Develop Reference

Too large files when downloading Piktogramms

I'm trying to download some images provided by a hoster. This is the method I use:
public static void downloadImage(String imageLink, File f) throws IOException
{
URL url = new URL(imageLink);
byte[] buffer = new byte[1024];
BufferedInputStream in = new BufferedInputStream(url.openStream(), buffer.length);
BufferedOutputStream out = new BufferedOutputStream(new FileOutputStream(f), buffer.length);
while (in.read(buffer) > 0)
out.write(buffer);
out.flush();
out.close();
in.close();
}
However, the file turn out too big. 5MB for a 80x60 jpg is too much in my opinion.
What could be the cause of this?

You are doing things wrong here: read() returns the number of bytes that were really read; thus you have to write exactly that number from your buffer array into your output stream.
Your code is corrupting your output; and simply writing out a buffer array ... that mostly consists of 0s!
Instead do something like:
int bytesRead;
while ( ( bytesRead = in.read(buffer)) > 0) {
byte outBuffer[] = new byte[bytesRead];
... then use arraycopy to move bytesRead bytes
out.write(outBuffer);
}
( this is meant as inspiration to get you going, more pseudo like than real code )

java: read large binary file

I need to read out a given large file that contains 500000001 binaries. Afterwards I have to translate them into ASCII.
My Problem occurs while trying to store the binaries in a large array. I get the warning at the definition of the array ioBuf:
"The literal 16000000032 of type int is out of range."
I have no clue how to save these numbers to work with them! Has somebody an idea?
Here is my code:
public byte[] read(){
try{
BufferedInputStream in = new BufferedInputStream(new FileInputStream("data.dat"));
ByteArrayOutputStream bs = new ByteArrayOutputStream();
BufferedOutputStream out = new BufferedOutputStream(bs);
byte[] ioBuf = new byte[16000000032];
int bytesRead;
while ((bytesRead = in.read(ioBuf)) != -1){
out.write(ioBuf, 0, bytesRead);
}
out.close();
in.close();
return bs.toByteArray();
}

The maximum Index of an Array is Integer.MAX_VALUE and 16000000032 is greater than Integer.MAX_VALUE
Integer.MAX_VALUE = 2^31-1 = 2147483647
2147483647 < 16000000032
You could overcome this by checking if the Array is full and create another and continue reading.
But i'm not quite sure if your approach is the best way to perform this. byte[Integer_MAX_VALUE] is huge ;)
Maybe you can split the input file in smaller chunks process them.
EDIT: This is how you could read a single int of your file. You can resize the buffer's size to the amount of data you want to read. But you tried to read the whole file at once.
//Allocate buffer with 4byte = 32bit = Integer.SIZE
byte[] ioBuf = new byte[4];
int bytesRead;
while ((bytesRead = in.read(ioBuf)) != -1){
//if bytesRead == 4 you read 1 int
//do your stuff
}

If you need to declare a large constant, append an 'L' to it which indicates to the compiler that is a long constant. However, as mentioned in another answer you can't declare arrays that large.
I suspect the purpose of the exercise is to learn how to use the java.nio.Buffer family of classes.

I made some progress by starting from scratch! But I still have a problem.
My idea is to read up the first 32 bytes, convert them to a int number. Then the next 32 bytes etc. Unfortunately I just get the first and don't know how to proceed.
I discovered following method for converting these numbers to int:
public static int byteArrayToInt(byte[] b){
final ByteBuffer bb = ByteBuffer.wrap(b);
bb.order(ByteOrder.LITTLE_ENDIAN);
return bb.getInt();
}
so now I have:
BufferedInputStream in=null;
byte[] buf = new byte[32];
try {
in = new BufferedInputStream(new FileInputStream("ndata.dat"));
in.read(buf);
System.out.println(byteArrayToInt(buf));
in.close();
} catch (IOException e) {
System.out.println("error while reading ndata.dat file");
}

Reading a binary input stream into a single byte array in Java

The documentation says that one should not use available() method to determine the size of an InputStream. How can I read the whole content of an InputStream into a byte array?
InputStream in; //assuming already present
byte[] data = new byte[in.available()];
in.read(data);//now data is filled with the whole content of the InputStream
I could read multiple times into a buffer of a fixed size, but then, I will have to combine the data I read into a single byte array, which is a problem for me.

The simplest approach IMO is to use Guava and its ByteStreams class:
byte[] bytes = ByteStreams.toByteArray(in);
Or for a file:
byte[] bytes = Files.toByteArray(file);
Alternatively (if you didn't want to use Guava), you could create a ByteArrayOutputStream, and repeatedly read into a byte array and write into the ByteArrayOutputStream (letting that handle resizing), then call ByteArrayOutputStream.toByteArray().
Note that this approach works whether you can tell the length of your input or not - assuming you have enough memory, of course.

Please keep in mind that the answers here assume that the length of the file is less than or equal to Integer.MAX_VALUE(2147483647).
If you are reading in from a file, you can do something like this:
File file = new File("myFile");
byte[] fileData = new byte[(int) file.length()];
DataInputStream dis = new DataInputStream(new FileInputStream(file));
dis.readFully(fileData);
dis.close();
UPDATE (May 31, 2014):
Java 7 adds some new features in the java.nio.file package that can be used to make this example a few lines shorter. See the readAllBytes() method in the java.nio.file.Files class. Here is a short example:
import java.nio.file.FileSystems;
import java.nio.file.Files;
import java.nio.file.Path;
// ...
Path p = FileSystems.getDefault().getPath("", "myFile");
byte [] fileData = Files.readAllBytes(p);
Android has support for this starting in Api level 26 (8.0.0, Oreo).

You can use Apache commons-io for this task:
Refer to this method:
public static byte[] readFileToByteArray(File file) throws IOException
Update:
Java 7 way:
byte[] bytes = Files.readAllBytes(Paths.get(filename));
and if it is a text file and you want to convert it to String (change encoding as needed):
StandardCharsets.UTF_8.decode(ByteBuffer.wrap(bytes)).toString()

You can read it by chunks (byte buffer[] = new byte[2048]) and write the chunks to a ByteArrayOutputStream. From the ByteArrayOutputStream you can retrieve the contents as a byte[], without needing to determine its size beforehand.

I believe buffer length needs to be specified, as memory is finite and you may run out of it
Example:
InputStream in = new FileInputStream(strFileName);
long length = fileFileName.length();
if (length > Integer.MAX_VALUE) {
throw new IOException("File is too large!");
}
byte[] bytes = new byte[(int) length];
int offset = 0;
int numRead = 0;
while (offset < bytes.length && (numRead = in.read(bytes, offset, bytes.length - offset)) >= 0) {
offset += numRead;
}
if (offset < bytes.length) {
throw new IOException("Could not completely read file " + fileFileName.getName());
}
in.close();

Max value for array index is Integer.MAX_INT - it's around 2Gb (2^31 / 2 147 483 647).
Your input stream can be bigger than 2Gb, so you have to process data in chunks, sorry.
InputStream is;
final byte[] buffer = new byte[512 * 1024 * 1024]; // 512Mb
while(true) {
final int read = is.read(buffer);
if ( read < 0 ) {
break;
}
// do processing
}

How to write file data correctly?

My application is unable to transfer data properly over a socket connection and write it to a file properly. Files over about 65,535 bytes get corrupted and are no longer recognized by the programs designed to run them.
I have been able to send small .doc and .txt files successfully, but .mp3 .wmv .m4a .avi and just about anything else does not work. Neither do larger docs.
I have looked all over the internet for a solution to this problem. I have repeatedly tweaked the I/O code to fix the problem but it still doesn't work! Here is the I/O code in the super class that handles sending and receiving files. If you need anymore information/other parts of code, let me know.
protected void sendFile() throws IOException {
byte[] bytes = new byte[(int) file.length()];
buffin = new BufferedInputStream(new FileInputStream(file));
int bytesRead = buffin.read(bytes,0,bytes.length);
System.out.println(bytesRead);
out = sock.getOutputStream();
out.write(bytes,0,fileBytes);
out.flush();
out.close();
}
protected void receiveFile() throws IOException {
byte[] bytes = new byte[fileBytes];
in = sock.getInputStream();
for(int i=0;i<fileBytes;i++) {
in.read(bytes);
}
fos = new FileOutputStream("/Datawire/"+fileName);
buffout = new BufferedOutputStream(fos);
buffout.write(bytes,0,fileBytes);
buffout.flush();
buffout.close();
}
UPDATED CODE (that works):
protected void sendFile() throws IOException {
if((file.length())<63000) {
byte[] bytes = new byte[(int)file.length()];
buffin = new BufferedInputStream(new FileInputStream(file));
buffin.read(bytes,0,bytes.length);
out = sock.getOutputStream();
out.write(bytes,0,bytes.length);
out.close();
} else {
byte[] bytes = new byte[32000];
buffin = new BufferedInputStream(new FileInputStream(file));
out = sock.getOutputStream();
int bytesRead;
while((bytesRead = buffin.read(bytes))>0) {
out.write(bytes,0,bytesRead);
}
out.close();
}
}
protected void receiveFile() throws IOException {
if(fileBytes<63000) {
byte[] bytes = new byte[32000];
in = sock.getInputStream();
System.out.println(in.available());
in.read(bytes,0,fileBytes);
fos = new FileOutputStream("/Datawire/"+fileName);
buffout = new BufferedOutputStream(fos);
buffout.write(bytes,0,bytes.length);
buffout.close();
} else {
byte[] bytes = new byte[16000];
in = sock.getInputStream();
fos = new FileOutputStream("/Datawire/"+fileName);
buffout = new BufferedOutputStream(fos);
int bytesRead;
while((bytesRead = in.read(bytes))>0) {
buffout.write(bytes,0,bytesRead);
}
buffout.close();
}
}

The issue is that you are sending only chunks of it. That is, you are only sending 64k of the file ever. If the file is ever larger then 64k the other end will never see it.
You want to continously read from the BufferedInputStream until the read() returns either less then the length or -1.

Your code is completely wrong. This is how to copy a stream in Java:
int count;
byte[] buffer = new byte[8192]; // more if you like but no need for it to be the entire file size
while ((count = in.read(buffer)) > 0)
{
out.write(buffer, 0, count);
}
You should use this both when sending the file and when receiving the file. At present your sending method hopes that the entire file fits into memory; fits into INTEGER_MAX bytes; and is read in one chunk by the read method, without even checking the result. You can't assume any of those things. Your receive method is complete rubbish: it just keeps overwriting the same array, again without checking any read() results.
EDIT: Your revised code is just as bad, or worse. You are calling read() to check for EOS and then throwing that byte away, and then calling read() again and throwing away the read count it returns. You pointlessly have a different path for files < 64000, or 63000, or whatever it is, that has zero benefit except to give you two code paths to test, or possibly four, instead of one. The network only gives you 1460 bytes at a time at best anyway so what is the point? You already have (a) a BufferedInputStream with a default buffersize of 8192, and (b) my code that uses a byte[] buffer of any size you like. My code above works for any amount of data in two lines of executable code. Yours is 20. QED.

I suggest that you use some good library to read and write file contents as well as socket read/write. For example Apache Commons IO. If you insist on writig code yourself, do it smaller chunks rather than the whole file at once.

You have to consider that InputStream.read returns the number of bytes read which may be less than the total number of bytes in the file.
You would probably be better off just letting something like CopyUtils.copy take care of this for you.

You need to loop until bytesRead < 0. You need to make sure that fileBytes is => than the transferred file.
protected void receiveFile() throws IOException {
byte [] bytes = new byte [fileBytes];
InputStream is = sock.getInputStream();
FileOutputStream fos = new FileOutputStream("/Datawire/"+fileName);
BufferedOutputStream bos = new BufferedOutputStream(fos);
int bytesRead = is.read(bytes,0,bytes.length);
int current = bytesRead;
do {
bytesRead =
is.read(bytes, current, (bytes.length-current));
if(bytesRead >= 0) current += bytesRead;
} while(bytesRead > -1);
bos.write(bytes, 0 , current);
bos.flush();
bos.close();
}

java servlet serving a file over HTTP connection

I have the following code(Server is Tomcat/Linux).
// Send the local file over the current HTTP connection
FileInputStream fin = new FileInputStream(sendFile);
int readBlockSize;
int totalBytes=0;
while ((readBlockSize=fin.available())>0) {
byte[] buffer = new byte[readBlockSize];
fin.read(buffer, 0, readBlockSize);
outStream.write(buffer, 0, readBlockSize);
totalBytes+=readBlockSize;
}
With some files of type 3gp
When i attach the debugger, in line:
outStream.write(buffer, 0, readBlockSize);
it breaks out the while with the following error;
ApplicationFilterChain.internalDoFilter(ServletRequest, ServletResponse) line:299
And the file is not served.
Any clues?
Thanks
A.K.

You can't guarantee that InputStream.read(byte[], int, int) will actually read the desired number of bytes: it may read less. Even your call to available() will not provide that guarantee. You should use the return value from fin.read to find out how many bytes were actually read and only write that many to the output.
I would guess that the problem you see could be related to this. If the block read is less than the available size then your buffer will be partially filled and that will cause problems when you write too many bytes to the output.
Also, don't allocate a new array every time through the loop! That will result in a huge number of needless memory allocations that will slow your code down, and will potentially cause an OutOfMemoryError if available() returns a large number.
Try this:
int size;
int totalBytes = 0;
byte[] buffer = new byte[BUFFER_SIZE];
while ((size = fin.read(buffer, 0, BUFFER_SIZE)) != -1) {
outStream.write(buffer, 0, size);
totalBytes += size;
}

Avoiding these types of problems is why I start with Commons IO. If that's an option, your code would be as follows.
FileInputStream fin = new FileInputStream(sendFile);
int totalBytes = IOUtils.copy(fin, outStream);
No need reinventing the wheel.

It is possible that the .read() call returns less bytes than you requested. This means you need to use te returnvalue of .read() as argument to the .write() call:
int bytesRead = fin.read(buffer, 0, readBlockSize);
outStream.write(buffer, 0, bytesRead);
apart from this, it is better to pre-allocate a buffer and use it (your could could try to use a 2Gb buffer if your file is large :-))
byte[] buffer = new byte[4096]; // define a constant for this max length
while ((readBlockSize=fin.available())>0) {
if (4096 < readBlockSize) {
readBlockSise = 4096;
}