Develop Reference

Java ByteBuffer for Zipfile

I have a binary file that contains big endian data. I am using this code to read it in
FileChannel fileInputChannel = new FileInputStream(fileInput).getChannel();
ByteBuffer bb = ByteBuffer.allocateDirect((int)fileInputChannel.size());
while (bb.remaining() > 0)
fileInputChannel.read(bb);
fileInputChannel.close();
bb.flip();
I have to do something identical for zip files. In other words decompress the file from a zip file and order it. I understand I can read it in via ZipInputStream but then I have to provide the coding for the "endianness". With ByteBuffer you can use ByteOrder.
Is there an NIO alternative for zip files ?

If you have your ZipInputStream, just use Channels.newChannel to convert it to a Channel then proceed as you wish. But you should keep in mind that it might be possible that a ZipInputStream can’t predict its uncompressed size so you might have to guess the appropriate buffer size and possibly re-allocate a bigger buffer when needed. And, since the underlying API uses byte arrays, there is no benefit in using direct ByteBuffers in the case of ZipInputStream, i.e. I recommend using ByteBuffer.allocate instead of ByteBuffer.allocateDirect for this use case.
By the way you can replace while(bb.remaining() > 0) with while(bb.hasRemaining()). And since Java 7 you can use FileChannel.open to open a FileChannel without the detour via FileInputStream.

FileChannel.open() vs RandomAccessFile in Jdk 7

I would like to know the difference between the following:
FileChannel fc = FileChannel.open();
RandomAccessFile ra = new RandomAccessFile("RandomFile", "rw");
Since Java 7 the class FileChannel implements SeekableByteChannel therefore has all it needs in order to randomly access the file.
Can we say that the 2 are totally the same?

FileChannel has many more features since it is also GatheringByteChannel, InterruptibleChannel, ScatteringByteChannel. Besides it can lock files, transfer files, work with direct byte buffers, see API

Java file IO truncated while reading large files using BufferedInputStream

I have a function in which I am only given a BufferedInputStream and no other information about the file to be read. I unfortunately cannot alter the method definition as it is called by code I don't have access to. I've been using the code below to read the file and place its contents in a String:
public String[] doImport(BufferedInputStream stream) throws IOException, PersistenceException {
int bytesAvail = stream.available();
byte[] bytesRead = new byte[bytesAvail];
stream.read(bytesRead);
stream.close();
String fileContents = new String(bytesRead);
//more code here working with fileContents
}
My problem is that for large files (>2Gb), this code causes the program to either run extremely slowly or truncate the data, depending on the computer the program is executed on. Does anyone have a recommendation regarding how to deal with large files in this situation?

You're assuming that available() returns the size of the file; it does not. It returns the number of bytes available to be read, and that may be any number less than or equal to the size of the file.
Unfortunately there's no way to do what you want in just one shot without having some other source of information on the length of the file data (i.e., by calling java.io.File.length()). Instead, you have to possibly accumulate from multiple reads. One way is by using ByteArrayOutputStream. Read into a fixed, finite-size array, then write the data you read into a ByteArrayOutputStream. At the end, pull the byte array out. You'll need to use the three-argument forms of read() and write() and look at the return value of read() so you know exactly how many bytes were read into the buffer on each call.

I'm not sure why you don't think you can read it line-by-line. BufferedInputStream only describes how the underlying stream is accessed, it doesn't impose any restrictions on how you ultimately read data from it. You can use it just as if it were any other InputStream.
Namely, to read it line-by-line you could do
InputStreamReader streamReader = new InputStreamReader(stream);
BufferedInputReader lineReader = new BufferedInputReader(streamReader);
String line = lineReader.readLine();
...
[Edit] This response is to the original wording of the question, which asked specifically for a way to read the input file line-by-line.

Does FileOutputStream truncate an existing file

Does
final OutputStream output = new FileOutputStream(file);
truncate the file if it already exists? Surprisingly, the API documentation for Java 6 does not say. Nor does the API documentation for Java 7. The specification for the language itself has nothing to say about the semantics of the FileOutputStream class.
I am aware that
final OutputStream output = new FileOutputStream(file, true);
causes appending to the file. But appending and truncating are not the only possibilities. If you write 100 bytes into a 1000 byte file, one possibility is that the final 900 bytes are left as they were.

FileOutputStream without the append option does truncate the file.
Note that FileOutputStream opens a Stream, not a random access file, so i guess it does make sense that it behaves that way, although i agree that the documentation could be more explicit about it.

I tried this on Windows 2008 x86 and java 1.6.0_32-b05
I created 2 processes which wrote continually to the same file one 1Mb of the character 'b' and the other 4Mb of the character 'a'. Unless I used
out = new RandomAccessFile(which, "rw");
out.setLength(0);
out.getChannel().lock();
I found that a 3rd reader process could read what appeared to be a File which started with 1Mb of 'b's followed by 'a's
I found that writing first to a temporary file and then renaming it
File.renameTo
to the File also worked.
I would not depend on FileOuputStream on windows to truncate a file which may be being read by a second process...
Not new FileOutputStream(file)
Nor FileOutputStream(file, false) ( does not truncate )
Nor
this;
out = new FileOutputStream(which, false);
out.getChannel().truncate(0);
out.getChannel().force(true);
However
out = new FileOutputStream(which, false);
out.getChannel().truncate(0);
out.getChannel().force(true);
out.getChannel().lock();
does work

FileOutputStream is meant to write binary data, which is most often overwritten.
If you are manipulating text data, you should better use a FileWriter which has convenient append methods.

Creating a File from byte array

So - I've got a third party library that needs a File as input. I've got a byte array.
I don't want to write the bytes to disk .. I'd like to keep this in memory. Any idea on how I can create a File from the provided byte array (without writing to disk)?

Sorry, not possible. A File is inherently an on-disk entity, unless you have a RAM disk - but that's not something you can create in Java.
That's exactly the reason why APIs should not be based on File objects (or be overloaded to accept an InputStream).

There's one possibility, but it's a real long-shot.
If the API uses new FileReader(file) or new FileInputStream(file) then you're hosed, but...
If it converts the file to a URL or URI (using toURL() or toURI()) then, since File is not final, you can pass in a subclass of File in which you control the construction of the URL/URI and, more importantly, the handler.
But the chances are VERY slim!

So I see there is an accepted answer (and this is old), but I found a way to do this. I was using the IDOL On Demand API and needed to convert a byte array to a File.
Here is an example of taking a byte array of an image and turning into a File:
//imageByte is the byte array that is already defined
BufferedImage image = null;
ByteArrayInputStream bis = new ByteArrayInputStream(imageByte);
image = ImageIO.read(bis);
bis.close();
// write the image to a file
File outputfile = new File("image.png");
ImageIO.write(image, "png", outputfile);
And so outputfile is a File that can be used later in your program.