Computer Science homework help. Department of Computer Science and Information Technology
La Trobe University
CSE1OOF/4OOF Semester 1, 2020
Assignment Part B
First and Final date for SUBMISSION Tuesday 12th May at 10.00 am
Delays caused by computer downtime cannot be accepted as a valid reason for a late submission
without penalty. Students must plan their work to allow for both scheduled and unscheduled
downtime. There are no days late or extensions on this assignment as execution test marking will
begin on the 13th
This is an individual Assignment. You are not permitted to work as a Pair Programming partnership
or any other group when writing this assignment.
Copying, Plagiarism: Plagiarism is the submission of somebody else’s work in a manner that gives the
impression that the work is your own. The Department of Computer Science and Information
Technology treats academic misconduct seriously. When it is detected, penalties are strictly imposed.
Refer to the subject guide for further information and strategies you can use to avoid a charge of
Submission Details: Full instructions on how to submit electronic copies of your source code files
from your latcs8 account are given at the end. If you have not been able to complete a program that
compiles and executes containing all functionality, then you should submit a program that compiles
and executes with as much functionality as you have completed. (You may comment out code that
does not compile.)
Note you must submit electronic copies of your source code files using the submit command on
latcs8. Ensure you submit the required file. For example, the file HexEditor.java would be submitted
with the command:
> submit OOF HexEditor.java
PLEASE NOTE: While you are free to develop the code for this progress check on any operating
system, your solution must run on the latcs8 system.
Marking Scheme: This assignment is worth 10% of your final mark in this subject.
Implementation (Execution of code) 90%, explanation of code 10%
You may be required to attend a viva voce (verbal) assessment
(to be used as a weighting factor on the final mark).
Do NOT use the LMS to submit your files, use latcs8 only
OOF Assignment Part B – due: 10:00 am Tues 12th May this is the first and final hand in date p. 2 of 22
Return of Mark sheets:
The face to face execution test marking in the lab (constitutes a return of the assignment mark,
subject to the conditions on the last page of this document.
Please note carefully:
The submit server will close exactly 10:00 am on the due date.
After the submit server has closed, NO assignments can be accepted.
Please make sure that you have submitted all your assignment files before
the submit server closes.
There can be NO extensions or exceptions.
Your assignment will be marked in your normal lab just after the due date.
If you cannot attend the lab you have signed up for on Allocate, please
email email@example.com to arrange another time.
90% for the code executing correctly
10%, you will be asked to explain (and / or alter) parts of your code.
You must attend the lab you have signed up for on Allocate.
Non-attendance at the lab you have signed up for on Allocate will
also result in your assignment being awarded 0, except as detailed
OOF Assignment Part B – due: 10:00 am Tues 12th May this is the first and final hand in date p. 3 of 22
Please also note carefully that whilst we encourage innovation and exploring java beyond what has
been presented in the subject to date, above all, we encourage understanding.
Code and techniques that are outside the material presented will not be examined, of course.
You are free to solve the Tasks below in any way, with one condition.
Any assignment that uses code that is outside what has been presented to this point must be fully
explained at the marking execution test. Not being able to fully explain code outside what has been
presented in the subject so far will result in the assignment being awarded a mark of 0, regardless of
the correctness of the program.
Submitting an assignment with code outside what has been presented so far and not attending the
marking execution test will result in an automatic mark of 0, regardless of the correctness of the
All assignments in OOF are marked, in the lab, in an execution test. This
means that we mark running code. Your code must compile and display a
result to the screen. Regrettably, we don’t have the time or resources to
look at code. The smallest amount of code that produces and displays a
correct result will gain more marks than lots of code that doesn’t compile,
run or display something to the screen.
Using code not taught in OOF – READ THIS
How this assignment is marked
OOF Assignment Part B – due: 10:00 am Tues 12th May this is the first and final hand in date p. 4 of 22
Working with files.
It is important for any good programmer to understand the contents of a file and how files work.
So far in OOF we have looked at text files, that is files that can be opened in Vim or Notepad.
However, most files you encounter are not text files, they are binary files. Think about things like PNG
ZIP or MP3, none of these can open opened in Notepad or Vim.
The .class files that javac produces are an example of a binary file. Try opening them in Vim, are they
nice to read?
Since you cannot view binary files using a text editor, we need a special tool to look at the internals of
a file. This is known as a Hex Editor and is a valuable tool for any programmer.
Below is an example of a Hex Editor for windows.
Notice that it has two panels, the left side is the bytes of the file in hexadecimal. The right side is the
same data but shown as ASCII characters. You may notice that many of the bytes are not valid ASCII
characters and are replaced with placeholder characters or spaces.
The far-left panel is the address of the first byte of each row.
The highlighted row “60” indicates that the first byte of that row (value A7) has address 60.
Remember though that all these numbers are in hexadecimal, so “60” in hex is equal to “96” in
decimal. Typically, programmers will prefix a number with “0x” to indicate that it is a hex number.
The highlighted cell (value “2F”) is at address 0x64, remember we count from zero.
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You need to implement a basic hex editor in Java that will run inside the PuTTY window.
A hex editor is a tool that you use throughout your programming career, you may find yourself using
this tool long after you finish OOF. So, take the opportunity to design it well.
Your hex editor will be able to:
1) Open a file of any size and not crash.
2) Show both hex and ASCII representation of the data in a neat well formatted grid.
3) Be able to jump to a given address using a command.
4) Be able to get and set the value of a byte at a given address using a command.
When you run the program, you will provide it with the name of a file.
Then the program will show the first 256 bytes of the file, then prompt the user for a command.
The valid commands will be:
1) “exit” to exit the program.
2) “goto” to select a specific byte.
3) “jump” to select a byte relative to the currently selected byte.
4) “set” to set a byte
5) “hex” to switch to hex view.
6) “ascii” to switch to ASCII view (more on this later)
7) “truncate” to delete all bytes after the selected byte.
8) “next” to move to the next page of bytes
9) “previous” to move to the previous page of bytes.
10) “find” to find a string in the data.
Tip: only look at the first letter of the commands so that the user can use “e” for exit.
String.startsWith() may be handy for this.
The program should produce a message at the bottom of the screen saying what the last command
did, for example if the user used “ascii” the program should say “Switched to ASCII Mode”
How do we even begin to do this?
First you need to have some understanding on how files work and how numbers are represented.
I have provided some reading on the next few pages.
→ Please read the entire assignment before you start.
I have also provided some sample files in CSILIB. Use this command.
Note that there is a dot at the end of this command
> cp /home/student/csilib/cse1oof/hexapp/*.dat .
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Background: What is a signed & unsigned integer?
Computers work entirely in binary (1 and 0 only) so there is no room for storing the minus sign.
So how do we store negative numbers using only 1s and 0s?
The obvious way to do it would be to just use the left most bit to indicate a negative or positive
Which is what computers do, but with some extra steps, this is called two’s complement encoding.
Imagine if you wanted to encode the number -4 into an 8-bit binary number.
First write out 4 in binary using 8 bits like so 0000 0100
Then, flip all the bits 1111 1011
Then, plus 1 1111 1100 <- This is what a -4 byte looks like
This seems more complicated than it needed to be but there is a reason for it.
Consider what would happen if we added the number binary number 1111 1100 (negative 4) to the
number 0000 1010 (positive 10)
With two’s complement you don’t ever need to do subtraction, just convert the number to negative
and use addition. -4 + 10 = 6.
How do you know if a binary number is a two’s complement encoded negative number or just a
normal binary number? The answer is: you don’t.
If you are given a binary value like 1111 1100 it could be -4 or it could be 252.
It depends on if you interpret the number as signed, or unsigned.
When working with signed numbers, just remember this:
If the left most bit is a 1, then the number is negative, if it’s a zero, its positive.
In Java everything is signed by default (in the above example it would output -4)
Try running this code to see for yourself.
for(int i=0; i < 256; i++)
System.out.printf(” Integer: %d, Byte: %d\n”, i, (byte)i);
You will see two columns with identical numbers, until about halfway.
The divergence starts just after 127, why?
Consider what the number 128 looks like in a bye vs an int.
0000 0000 0000 0000 0000 0000 1000 0000 32 bit integer of 128
1000 0000 8 bit byte of 128
The value is identical, but ask yourself “what is the left most bit”
Remember: If the left most bit is a 1, then the number is negative.
1111 1100 Binary of -4 (Two’s complement)
+ 0000 1010 Binary of 10
Wait, that’s 9 bits, remove the extra bit.
This is the binary value for 6
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Background: Random Access Files.
The Scanner class is useless for a hex editor application, because it only reads forwards.
You need a way to jump to any random location in a file.
Read the API documentation for RandomAccessFile and File
The methods in RandomAccessFile we are interested in are:
➢ int read()
➢ void write(int)
➢ int length()
➢ void seek(long)
When working with files it is important to know that internally there is a “file cursor” which is an
integer that represents your current position within the file. Each time you call read() or write() the
file cursor moves forwards.
Here is an example:
File myFile = new File(“a.dat”);
RandomAccessFile raf = new RandomAccessFile(myFile, “rw”);
int byte0 = raf.read(); // read first byte and move cursor forwards.
int byte1 = raf.read(); // read second byte and move cursor forwards.
int byte2 = raf.read(); // read third byte and move cursor forwards.
raf.seek(0); // move cursor back to the start of the file.
The “rw” String indicates that we want to open the file for both reading and writing.
We can use seek() to move to any address within the file, even if the file is 10TiB in size.
We will only need to read small sections of the file at a time.
You may be wondering, if read() returns a byte, why is the return type an integer?
This is mainly done for convenience, using an integer allows us to avoid dealing with negative valued
bytes and allows the API to return special values like -1 to indicate the end of the file has been
reached. This method will always return a value between 0 and 255, or -1 indicate end of file.
Also notice that seek takes a long, consider for a moment that an int is only 32 bits long.
A 32 bit number ranges from -2147483648 to 2147483647. How many gigabytes is that?
Since you cannot have negative file size, the maximum file size we could support with 32 bits is
2147483647 bytes, or 2GiB.
You should use longs instead of ints when working with the file length to allow for large files.
Note that when you write an integer literal in Java it defaults to an Integer.
This is an issue if you are working with big numbers, so if you need a long literal you put an L on the
end like so: long a = 2147483647L;
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Background: What exactly does 2GiB mean?
What is a GiB?
It’s a “Gibibyte” which is the technically correct way of
referring to 1073741824 bytes.
The metric prefix “kilo” means 1000 since the metric system
is based around powers of 10.
However, computers work in powers of 2, so a “kilobyte” was
defined as 1024 byes, not 1000.
This causes immense confusion; some software uses 1000
others use 1024 for measuring sizes.
So, to avoid confusion we will use “Kibibytes” which are
defined in powers of 2 like so.
1KiB = 210
1MiB = 220
1GiB = 230
Create a class HexEditor and create a method
public static void repeat(String str, int count)
it will print out the string str count times. For example
repeat(“hello”,3) -> “hellohellohello”
This will come in handy later.
Create a main() method and test your code.
Create a static method String formatSize(long sizeInBytes)
This method needs to return the most appropriate representation for the provided file size (in bytes)
to a human readable format with at least 2 decimal places (except for bytes).
2147483647 2.00 GiB
123 123 bytes
83647 81.69 KiB
9585631 9.14 MiB
188900977659375 171.80 TiB
Make sure you toughly test this method.
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Task 3 – Create the user interface
Consider how the interface will look, for now lest assume that we will have 16 bytes per row with 16
rows on screen at a time, all in hexadecimal. Tip: instead of hard coding the value “16” everywhere,
use a variable instead so that you can change it later.
Use some for loops and System.out.printf() to construct the following output, make sure to use the
padding feature of printf() to that numbers like “1” get padded to “ 1“ the extra spaces will make it
look neater later.
How do I convert an int/long to a hex string?
You could use printf like so: System.out.printf(“%X”, 1234);
Or you could use String str = Long.toHexString(1234);
Make sure to include a header row and column with a hex count.
Put this code into a method, name it something like “printHexTable”.
It should take parameters, the RandomAccessFile object, and a “rownumber” variable.
The row number will indicate what the first row will be, so for example if I passed in a rownum of 7
the output should look like this
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Task 4 – Using command line args.
Consider the following Java program
public class MyProgram
public static void main(String args)
System.out.println(“Number of args: “+args.length);
System.out.println(“First Arg: “+args);
System.out.println(“Second Arg: “+args);
When you run this program from the command line you can pass it command line arguments like so
> java MyProgram hello world
Number of args: 2
First Arg: hello
Second Arg: world
As you can see you can access the command line arguments “hello” and “world” using the args
Edit your program to accept a file name as a command line argument.
Do not ask the user for a file name using Scanner!
Make sure to check that the user has provided the correct number of arguments, if the user provides
the wrong number of arguments you must show an error that says what the program does, and what
the correct argument format is.
You must also check if the file exists before creating the RandomAccessFile you can do this using the
File.exists() method. If the file does not exist, ask the user if they would like to create it, if they say
yes, then use File.createNewFIle() otherwise, exit.
Do not assume anything about the name of the file, it could be anything and contain any characters.
Do not hard code the name of the file.
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Task 5 – Populating the table.
Once you have the table drawing nicely, we can think about populating it with data.
The first thing you need to do is seek() to the correct location. If we wanted to start from row 7, then
we would need to seek to position 7*16 = 112.
Once you have used seek() you can call read() and printf() the value inside the for loops, the bytes
should be zero padded two characters ie “00” to “FF”
Note that read() returns -1 if you try to read past the end of the file, you should detect this and just
print out some spaces.
At this point you will want to have some files to read so I have provided some use this command to
copy them from CSILIB. Note that there is a dot at the end of this command.
> cp /home/student/csilib/cse1oof/hexapp/*.dat .
Try viewing “a.dat” in your program. you should get something like this
The file is less than 256 bytes long, which is why it stops part way.
You must also show the size of the file in both hex and decimal at the bottom of the screen.
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Task 6 – Unicode Box Characters
In the example above I have just used basic ASCII characters to draw the borders of the table.
If you look closely you can see that it’s not very nice, there are gaps between the dashes and pipes.
It would be nice if they were slightly longer, so it could form a continuous line.
Thankfully, there are specialised Unicode characters for this purpose.
Consider the difference between
————– vs ────────────
The left is just normal HYPHEN-MINUS (0x2D).
The right is BOX DRAWINGS LIGHT HORIZONTAL (0x2500).
You can get quite fancy with these, like so.
║ Hello OOF ║
║ test 123 ║
Note that these only work in monospaced fonts, such as courier new which is what PuTTY uses.
Your task is to make the grid look nicer by using box chatacters of your choosing.
You can just copy and paste the characters into the Java string literals from the wiki page.
Use right click to paste into PuTTY.
If you are an ASCII purist who objects to having any non-ASCII characters in their source code for fear
of weird encoding problems, you may also use the hex escape sequence like so.
Check out some websites that have Unicode information such as
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Task 7 – Interactive Design
When the program starts, display the hex data shown in Task 3 and then use Scanner.nextLine() to
read input from the user’s keyboard.
If the user enters any non-valid command, show a list of valid commands.
The program must continuously display the bytes and then ask for a command using a while loop.
Only when the user enters the exit command should the program terminate.
The goto command must function as follows
“goto 1234 h” means go to byte at address hexadecimal 1234.
“goto 1234 d” means go to byte at address decimal 1234.
“goto -1234 d” means go to byte at address (file size -1234) decimal.
“goto -1234 h” means go to byte at address (file size -1234) hexadecimal.
If the user did not provide a “h” or “d” suffix, assume they meant hexadecimal.
The jump command allows moving the selection forwards or backwards relative to the currently
selected byte. “jump 15” or “jump 15 h” means just forwards 0x15 bytes. “jump 15 d” means jump
forwards 15 decimal bytes. Negative numbers move backwards.
You must calculate what the rownum needs to be so that you can pass it to the draw method.
The selected byte should appear near the middle of the screen, basically you should always be able to
see the bytes before and after the selected byte. Make sure to adjust the row accordingly.
When using the next and previous page commands, make sure that the selected byte is always on
screen. You should keep it in the same relative position on the screen.
Show the selected byte by putting brackets around it.
At the bottom of the screen show the following data:
➔ The address of the byte in hex, decimal, and formatted using formatSize()
➔ The value of the byte in hex (0 to FF), unsigned decimal (0 to 255) and signed decimal (-127 to 128)
➔ The size of the file in hex, decimal, and formatted using formatSize()
You can omit formatSize if the size is less than 1024 bytes.
Make sure you support selecting bytes that are beyond the end of the file.
When the selected byte is changed, make sure to redraw the table.
Do not allow the user to select an address that is negative.
The program must say what the previous command did for example “jump 14” should say “jumped
0x14 bytes forward”
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Task 8 – The set & truncate command.
The set command will accept ether “set 12 d” for decimal, or “set 3A h” for hexadecimal
and will write that byte to the currently selected address. The default is hex mode.
The user may also enter a single quoted ASCII character like “set ‘a'”
Values that are over 255 (0xFF) are not allowed.
The set command needs to seek() to the correct location and use the write() method to set the byte.
Note that write() saves the change immediately to the file. You must also move the currently
selected byte forward by 1.
The user can select and set a byte beyond the end of the file, when you write a byte to such a
location the OS will automatically backfill the intervening bytes with zeros.
The “truncate” command will just call the setLength() method in RandomAccessFile
Converting Strings to Integers
All the user input should be taken as a string since it may contain hex characters.
To convert a string to an integer use Integer.parseInt(String) like so:
String myStr = “123”;
int myInt = Integer.parseInt(myStr);
This also works with hex, but you need to remove the 0x part and add an extra argument.
String myStr = “1ABC”;
int myInt = Integer.parseInt(myStr, 16);
Your program needs to support large (>2GiB) files, which means the address may be bigger than the
maximum value an integer can store. You will need to use longs instead.
String myStr = “1ABC”;
long myInt = Long.parseLong(myStr, 16);
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Task 9 – ASCII Mode
When this option is selected the bytes of the file will be shown as characters rather than hex
numbers. The program will stay in ASCII mode until the user changes back to hex mode.
However, care must be taken to not confuse PuTTY, if you just print out control characters like 0x0A
(newline) or 0x09 (tab) you will cause the formatting of your grid to break.
To deal with this, all control characters will not be printed and will instead be replaced by a symbol.
The characters you use to replace them with are up to you, but they must be visually distinct from
normal ASCII characters. And work in PuTTY. Things like line feeds could be shown as “LF” or “\n”
You must also show the space character (0x20) visually. So that it can be seen.
Use the Unicode 0x25CF ● character since it is visually distinct from all other characters.
Most of the control characters are below 0x20, but don’t forget about 0x7F.
Make sure to assign a character for 0x00.
You should move the logic for this into a separate method to keep the draw method short.
The file b.dat contains every possible ASCII value, use this file to check that none of the bytes cause
PuTTY to misbehave.
Task 10 – Read Only Files.
You may encounter a file that you do not have write access to.
If you attempt to open such as file in “rw” mode, then your program will crash.
The File class has a method boolean canWrite()
Use this method to determine if the file is writable before you create the RandomAccessFile.
If it is, then open the file in “rw” mode, else open in “r” mode and prevent the user from using the set
or truncate command.
The program should display “READ ONLY” for files that are read only.
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Task 11 – Large file support
If your program is using longs instead of ints for seek() etc the you should have no problem with a file
over 2GiB in size.
There is a file located at /home/student/csilib/cse1oof/hexapp/bigfile.zip which is more than 2GiB.
Open it in your hex editor (do not copy it, it’s too big)
> java HexEditor /home/student/csilib/cse1oof/hexapp/bigfile.zip
Once you have it open, try to goto position 0x80000000 (2GiB) see if your program handles it.
Then try 0x86666666 (2.1GiB) and see if it still works.
Try going to a location like 0xFFFFFF, you may notice that your string padding
alignment on the left column didn’t work so well.
This is caused by the fact you used something like printf(“%4X”) to do the alignment
which fails once you get more than 4 digits.
Instead of hard coding a number such as “4”, you should calculate how many digits
you will need and use the repeat() method you crated in task 1 to output the correct number of
It should expand as required.
Pro-tip: To calculate the number of spaces you need before you start printing the table. You could do
some math magic with log() or you could be lazy and convert the last row header to a hex string and
get the length of that string.
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Task 12 – Search System
When working with large files it is helpful to be able to find interesting parts.
The “find” command will take an ASCII string and try to find it in the file.
This command will not support Unicode characters.
How to do this:
First when you take the users command, they could enter “find hello world”
You need to remove the command part, so it’s just “hello world”
Create a method for searching that takes the RandomAccess file, string to search, and the currently
selected byte address. Also, the search should be case sensitive.
The search process should start from the byte after the currently selected byte, so use seek() to move
here. Form there use a loop to move though the file and check to see if the byte from the file
matches the first character of the search (in our case “h”’) if it matches, try to compare the second
character “e” if that matches then try the third “l” and so on.
If you find all the characters matching, then move the selected byte to the first letter of the match.
If not, leave the selected byte where it was and tell the user that the search failed.
Remember that read() returns -1 when it hits the end of the file. Show an error if you reach the end
of the file and were unable to find the search string.
If the user enters the find command with no string, rerun the previous search again.
If there was no previous search, show an error.
Optional Bonus 1
It is annoying to need to type in long address like 0x80000000
Make you program support entering commands like “goto 1GiB”
It should support decimals, like “goto 2.4GiB” and support shorthand like “g 3.1m” for 3.1MiB
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Optional Bonus 2 (hard)
Many binary files will need to store integer values which are bigger than 255.
In Java the integer 1234 would be stored in memory as 4 separate bytes (00, 00, 04, D2)
This encoding is called “Big Endian” because the most significant byte is first.
There is another style “Little Endian” which stores the bytes backwards (D2, 04, 00, 00)
Little endian is the “normal” way to do it, (homework: investigate why this is).
You task is to add 2 commands to your program
1) “be X” where X is a number, switch to big endian of size x.
2) “le X” which X is a number, switch to little endian of size x.
The value for X can be between 1 and 8.
Once the user selects a mode the program must stay in that mode until the user changes the mode.
Byte mode is the default normal mode where you can select 1 byte at a time.
For modes with X greater than 1, you need to allow the user to select X bytes at at time.
For example, “goto 0x23” then “BE 2” should show this =>
At the bottom of the screen the big-endian value for a 2-byte
integer should be shown in both hex and decimal. In this
example the value of (68,20) in big endian is 0x6820 which is
26656 in decimal.
If the user were to enter the command “LE 2” then the value for (68,20) in little endian is 0x2068 or
8296 in decimal. And this would be shown at the bottom of the screen.
This must work for all values of X between 1 and 8.
Obviously selecting BE1 or LE1 is the same thing (normal single byte mode).
The “set” command must also be able to support the new values. For example, if the program is in LE
2 mode then the maximum value that can be set is 0xFFFF or 65535.
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Task 13 – Using the tool
I have provided you with a compiled C program (not source code) named “hackme.elf”
> cp /home/student/csilib/cse1oof/hexapp/hackme.elf .
This program checks to see if your username matches “hackerman1337” and if so, it allows you entry.
If your username does not match, you get an access denied message.
Try running it yourself.
You need to edit the program to gain access.
Since you don’t have the source code you can’t just recompile it.
The program must have the phrase “hackerman1337” in it somewhere.
Open the file in your hex editor and try to locate the string “hackerman1337”
If your search system is working this should be easy.
Once you have located it, change it to your username, then run the program.
Some background about C strings.
In the C programming language, strings don’t have a length. Instead they mark the end of a string
using a NUL byte (0x00) so when you edit the string make sure to put a 0x00 at the end.
Submit the modified elf file when you submit your assignment.
Consider some extra features that you could add to your hex editor such as:
➢ Search for a UTF16 & UTF8 strings.
➢ Open and file and append the contents of a second file to the end of the first.
➢ Remove bytes from the front or middle of the file.
➢ Move bytes from one part of the file to another.
Change the width to something other than 16
OOF Assignment Part B – due: 10:00 am Tues 12th May this is the first and final hand in date p. 22 of 22
Electronic Submission of the Source Code
• Submit all the Java files that you have developed in the tasks above.
• The code must run under Unix on the latcs8 machine.
• You submit your file from your latcs8 account. Make sure you are in the same directory as the file
you are submitting. Submit the file using the submit command.
submit OOF HexEditor.java
submit OOF hackme.elf
After submitting the files, you can run the following command that lists the files submitted from your
You can submit the same filename as many times as you like before the assignment deadline; the
previously submitted copy will be replaced by the latest one.
Please make sure that you have read page 2 about the submission close off date and time and the
compulsory requirement to attend the execution test.
Failure to do both things will result in your assignment be awarded a mark of 0, regardless of the
correctness of the program.
Execution test marks are provisional and subject to final plagiarism checks and checks on the
compliance of your code to this assignment document.
As such final assignment marks may be lower or withdrawn completely.
Don’t let anyone look at your code and don’t give anyone a copy of your code. The plagiarism checker
will pick up that the assignments are the same and you will both get 0 (it doesn’t matter if you can
explain all your code).
And a final, final, final note, “eat the dragon in little bits”. Do a little bit every night; before you know
it you will be finished. The assignment is marked with running code, so you are better to have 2 or 3
parts completed that compile and run, rather than a whole lot of code that doesn’t compile.
The execution test is done on latcs8 so please make sure that your code
runs on latcs8 before you submit.