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FileOutputStream - sometime writing data and sometime not

I am trying to understand few java file concepts. So I tried with below program to understand FileOutputStream
FileOutputStream out = new FileOutputStream("test.txt");
int i = 1;
out.write(i);
out.flush();
out.close();
Some binary data has been writen in the file.
But the for the same program when I change the value of i from 1 to 10. I don't see anything in my output file. Can someone explain me why with some internal details.

int i = 1;
out.write(i);
That writes 0x1 to the file.
out.flush();
out.close();
The flush() is redundant.
Some binary data has been writen in the file.
Correct.
But the for the same program when I change the value of i from 1 to 10. I don't see anything in my output file.
Yes you do. You see 0xa, which is a line feed character.

JAVA: Stream any file to browser correctly

So I have created my own personal HTTP Server in Java from scratch.
So far it is working fine but with one major flaw.
When I try to pass big files to the browser I get a Java Heap Space error. I know how to fix this error through the JVM but I am looking for the long term solution for this.
//declare an integer for the byte length of the file
int length = (int) f.length();
//start the fileinput stream.
FileInputStream fis = new FileInputStream(f);
//byte array with the length of the file
byte[] bytes = new byte[length];
//write the file until the bytes is empty.
while ((length = fis.read(bytes)) != -1 ){
write(bytes, 0, length);
}
flush();
//close the file input stream
fis.close();
This way sends the file to the browser successfully and streams it perfectly but the issue is, because I am creating a byte array with the length of the file. When the file is very big I get the Heap Space error.
I have eliminated this issue by using a buffer as shown below and I dont get Heap Space errors anymore. BUT the way shown below does not stream the files in the browser correctly. It's as if the file bytes are being shuffled and are being sent to the browser all together.
final int bufferSize = 4096;
byte buffer[] = new byte[bufferSize];
FileInputStream fis = new FileInputStream(f);
BufferedInputStream bis = new BufferedInputStream(fis);
while ( true )
{
int length = bis.read( buffer, 0, bufferSize );
if ( length < 0 ) break;
write( buffer, 0, length );
}
flush();
bis.close();
fis.close();
NOTE1:
All the correct Response Headers are being sent perfectly to the browser.
Note2:
Both ways work perfectly on a computer browser but only the first way works on a smartphone's browser (but sometimes it gives me Heap Space error).
If someone knows how to correctly send files to a browser and stream them correctly I would be a very very happy man.
Thank you in advance! :)

When reading from a BufferedInputStream you can allow its' buffer to handle the buffering, there is no reason to read everything into a byte[] (and certainly not a byte[] of the entire File). Read one byte at a time, and rely on the internal buffer of the stream. Something like,
FileInputStream fis = new FileInputStream(f);
BufferedInputStream bis = new BufferedInputStream(fis);
int abyte;
while ((abyte = bis.read()) != -1 ){
write(abyte);
}

Emm... As I can see it, you try to use chunks in your code anyway,
as I can remember, even the apache HttpClient+FileUpload solution has file size limit about <=2.1GB or something (correct me if I am wrong) so it is a bit hard thing...
I haven't tried the solution yet but as a test you can use java.io.RandomAccessFile in combination with File(Input/Output)Stream on the client and server not to read and write the whole file at a time but sequence of lets say <=30MB blocks for example to avoid the annoying outofmemory errors ; An example of using RandomAccessFile can be found here https://examples.javacodegeeks.com/core-java/io/randomaccessfile/java-randomaccessfile-example/
But still you give less details :( I mean is your client suppose to be a common Java application or not?
If you have some additional information please let me know
Good luck :)

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).

Why does my untar not contain the last bytes

I've written a rest resource that serves a .tar.gz file. It's working OK. I've tried requesting it, saving the data, unpacking it (with tar xzvf [filename]) and I get the correct data.
However, I'm trying to use java.util.zip.GZIPInputStream and org.apache.tools.tar.TarInputStream to unzip and untar a .tar.gz that I'm serving in a JUnit test, to verify that it's working automatically. This is the code in my unit test with some details removed:
HttpResponse response = <make request code here>
byte[] receivedBytes = FileHelper.copyInputStreamToByteArray(response.getEntity().getContent(), true);
GZIPInputStream gzipInputStream = new GZIPInputStream(new ByteArrayInputStream(receivedBytes));
TarInputStream tarInputStream = new TarInputStream(gzipInputStream);
TarEntry tarEntry = tarInputStream.getNextEntry();
ByteArrayOutputStream byteArrayOutputStream = null;
System.out.println("Record size: " + tarInputStream.getRecordSize());
while (tarEntry != null) // It only goes in here once
{
byteArrayOutputStream = new ByteArrayOutputStream();
tarInputStream.copyEntryContents(byteArrayOutputStream);
tarEntry = tarInputStream.getNextEntry();
}
byteArrayOutputStream.flush();
byteArrayOutputStream.close();
byte[] archivedBytes = byteArrayOutputStream.toByteArray();
byte[] actualBytes = <get actual bytes>
Assert.assertArrayEquals(actualBytes, archivedBytes);
The final assert fails with a difference at byte X = (n * 512) + 1, where n is the greatest natural number such that n * 512 <= l and l is the length of the data. That is, I get the the biggest possible multiple of 512 bytes of data correctly, but debugging the test I can see that all the remaining bytes are zero. So, if the total amount of data is 1000 bytes, the first 512 bytes in archivedBytes are correct, but the last 488 are all zero / unset, and if the total data is 262272 bytes I get the first 262144 (512*512) bytes correctly, but the remaining bytes are all zero again.
Also, the tarInputStream.getRecordSize() System out above prints Record size: 512, so I presume that this is somehow related. However, since the archive works if I download it, I guess the data must be there, and there's just something I'm missing.
Stepping into the tarInputStream.copyEntryContents(byteArrayOutputStream) with the 1000 byte data, in
int numRead = read(buf, 0, buf.length);
the numRead is 100, but looking at the buffer, only the first 512 bytes are non-zero. Maybe I shouldn't be using that method to get the data out of the TarInputStream?
If anyone knows how it's supposed to work, I'd be very grateful for any advice or help.

You can specify the output block size to be used when you create a tar archive. Thus the size of the archive will be a multiple of the block size. As the archive size doesn't normally fit in a whole number of blocks, zeros are added to the last block of data to make it of the right size.

It turned out that I was wrong in my original question, and the error was in the resource code. I wasn't closing the entry on the TarOutputStream when writing to it. I guess this was not causing any problems when requesting it manually from the server, maybe because the entry was closed with the connection or something, but working differently when being requested from a Unit test... though I must admit that doesn't make a whole lot of sense to be :P
Looking at the fragment of my writing code below, I was missing line 3.
1: tarOutputStream.putNextEntry(tarEntry);
2: tarOutputStream.write(fileRawBytes);
3: tarOutputStream.closeEntry();
4: tarOutputStream.close();
I didn't even know there was such a thing as a "closeEntry" on the TarOutputStream... I do now! :P

Obtaining the wav file from ServletInputStream

Our current project requires us to send an audio file to the server and then use the audio file for further computation.
Using the Java sound api, I was able to capture the recording and save it as a wav file in my system. Then in order to pass the audio wav to the server, I am using Apache Commons HttpClient to post a request to the server. (I am using InputstreamEntity provided by apache and sending the data as a chunk).
The problem appears when i am trying to recreate/retrieve the wav file on the server. I understand that I would have to use the AudioSystem.write API to create the wav file (exactly as what was done on my system). However what I observe is that althought the file gets created , it does not play (I am using vlc media player to test it FYI). I have searched in Google for sample codes and have tried to implement it, but is unable to play it once the file gets created.
The sample code snippets indicates the approaches i have tried:
//******************************************************************
try {
InputStream is = request.getInputStream();
FileOutputStream fs = new FileOutputStream("output123.wav");
byte[] tempbuffer = new byte[4096];
int bytesRead;
while((bytesRead=is.read(tempbuffer))!=-1)
{
fs.write(tempbuffer, 0,bytesRead);
}
is.close();
fs.close();
AudioInputStream inputStream =AudioSystem.getAudioInputStream(newFile("output123.wav"));
int numofbytes = inputStream.available();
byte[] buffer = new byte[numofbytes];
inputStream.read(buffer);
int bytesWritten = AudioSystem.write(inputStream, AudioFileFormat.Type.WAVE,new File("outputtest.wav"));
System.out.println("written"+bytesWritten);
Approach 2
InputStream is = request.getInputStream();
System.out.println("inputStream obtained : "+is.toString());
ByteArrayInputStream bais = null;
byte[] audioBuffer = IOUtils.toByteArray(is);
System.out.println(" is audioBuffer empty? : length = ? "+audioBuffer.length);
try {
AudioFileFormat ai = AudioSystem.getAudioFileFormat(is);
System.out.println("ai bytelength ? "+ai.getByteLength());
System.out.println("ai frame length = "+ai.getFrameLength());
Set<Map.Entry<String,Object>> audioProperties = ai.getFormat().properties().entrySet();
System.out.println("entry set is empty ? "+audioProperties.isEmpty());
for(Map.Entry me : audioProperties){
System.out.println("key = "+me.getKey());
System.out.println("value ="+me.getValue());}
bais = new ByteArrayInputStream(audioBuffer);
AudioInputStream ais = new AudioInputStream(bais, new AudioFormat(8000,8,2,true,true), 2);
AudioSystem.write(ais, AudioFileFormat.Type.WAVE,new File("testtest.wav"));
//*************************************************************************************
The audioFormat properties all turned out to be null. Are these null values giving the problem? So while creating the wave file on the server, I tried to set the properties manually once again. But even then the wav file would not play.
I have also tried quite a few approaches already mentioned on this site, but somehow they aren't working. I am sure i am missing something, but I am unable to pinpoint the exact problem.
Would be really helpful, if you guys can point out how to go about the conversion from ServletInputStream to getting a wav.
P.S (1) I know the code is shabby, because i have been under a trial and error situation for quite some time now. But I will give more details on the approaches if needed.
2) Apologise for the clumsiness, this happens to be my first post.. )

this is not how you copy a stream (from Approach 1). you have the correct code to copy a stream just above this.:
int numofbytes = inputStream.available();
byte[] buffer = new byte[numofbytes];
inputStream.read(buffer);

If all your server wants to do is get the data and write it to a file, then you do not need to use any of the audio API: simply treat the data as a stream of bytes.
So the part of approach 1 that is before any mention of AudioInputStream should be sufficient.

Although the approach chosen might not be the perfect solution, due to time constraints, I adopted a simpler approach. Using java.util.zip i simply zipped it up and sent it over to the server and then wrote a layer wherin the file gets unzipped . then i deleted the zip files. Seems like an immature solution (bcos the original challenge was to send the audio file). now i am incurring an overhead of zipping the files, but the file transfer would hapeen relatively faster. Thanks for your help guys.