Develop Reference

Getting a wrong FileChannel size

I'm trying to get the size of a file contained in assets.
I'm using a FileChannel because I need a FileChannel later.
The file myfile.txt contains 7 bytes.
Here is my code:
AssetManager amgr;
AssetFileDescriptor afd;
FileChannel fchIn;
FileInputStream fis;
amgr=context.getAssets();
afd=amgr.openFd("myfile.txt");
fis=afd.createInputStream();
fchIn=fis.getChannel();
Log.d("mytag", fchIn.size());
Log.d("mytag", fis.available());
And the output is:
7237492
7
Why is the size returned by the FileChannel.size() method wrong ?
Thanks for your help

FileInputStream.getChannel() documentation says it "Returns a read-only FileChannel that shares its position with this stream." You are assuming that the channel begins and ends exactly at the boundaries of your file, which is the only way its total size() would match what you expect, but the documentation does not make that guarantee.
Also note that FileInputStream.available() is not documented to mean the same thing as total size -- technically, it is the amount of data available to read without needing to load/buffer more from the source. Unless the entire source has already been read into memory, it may have almost nothing to do with the actual file size.
I expect that the underlying FileChannel has access to a large range of bytes for multiple files, like all of your assets combined (and hence the large size), and you'd need to reference the AssetFileDescriptor's getStartOffset() and getLength() methods to know the actual position of the associated file's bytes within the channel. My guess is that AssetFileDescriptor.createInputStream() is giving you an object that already knows the position and range and takes care of that for you. If you're going to use the raw channel, then honor the information in the file descriptor. Also note that, per the documentation, if you move the position of the FileInputStream (e.g. by reading bytes) then you also move the position of the channel at the same time, so be careful if you're trying to use both.

Avoid obtaining same InputStream more than once

I can see there are a number of posts regarding reuse InputStream. I understand InputStream is a one-time thing and cannot be reused.
However, I have a use case like this:
I have downloaded the file from DropBox by obtaining the DropBoxInputStream using the DropBox's Java SDK. I then need to upload the file to another system by passing the InputStream. However, as part of the download, I have to provide the MD5 of the file. So I have to read the file from the stream before uploading the file. Because the DropBoxInputStream I received can only be used once, I have to get another DropBoxInputStream after I have calculated the MD5 and before uploading the file. The procedure is like:
Get first DropBoxInputStream
Read from the DropBoxInputStream and calculate MD5
Get the second DropBoxInputStream
Upload file using the MD5 and the second DropBoxInputStream.
I am thinking that, if there are many way for me to "cache" or "backup" the InputStream before I calculate the MD5 so that I can save step 3 of obtaining the same DropBoxInputStream again?
Many thanks
EDIT:
Sorry I missed some information.
What I am currently doing is that I use a MD5DigestOutputStream to calculate MD5. I stream data across the MD5DigestOutputStream and save them locally as a temp file. Once the data goes through the MD5DigestOutputStream, it will calculate the MD5.
I then call a third party library to upload the file using the calculated md5 and a FileInputStream which reads from the temp file.
However, this requires huge disk space sometime and I want to remove the needs to use temp file. The library I use only accepts a MD5 and InputStream. This means I have to calculate the MD5 on my end. My plan is to use my MD5DigestOutputStream to write data to /dev/null (not keeping the file) so that I can calculate theMD5, and get the InputStream from DropBox again and pass that to the library I use. I assume the library will be able to get the file directly from DropBox without the need for me to cache the file either in the memory of at the disk. Will it work?

Input streams aren't really designed for creating copies or re-using, they're specifically for situations where you don't want to read off into a byte array and use array operations on that (this is especially useful when the whole array isn't available, as in, for e.g. socket comunication). You could buffer up into a byte array, which is the process of reading sections from the stream into a byte array buffer until you have enough information.
But that's unnecessary for calculating an md5. Notice that InputStream is abstract, so it needs be implemented in an extended class. It has many implementations- GZIPInputStream, fileinputstream etc. These are, in design pattern speak, decorators of the IO stream: they add extra functionality to the abstract base IO classes. For example, GZIPInputStream gzips up the stream.
So, what you need is a stream to do this for md5. There is, joyfully, a well documented similar thing: see this answer. So you should just be able to pass your dropbox input stream (as it will be itself an input stream) to create a new DigestInputStream, and then you can both take the md5 and continue to read as before.
Worried about type casting? The idea with decorators in Java is that, since the InputStream base class interfaces all the methods and 'beef' you need to do your IO, there's no harm in passing instances of objects inheriting from InputStream in the constructor of each stream implementation, and you can still do the same core IO.
Finally, I should probably answer your actual question- say you still want to "cache" or "backup" the stream anyway? Well, you could just write it to a byte array. This is well documented, but can become faff when your streams get more complicated. Alternatively, try looking at a PushbackInputStream. Here, you can easily write a function to read off n bytes, perform and operation on them, and then restore them to the stream. Generally good to avoid these implementations of streams in Java, as it's bad for memory use, but no worse than buffering everything up which you'd otherwise have to do.
Or, of course, I would have a go with DigestInputStream.
Hope this helps,
Best.

You don't need to open a new InputStream from DropBox.
Once you have read the file from DropBox, you have it locally. So it is either in memory (in a byte array) or you stored it in a local file. Now you can create an InputStream that reads the data from memory (ByteArrayInputStream) or disk (FileInputStream) in order to upload the file.
So instead of caching the InputStream (which you can't) you cache the contents (which you can).

Java: Handling large files in input streams

I have a Java app that fetches a relatively small .zip file using a URL, saves it in a temp directory, unzips it onto the local machine and makes some changes to one of the files. This all works great.
However, I am accessing the .zip file via a BufferedInputStream in the following way:
Url url = "http://somedomain.com/file.zip";
InputStream is = new BufferedInputStream(url.openStream(), 1024);
My concern is that this app will actually be used to transfer very large zip files and I was wondering if a BufferedInputStream is actually the best way to do this, or whether I would just end up throwing some type of OutOfMemoryException?
So my question is, will a BufferedInputStream be suitable for this job, or should I be going about it in a completely different way?

BufferedInputStream doesn't load all the file into memory, it only uses an internal buffer, in your case of size 1024 bytes = 1kb. It never gets larger than that. You could actually increase the value if you aren't going to have many streams at once.
Edit: what you are thinking about maybe is a ByteArrayOutputStream, where data is saved in memory.

It depends on what you do with the content you read. If you read everything in memory, it will fail. If you write it to another stream, then it will be fine. Use BufferedInputStream

From the official Java Tutorials - Buffered Streams:
The Java platform implements buffered I/O streams. Buffered input
streams read data from a memory area known as a buffer; the native
input API is called only when the buffer is empty. Similarly, buffered
output streams write data to a buffer, and the native output API is
called only when the buffer is full.
There is another great SUN article.
So the answer is: BufferedInputStream is suitable for this kind of job in the sense of performance.
And yes, the memory consumption isn't so much dependent on the type of the input stream....

Reading a gz file and keeping track of position in file

So, here is the situation:
I have to read big .gz archives (GBs) and kind of "index" them to later on be able to retrieve specific pieces using random access.
In other words, I wish to read the archive line by line, and be able to get the specific location in the file for any such line. (so that I can jump directly to these specific locations upon request). (PS: ...and it's UTF-8 so we cannot assume 1 byte == 1 char.)
So, basically, what I just need is a BufferedReader which keeps track of its location in the file. However, this doesn't seem to exist.
Is there anything available or do I have to roll my own?
A few additional comments:
I cannot use BufferedReader directly since the file location corresponds to what has been buffered so far. In other words, a multiple of the internal buffer size instead of the line location.
I cannot use InputStreamReader directly for performance reasons. Unbuffered would be way to slow, and, btw, lacks convenience methods to read lines.
I cannot use RandomAccessFile since 1. it's zipped, and 2. RandomAccessFile uses "modified" UTF-8
I guess the best would be use a kind of of buffered reader keeping track of file location and buffer offset ...but this sounds quite cumbersome. But maybe I missed something. Perhaps there is already something existing to do that, to read files line by lines and keep track of location (even if zipped).
Thanks for tips,
Arnaud

I think jzran could be pretty much what you're looking for:
It's a Java library based on the
zran.c sample from zlib.
You can preprocess a large gzip
archive, producing an "index" that can
be used for random read access.
You can balance between index size and
access speed.

What you are looking for is called mark(), markSupported() and skip().
This methods are declared both in InputStream and Reader, so you are welcome to use them.

GZIP compression does not support seeking. Previous data blocks are needed to build compression tables...

Java loading binary files

Please show me the best/fast methods for:
1) Loading very small binary files into memory. For example icons;
2) Loading/reading very big binary files of size 512Mb+. Maybe i must use memory-mapped IO?
3) Your common choice when you do not want to think about size/speed but must do only thing: read all bytes into memory?
Thank you!!!
P.S. Sorry for maybe trivial question. Please do not close it;)
P.S.2. Mirror of analog question for C#;

For memory mapped files, java has a nio package: Memory Mapped Files
Check out byte stream class for small files:Byte Stream
Check out buffered I/O for larger files: Buffered Stream

The simplest way to read a small file into memory is:
// Make a file object from the path name
File file=new File("mypath");
// Find the size
int size=file.length();
// Create a buffer big enough to hold the file
byte[] contents=new byte[size];
// Create an input stream from the file object
FileInputStream in=new FileInutStream(file);
// Read it all
in.read(contents);
// Close the file
in.close();
In real life you'd need some try/catch blocks in case of I/O errors.
If you're reading a big file, I would strongly suggest NOT reading it all into memory at one time if it can possibly be avoided. Read it and process it in chunks. It's a very rare application that really needs to hold a 500MB file in memory all at once.
There is no such thing as memory-mapped I/O in Java. If that's what you need to do, you'd just have to create a really big byte array.