function B to Function A as InputStream without reading it on first hand?
I did something like this:
File file = new File("c:\\file.txt");
DataInputStream stream= new DataInputStream(new FileInputStream(file));
This generated the exception below:
java.io.WriteAbortedException: writing aborted; java.io.NotSerializableException: java.io.DataInputStream
EDIT:
loadFile() is passing the InputStream as RMI response.

The following should work just fine
processStream(new FileInputStream(file));
You should only not attempt to serialize an InputStream instance by ObjectOutputStream like as
objectOutputStream.writeObject(inputStream);
which you're apparently doing in processStream() method. That's namely exactly what the exception is trying to tell you. How to solve it properly depends on the sole functional requirement which you omitted from the question.
Update as per the comment
I am passing the InputStream as an RMI response.
There's the problem. You cannot pass non-serializable objects around as RMI response, let alone unread streams. You need to read the InputStream into a ByteArrayOutputStream the usual IO way and then use its toByteArray() to get a byte[] out of it and pass that instead. Something like:
InputStream input = new FileInputStream(file);
ByteArrayOutputStream output = new ByteArrayOutputStream();
byte[] buffer = new byte[8192];
for (int length = 0; (length = input.read(buffer)) > 0;) {
output.write(buffer, 0, length);
}
byte[] bytes = output.toByteArray(); // Pass that instead to RMI response.
Be careful with large files though. Every byte of a byte[] eats one byte of JVM's memory.

That exception seems to indicate that you are calling the processStream method on a remote object using something like RMI? if that is the case, you will need to re-visit what you are doing. sending streams of data over RMI is not an easy thing to do. if you are guaranteed to be using small files, you could copy the file data to a byte[] and pass that to the remote method call. if you need to process larger files, however, that will most likely cause memory issues on the client and/or server. in that case, you should use something like rmiio, which provides utilities for streaming data over RMI.

You could just pass the FileInputStream ?
processStream(new FileInputStream(yourFile));
The reason you are getting the exception is because DataInputStream is intended to read primitive Java types