eantoranz

raid5recovery.txt

I just thought this was interesting--so copying here. Not my words:

In this article I (Edmundo Carmona) will describe an experience I had that began with the failure of some RAID5 disks at the Hospital of Pediatric Especialties, where I work. While I wouldn’t wish such an event on my worst enemy, it was something that made me learn about the power of knowledge—a deep knowledge, which is so important in the hacking culture.

Are you in dire straights with your hard drive?

We at Free Software Magazine had a major hardware crash. The good guys at DTI DATA performed hard drive recovery and saved our magazine!
Friday, April 29, 2005

This article has downloads!

A 5-disk (18GB each) RAID5 was mounted on a HP Netserver Rack Storage/12. Due to a power outage yesterday, it would no longer recognize the RAID. As a matter of fact, there were two more RAIDs on the rack that were recovered... but this one (holding about 60GB of data) just wouldn’t work.

The IT manager decided to call in some “gurus” to try to get the data back on-line. I (the only GNU/Linux user at the IT department) thought that something could be done with GNU/Linux. My first thought was: “If I get images of the separate disks, maybe I can start a software RAID on GNU/Linux. All I need is enough disk space to handle all of the images”. I told my crazy (so far) idea to the IT manager and he decided to give it a try... but only once the gurus gave up.
Monday, May 2, 2005

The gurus are still trying to get the data back on-line.
Tuesday, May 3, 2005

The gurus are still trying to get the data back on-line.
Wednesday, May 4, 2005

These guys are stubborn, aren’t they?
Thursday, May 5, 2005

The IT manager called me late in the afternoon. I was given the chance to Save the Republic. One of the disks of the array had been removed. I put the disks on a computer as separate disks (no RAID), booted with Knoppix (the environment of the IT department is Windows based, apart for my desktop, which has the XP that came with the HP box and Mandriva, which is where the computer normally stays) and made the four images of the four disks left from the original five:

# for i in a b c d; do dd if=/dev/sd$i of=image$i.dat bs=4k; done

I got all the files in a single HD and left the office.
Friday, May 6, 2005

I wanted to start a software RAID, fooling the kernel into thinking that the files where HDs. Just having the images was not enough to bring the RAID on-line. RAID5 has a number of options: algorithm (left/right parity, synchronous/asynchronous), chunk (strip) size, but most important: the order of the images in the RAID. I had to tell the kernel how the RAID controller had mounted them so it could replicate the RAID.

I had already been given the hint that the chunks were 64KB long. By the end of the day, the software RAID idea hadn’t worked at all. I started thinking about rebuilding the data the “hard” way: Making a single image of the RAID from the separate images.
Weekend, May 7 and May 8, 2005

I did some research during the weekend, plus a little study of the images. The images didn’t look encrypted at all. The first “chunk” of the four images looked like garbage, but one of the disks showed a Partition Table right on the second chunk and the other chunks appeared to have other kind of data:

# fdisk -lu discoa1

You must set cylinders.

You can do this from the extra functions menu.



Disk discoa1: 0 MB, 0 bytes

255 heads, 63 sectors/track, 0 cylinders, total 0 sectors

Units = sectors of 1 * 512 = 512 bytes



  Device Boot   Start     End   Blocks  Id System

discoa1p1       63  142175249  71087593+  7 HPFS/NTFS

Partition 1 has different physical/logical endings:

   phys=(1023, 254, 63) logical=(8849, 254, 63)

fdisk was complaining because it was a 64KB file, not the expected 72GB one (written in the partition table). I studied the images and noticed that the data chunks and the parity chunks were distinguishable from each other, and that they seemed to follow a plain RAID5 distribution and algorithm... I was hopeful.
				
Disk 1 	Disk 2 	Disk 3 	Disk 4 	Disk 5
1 	2 	3 	4 	P
5 	6 	7 	P 	8
9 	10 	P 	11 	12
13 	P 	14 	15 	16
P 	17 	18 	19 	20
21 	22 	23 	24 	P
25 	26 	27 	P 	28

Table 1 - RAID5’s chunk disposition (in a 5-disk array)

I made a java class that could rebuild the RAID content from the separate images (Had I used C/C++, I would still be coding!). It was all about placing the right chunk from the right disk (image of disk) at the right place of the final image. I was missing one image, but it could be calculated with the help of the parity chunks spread all over the disks (see Textbox 1). The class was no big deal: selecting the right chunks from the disks, and using XORs to calculate the missing chunks. I guess it took about three or four hours at most to code it. I was finally ready to give it a try. The problem I hit was that while testing the software RAID at home I had damaged the images. So, I have to wait until Monday to test the class with the images of the RAID.

RAID stands for Redundant Array of Independent Disks. All it does is make a number of disks “look” like they are one to improve throughput or fault-tolerance. There are a number of ways to put them together. Some of them are:

Mirroring: in this case, each disk has exactly the same content. Size of the array: the size of the smallest disk. Redundancy: There must be at least a disk working for the data to remain intact.

Linear: one disk follows the other. The size of each disk doesn’t matter at all. Size of the array: the sum of the size of the separate disks. Redundancy: If you remove one disk, you will lose the information on that disk and potentially all of the data in the array.

RAID5: The information is spread in all of the available devices in a manner different from linear. Size of the array: the size of the smallest disk multiplied by the number of the disks minus one. Redundancy: At most one disk can be removed/replaced from the array without data loss. Instead of having disks that follow each other, the information is written in “chunks” of data, one disk at a time (see Table 1).

In Table 1, the numbers represent the order in which the chunks are written on the disks (in this example, it’s left parity, asynchronous). There is a parity chunk per every n – 1 chunks of data. That is done for redundancy.

It works like this: parity is calculated by XORing the n – 1 chunks of data in a row. This logical operator has a very interesting property for redundancy. If you remove one of the data chunks and use the parity chunk instead for the XOR operation, you will get the missing chunk of data:

a xor b xor c xor d xor e = p

If you remove c, then:

a xor b xor d xor e xor p = c

What does this mean for the RAID? It means that if you remove a whole disk from the array, the RAID can still work... though with a little overhead to calculate the missing chunks. Furthermore, if you replace a missing disk with a new one, the data that was in the removed disk can be rewritten to the new disk. There will be no loss of data (provided that no more than a single disk is missing at any given moment).

The process of making a RAID image wasn’t complicated. I started the Java class by telling it the conditions of the run: algorithm, images, order of the disks, chunk size, skipped chunks (remember there were 64KB of garbage at the beginning of every image), and output file.

    I started thinking about rebuilding the data the “hard” way: making a single image of the RAID from the separate images

Monday, May 9 2005

I made some attempts at rebuilding the RAID content. Each try took roughly two or three hours. After a run, I had a RAID.dat file (about 72GB in size) that was the “supposed” image of a HD, just like doing:

# dd if=/dev/hda of=ata.dat

Please notice the lack** **of partition number in the input file (a raw HD block device).

Then I had to use that image as a hard drive. First, I had to use fdisk to know the “partitioning” of the hard drive (it had no problem handling the file at all). At that point, just as I had thought, I discovered that the file was the image of a HD and I could see a partition starting from sector 63. I was more than happy! There were no complaints from fdisk this time. Unfortunately, I can’t give you console output from now on, because the files have already been erased. Instead, I’ll show the commands that were involved:

# fdisk -lu RAID.dat

Then mounting. How could I make the kernel think that this file was a hard drive? Well... it took me some more research to learn that losetup is used to link loop devices to files. It felt like the solution was at hand! I had to link the file to a loop device starting from byte 32256 (I had to skip the first 63 sectors 512 bytes each, according to fdisk):

#losetup -o 32256 /dev/loop0 RAID.dat

It linked, no problem! Then mounted:

#mount -t ntfs /dev/loop0 /mnt/tmp

There was no complaint when mounting. All of the pieces were fitting together after all.

I just forgot to take into consideration one very important factor in the IT world: Murphy’s Law. The RAID was not going to give itself away so easily after all.

When I ran ls, in the mount point, I could see a few of the directories, but the information wasn’t usable. I couldn’t cd to those directories and dmesg said there were problems with the NTFS indexes. I guessed I must have made a mistake ordering the disks... or used the wrong algorithm. I tried twice (with different options), but failed.
Tuesday, May 10 2005

I had left another attempt working when I left the office. That one failed too. I was getting frustrated at the time. Three of the developers at the IT department offered their help and started analysing the whole thing with me. I made another class that rebuilt the missing image, which I felt would help us in the analysis—no matter what the algorithm, order or strip size, according to RAID theory, the missing image’s content would always be the same.

We noticed that I had indeed made a mistake when ordering the disks! (Hey, I can’t always be right, can I?) We studied the images a little further to make sure, and started the whole thing again. It was already getting late, so we had to wait until the next morning to see the results.

    I just forgot to take into consideration one very important factor in the IT world: Murphy’s Law

Wednesday, May 11 2005

First thing in the morning (and I didn’t sleep very well because of the wait), I did a ls and...

Eureka! It worked.

All of the directories were there (otherwise, I wouldn’t have written this article in the first place, right?). I tried to work with some of the files in the partition... and it was perfect.

I suddenly became the spoiled kid of the IT department! I got a big chocolate cake—that’s what I call a bargain!

Even better, the experience caused some of the guys from the IT department to install GNU/Linux on their own personal computers. That’s quite an achievement!
Conclusion

I want to finish saying that I did nothing miraculous... but definitely clever! I certainly used the resources I had at hand... plus Knoppix. I also got a lot of help from the GNU/Linux community (through www.linuxquestions.org mostly). Thank you people!

It’s very important that you make sure backups be made on a regular basis to avoid this kind of situation. I don’t think it’s likely you will find yourself in the same situation we got ourselves into. But, if you do find yourself in the same boat, I hope this information allows you to not lose the data the Microsoft way (just format the disk, and forget about your data). Don’t freak out, get a Knoppix CD (if you can get a GNU/Linux guru along with it, all the better!), and with a little programming you will most likely solve the problem.
Thanks

I’d like to thank Simon Carreno, Heberto Ramos and Javier Machado, for their help in analysing the way the images of the RAID had to be put together. I’d also like to thank the IT crew as a whole for their support.
License

Verbatim copying and distribution of this entire article are permitted worldwide, without royalty, in any medium, provided this notice is preserved.

https://www.linuxquestions.org/questions/linux-kernel-70/kernel-module-development-how-to-open-block-devices-and-use-them-open-seek-read-848766/page3.html



GENERAL
-------
This material I'm including in thsi jar is by no means a bullet proof project. It was intended to solve a
problem we had with a Array of Disks that its HP controller didn't want to recognize.

I'm doing this readme by memory, so there could be a mistake here or there.

The class that does the recovery job is Raid5Redcovery

That class will only get one argument: the name of a properties file having all the information
necessary for a run. For example:
chunk=65536
disks=6
1=./disk1
2=./disk2
4=./disk4
5=./disk5
6=./disk6
algorithm=left asymmetric
skip=2

In this example I just created:
6 disks. Image 3 is missing. Left Asymmetric algorithm and skip the first 2 chunks from all the images.

Recognizing the order of the images and the algorithm.
-----------------------------------------------------
Well... that took a little bit of over thinking to achieve it.

First I saw a Partition Table on one of the first chunks of the images. I saw it because of the classical 55aa signature.
When I used fdisk on that single chunk, I saw it really was a partition table:

# fdisk -lu discoa1
You must set cylinders.
You can do this from the extra functions menu.

Disk discoa1: 0 MB, 0 bytes
255 heads, 63 sectors/track, 0 cylinders, total 0 sectors
Units = sectors of 1 * 512 = 512 bytes

   Device Boot      Start         End      Blocks   Id  System
discoa1p1              63   142175249    71087593+   7  HPFS/NTFS
Partition 1 has different physical/logical endings:
     phys=(1023, 254, 63) logical=(8849, 254, 63)


As that HAS to be the first sector of the disk, then that chunk would have to be the first chunk of the first image (in case
of a left algorithm) or the second image (a right asymmetric algorithm. It can't be a right symmetric, because then the partition
table would be on the SECOND chunk of the first image).

The other thing was learning how to recognize parity chunks from data chunks. If you look at the chunks that I provide with this
zip file, you will see that data chunks they have a structure present for the NTFS file system.

a piece from discoc1:
000001F4   00 00 00 00  00 00 00 00  00 00 19 00  46 49 4C 45  2A 00 03 00  .......... ..FILE*...
00000208   EA 32 60 4E  04 00 00 00  02 00 02 00  30 00 03 00  58 02 00 00  .2`N...... ..0...X...

I had 5 disks running in the array, right? if you look at the chunks, the ones carrying data have this 
FILE* thing exactly at the same position. If you have 4 chunks having FILE* exactly at the same position.....
guess what a parity chunk will say in that position. ;-) You said 0s? Then you got it right. That's how we were 
able to see where the parity chunks were. That will set only two possible positions
for the images (depending on right or left algorithm).

If I had had an even number of disks in the array, it would have been a little more difficult to recognize the parity chunks
because they would have the same FILE* thing in exactly the same position in the parity chunks... We would have been forced to
analize the chunks with a little more attention (and having a missing chunk wouldn't have been of great help ;-))... 
but (thankfully) we didn't.

That leaves you with three possible attempts to "getting it right".

Doing a little more research of the chunks to see if you can discover the secuence of chunks with algorithm x, y or z, would
lead to discovering which of the three is THE algorithm.

Any other questions? email me: eantoranz at gmail dot com

Cheers!



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    under certain conditions; type `show c' for details.

The hypothetical commands `show w' and `show c' should show the appropriate
parts of the General Public License.  Of course, the commands you use may
be called something other than `show w' and `show c'; they could even be
mouse-clicks or menu items--whatever suits your program.

You should also get your employer (if you work as a programmer) or your
school, if any, to sign a "copyright disclaimer" for the program, if
necessary.  Here is a sample; alter the names:

  Yoyodyne, Inc., hereby disclaims all copyright interest in the program
  `Gnomovision' (which makes passes at compilers) written by James Hacker.

  <signature of Ty Coon>, 1 April 1989
  Ty Coon, President of Vice

This General Public License does not permit incorporating your program into
proprietary programs.  If your program is a subroutine library, you may
consider it more useful to permit linking proprietary applications with the
library.  If this is what you want to do, use the GNU Library General
Public License instead of this License.




/*
 * Copyright 2005 Edmundo Carmona
 * 
 * This file is part of the raid recovery project.
 * 
 * This project is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License 
 * as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version.
 * 
 * This project is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty 
 * of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
 * 
 * You should have received a copy of the GNU General Public License along with this project; if not, write to the 
 * Free Software Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA 
 * 
 * Created on May 9, 2005
 *
 */
package org.antix.raid;

import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Properties;

/**
 * @author antoranz
 *  
 */
public class Raid5Recovery {

	private final static int LEFT_ASYMMETRIC = 0;

	private final static int LEFT_SYMMETRIC = 1;

	private final static int RIGHT_SYMMETRIC = 2;

	private final static int RIGHT_ASYMMETRIC = 3;

	private int chunkSize;

	private int algorithm;

	private int disks;

	private FileInputStream[] inputs;

	private FileOutputStream output;

	private Properties properties;

	private Chunk[] chunks;

	private int skipChunks;

	private Raid5Recovery(int chunk, int disks, FileInputStream[] inputs,
			FileOutputStream output, int algorithm, int skipChunks) {
		chunkSize = chunk;
		this.disks = disks;
		this.inputs = inputs;
		this.output = output;
		this.algorithm = algorithm;
		this.skipChunks = skipChunks;
	}

	public static void main(String[] args) {
		Properties properties = new Properties();
		try {
			properties.load(new FileInputStream(args[0]));
		} catch (IOException e) {
			e.printStackTrace();
			System.err.println("Can't get properties due to "
					+ e.getMessage());
			System.exit(0);
		}

		int chunkSize = 0;
		try {
			chunkSize = Integer.parseInt(properties.getProperty("chunk"));
			System.out.println("Chunk size: " + chunkSize + " bytes");
		} catch (Exception e) {
			e.printStackTrace();
			System.err.println("Can't get chunk size due to "
					+ e.getMessage());
			System.exit(0);
		}

		// Hoy many disks are there?
		int disks = 0;
		int missingDisks = 0;
		try {
			disks = Integer.parseInt(properties.getProperty("disks"));
			System.out.println("Number of disks: " + disks);
		} catch (Exception e) {
			e.printStackTrace();
			System.err.println("Can't get the number of disks due to "
					+ e.getMessage());
			System.exit(0);
		}
		if (disks < 3) {
			System.err.println("There can't be less than 3 disks");
			System.exit(0);
		}

		FileInputStream[] readers = new FileInputStream[disks];

		for (int i = 0; i < disks; i++) {
			String disk = properties.getProperty(Integer.toString(i + 1));
			System.out.println("Disk " + (i + 1) + ": " + disk);
			if (disk == null || disk.length() == 0) {
				// This disk is missing
				missingDisks++;
				if (missingDisks > 1) {
					System.err.println("No more than one disk can be missing");
					System.exit(0);
				}
			} else {
				// Let's try to open the file
				try {
					readers[i] = new FileInputStream(disk);
				} catch (IOException e) {
					e.printStackTrace();
					System.err
							.println("Couldn't open specified file due to "
									+ e.getMessage());
					System.exit(0);
				}
			}
		}

		FileOutputStream output = null;
		try {
			String temp = properties.getProperty("output");
			System.out.println("output: " + temp);
			output = new FileOutputStream(temp);
		} catch (IOException e) {
			e.printStackTrace();
			System.err.println("Couldn't open output stream due to "
					+ e.getMessage());
			System.exit(0);
		}

		// algorithm
		int algorithm = LEFT_ASYMMETRIC;
		String temp = properties.getProperty("algoritmh");
		if (temp != null && temp.length() > 0) {
			try {
				temp = temp.toUpperCase();
				if (temp.equals("LEFT ASYMMETRIC")) {
					algorithm = LEFT_ASYMMETRIC;
				} else if (temp.equals("LEFT SYMMETRIC")) {
					algorithm = LEFT_SYMMETRIC;
				} else if (temp.equals("RIGHT SYMMETRIC")){
					algorithm = RIGHT_SYMMETRIC;
				} else if (temp.equals("RIGHT ASYMMETRIC")){
					algorithm = RIGHT_ASYMMETRIC;
				} else {
					System.out.println("Unknown algorithm: " + temp);
					System.exit(0);
				}
			} catch (Exception e) {
				e.printStackTrace();
				System.err.println("Couldn't determine algorithm due to "
						+ e.getMessage());
				System.exit(0);
			}
		}

		System.out.print("Algorithm: ");
		switch (algorithm) {
		case LEFT_ASYMMETRIC:
			System.out.println("Left Asymmetric");
			break;
		case LEFT_SYMMETRIC:
			System.out.println("Left Symmetric");
			break;
		case RIGHT_ASYMMETRIC:
			System.out.println("Right Asymmetric");
			break;
		case RIGHT_SYMMETRIC:
			System.out.println("Right Symmetric");
			break;
		default:
			System.out.println("Unknown (" + temp + ")");
			System.exit(0);
		}

		int skipChunks = 0;

		temp = properties.getProperty("skip");
		if (temp != null && temp.length() > 0) {
			try {
				skipChunks = Integer.parseInt(temp);
				System.out.println("skip " + skipChunks + " chunks");
			} catch (Exception e) {
				System.err
						.println("Don't understand hoy many chunks to skip ("
								+ temp + ")");
				System.exit(0);
			}
		}

		Raid5Recovery recovery = new Raid5Recovery(chunkSize, disks, readers,
				output, algorithm, skipChunks);
		recovery.recover();

		System.out.println("Finished!");

		// TODO close the files
	}

	private void recover() {
		chunks = new Chunk[disks];
		//int chunksRead = 0;

		for (int i = 0; i < disks; i++) {
			if (inputs[i] == null)
				continue;
			Chunk unChunk = new Chunk(chunkSize, inputs[i]);
			chunks[i] = unChunk;
			// I skip the chunks stipulated in the properties file
			for (int j = 0; j < skipChunks; j++) {
				try {
					if (unChunk.read() != chunkSize) {
						System.out
								.println("Not enouch data to read from disk  "
										+ i + 1);
						System.exit(0);
					}
				} catch (IOException e) {
					e.printStackTrace();
					System.err.println("Couldn't read from disk " + i + 1
							+ " due to " + e.getMessage());
					System.exit(0);
				}
			}
		}

		int row = 0;
		boolean cont = true;
		int bytesRead = Integer.MAX_VALUE;
		while (cont) {
			for (int i = 0; i < disks; i++) {
				// Read a chunk from all streams
				Chunk aChunk = chunks[i];
				if (aChunk != null) {
					// read
					try {
						bytesRead = Math.min(bytesRead, aChunk.read());
						// always have to get a full chunk
						cont = cont && bytesRead == chunkSize;
					} catch (IOException e) {
						e.printStackTrace();
						System.out.println("Couldn't read chunk " + row
								+ " from disk " + i + " due to " + e.getMessage());
						System.exit(0);
					}
				}
			}

			// have read all the chunks
			// Now, let's print
			if (bytesRead > 0)
				printRow(row, bytesRead);
			row++;
		}
	}

	/**
	 * Print this row of data (on the output stream)
	 * 
	 * @param row
	 * @param numberOfBytes number of bytes to print from the chunks
	 */
	private void printRow(int row, int numberOfBytes) {

		int parityDisk = 0;
		switch (algorithm) {
		case LEFT_ASYMMETRIC:
		case LEFT_SYMMETRIC:
			parityDisk = disks - row % disks - 1;
			break;
		case RIGHT_ASYMMETRIC:
		case RIGHT_SYMMETRIC:
			parityDisk = row % disks;
			break;
		}

		int from = 0;

		switch (algorithm) {
		case LEFT_SYMMETRIC:
		case RIGHT_SYMMETRIC:
			from = (disks - row % disks) % disks;
		}

		int to = from == 0 ? disks - 1 : from - 1;

		int i = from;
		byte[] data = null;
		boolean cont = true;
		while (cont) {

			// Take the data form the chucnk (if it's not empty)
			if (i != parityDisk) {
				Chunk aChunk = chunks[i];

				if (aChunk != null) {
					data = aChunk.getChunk();
				} else {
					// calculate the data from all teh chunks
					data = new byte[chunkSize];
					// just XORs
					for (int j = 0; j < disks; j++) {
						if (j == i)
							continue; // if it's the same image, skip it (it's empty)
						byte[] temp = chunks[j].getChunk();
						for (int w = 0; w < numberOfBytes; w++) { 
							data[w] ^= temp[w];
						}
					}
				}

				try {
					output.write(data, 0, numberOfBytes);
				} catch (IOException e) {
					e.printStackTrace();
					System.out
							.println("Couldn't write to output file on row "
									+ row
									+ "  of disk "
									+ i
									+ " due to "
									+ e.getMessage());
					System.exit(0);
				}
			} else {
				// If it's a parity disk, we won't "print" it
			}

			if (i == to) {
				cont = false;
			}

			i++;
			if (i >= disks)
				i = 0; // Reached the end. Let's go back
		}
	}

}







--- image recovery



/*
 * Copyright 2005 Edmundo Carmona
 * 
 * This file is part of the raid recovery project.
 * 
 * This project is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License 
 * as published by the Free Software Foundation; either version 2 of the License, or (at your option) any later version.
 * 
 * This project is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty 
 * of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
 * 
 * You should have received a copy of the GNU General Public License along with this project; if not, write to the 
 * Free Software Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA 
 * 
 * Created on May 10, 2005
 *
 */
package org.antix.raid;

import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;

/**
 * @author antoranz
 *  
 */
public class RaidImageRecovery {

	private Chunk[] inputs;

	private FileOutputStream output;

	private RaidImageRecovery(FileInputStream[] inputs, FileOutputStream output) {
		this.inputs = new Chunk[inputs.length];

		for (int i = 0; i < inputs.length; i++) {
			this.inputs[i] = new Chunk(64 * 1024, inputs[i]);
		}

		this.output = output;
	}

	/**
	 * Execution of this class.<p>
	 * 
	 * As arguments, the user has to specify the filenames of all the images
	 * that the user has and the name of the file for the missing image last.<p>
	 * 
	 * Example:<p>
	 * RaidImageRecovery ./image1 ./image2 ./image4 ./missingimage
	 * 
	 * @param args
	 */
	public static void main(String[] args) {
		// open all the files, the output is where the data will be written to

		// there can't be less than 3 disks

		if (args.length < 3) {
			System.out.println("Not enough disks");
			System.exit(0);
		}

		FileInputStream inputs[] = new FileInputStream[args.length - 1];

		for (int i = 0; i < args.length - 1; i++) {
			System.out.println("Input " + i + 1 + ": " + args[0]);
			try {
				inputs[i] = new FileInputStream(args[i]);
			} catch (IOException e) {
				e.printStackTrace();
				System.out.println("Couldn't open file due to "
						+ e.getMessage());
				System.exit(0);
			}
		}

		FileOutputStream output = null;
		try {
			System.out.println("Output: " + args[args.length - 1]);
			output = new FileOutputStream(args[args.length - 1]);
		} catch (IOException e) {
			e.printStackTrace();
			System.out.println("Couldn't open output file due to "
					+ e.getMessage());
			System.exit(0);
		}

		RaidImageRecovery raidRecovery = new RaidImageRecovery(inputs, output);
		try {
			raidRecovery.recover();
		} catch (IOException e) {
			e.printStackTrace();
			System.out.println("Couldn't complete recovery due to "
					+ e.getMessage());
			System.exit(0);
		}

	}

	private void recover() throws IOException {

		// let's read all inputs, until there's nothing left
		boolean cont = true;
		while (cont) {
			// read all input streams
			int bytesRead = Integer.MAX_VALUE;
			for (int i = 0; i < inputs.length; i++) {
				bytesRead = Math.min(inputs[i].read(), bytesRead);
				cont = cont && bytesRead != -1;
			}

			// calculate the missing chunk
			if (cont) {
				byte[] data = new byte[64 * 1024];
				// There's data left
				for (int i = 0; i < inputs.length; i++) {
					byte[] temp = inputs[i].getChunk();
					for (int j = 0; j < temp.length; j++) {
						data[j] ^= temp[j];
					}
				}

				output.write(data, 0, bytesRead);
			}
		}
	}
}