Lab Assignment 3:  The Buffer Bomb


This assignment helps you develop a detailed understanding of the calling stack organization on an x86 processor.  It involves applying a series of buffer overflow attacks on an executable file. 

Note:  In this lab, you will gain firsthand experience with one of the methods commonly used to exploit security weaknesses in operating systems and network servers.  Our purpose is to help you learn about the run-time operation of programs and to understand the nature of this form of security weakness so that you can avoid it when you write system code.  We vigorously condemn the use of these or any other form of attack to gain unauthorized access to any system resources.  There are criminal statutes governing such activities. 

Location of the Bomb

To obtain a bomb, point your browser at  (This will work only for browsers running on a csug machine.)  You’ll end up downloading a tar file which, when unpacked, will give you a set of three programs: 

Generates a “cookie” based on your team name. 
The code you will attack. 
A utility to help convert between string formats. 
All of these programs are compiled to run on the csug machines. 

Team Name and Cookie

You should create a team name for the one or two people in your group of the following form: 

where name is your username, if you are working alone, or
where name1 is the username of the first team member and name2 is the username of the second team member. 
You should choose a consistent ordering of the IDs in the second form of team name.  Teams “bob+jane” and “jane+bob” are considered distinct.  You must follow this scheme for generating your team name (with the plus sign and no spaces!)  Our grading program will only give credit to those people whose username can be extracted from the team names.

A cookie is a string of eight hexadecimal digits that is (with high probability) unique to your team.  You can generate your cookie with the makecookie program giving your team name as the argument.  For example: 

  unix> makecookie bob+jane
In three of your four buffer attacks, your objective will be to make your cookie show up in places where it ordinarily would not. 

The bufbomb Program

The bufbomb program reads a string from standard input with a function getbuf having the following C code: 

  int getbuf()
      char buf[NORMAL_BUFFER_SIZE];
      return 1;
The function Gets is similar to the standard library function gets.  It reads a string from standard input (terminated by ‘\n’ or end-of-file) and stores it (along with a null terminator) at the specified destination.  In this code, the destination is an array buf having sufficient space for 32 characters. 

Neither Gets nor gets has any way to determine whether there is enough space at the destination to store the entire string.  Instead, they simply copy the entire string, possibly overrunning the bounds of the storage allocated at the destination. 

If the string typed by the user to getbuf is no more than 31 characters long, it is clear that getbuf will return 1, as shown by the following execution example: 

  unix> bufbomb -u bovik
  Type string:  howdy doody
  Dud: getbuf returned 0x1
If we type a longer string, typically an error occurs:
  unix> bufbomb -u bovik
  Type string: This string is too long
  Ouch!: You caused a segmentation fault!
As the error message indicates, overrunning the buffer typically causes the program state to be corrupted, leading to a memory access error.  Your task is to be more clever with the strings you feed bufbomb so that it does more interesting things.  These are called exploit strings. 

Note:  This version of the lab has been specially modified to defeat the stack randomization techniques used by newer versions of Linux.  It works by using mmap() and a bit of in-line assembly code to move the stack pointed at by %esp to an otherwise unused part of your address space.  You may need to use gdb to figure out where that is. 

Bufbomb takes several command line arguments: 

Operate the bomb for the indicated team.  You should always provide this argument for several reasons: 
Print list of possible command line arguments
Operate in “Nitro” mode, as is used in Level 4 below. 
Submit your solution exploit string to the grading server. 
Your exploit strings will typically contain byte values that do not correspond to the ASCII values for printing characters.  The program hex2raw can help you generate these raw strings.  It takes as input a hex-formatted string.  In this format, each byte value is represented by two hex digits.  For example, the string “012345” could be entered in hex format as “30 31 32 33 34 35” since the ASCII code for decimal digit 0 is 0x30 and so forth. 

The hex characters you pass hex2raw should be separated by whitespace (blanks or newlines).  I recommend separating different parts of your exploit string with newlines while you're working on it.  hex2raw also supports C-style block comments, so you can mark off sections of your exploit string.  For example:

bf 66 7b 32 78 /* mov    $0x78327b66,%edi */
Be sure to leave space around both the starting and ending comment strings (/*, */) so they will be properly ignored. 

If you place a hex-formatted exploit string in the file exploit.txt, you can apply the raw string to bufbomb in at least two different ways: 

  1. You can set up a series of pipes to pass the string through hex2raw
      unix> cat exploit.txt | ./hex2raw | ./bufbomb -u bovik
  2. You can store the raw string in a file and use I/O redirection to supply it to bufbomb
      unix> ./hex2raw < exploit.txt > exploit-raw.txt
      unix> ./bufbomb -u bovik < exploit-raw.txt
This second approach can also be used when running bufbomb from within gdb
  unix> gdb bufbomb
  (gdb) run -u bovik < exploit-raw.txt
One important point:  your exploit string must not contain byte value 0x0A at any intermediate position, since this is the ASCII code for newline (‘\n’).  When Gets encounters this byte, it will assume you intended to terminate the string. 

When you have correctly solved one of the levels, say level 0:

    ../hex2raw < smoke-bovik.txt | ../bufbomb -u bovik 
    Userid: bovik
    Cookie: 0x1005b2b7
    Type string:Smoke!: You called smoke()
then you can submit your solution to the grading server using the -s option:
    ./hex2raw < smoke-bovik.txt | ./bufbomb -u bovik -s
    Userid: bovik
    Cookie: 0x1005b2b7
    Type string:Smoke!: You called smoke()
    Sent exploit string to server to be validated.
The server will test your exploit string to make sure it really works, and it will update the Buffer Lab scoreboard page indicating that your userid (listed by your cookie for anonimity) has completed this level. 

You can view the scoreboard by pointing your browser at

Unlike the bomb lab, there is no penalty for making mistakes in this lab.  Feel free to fire away at bufbomb with any string you like. 


Levels may be done in any order. 


As noted above, you may work in a group of up to 2 people. 

Any clarifications and revisions to the assignment will be posted to the Blackboard forum

Hand-in occurs automatically whenever you correctly solve a level and use the -s option.  Upon receiving your solution, the server will validate your string and update the Buffer Lab scoreboard Web page, which you can view by pointing your Web browser at  You should be sure to check this page after your submission to make sure your string has been validated.  (If you really solved the level, your string should be valid.) 

Note that each level is graded individually.  You do not need to do them in the specified order, but you will get credit only for the levels for which the server receives a valid message. 

Have fun!

“Trivia” Assignment

Before noon, Tuesday, Feb. 18, send email to containing answers to the following questions (a single email per team is acceptable).  Please use the subject

    [cs252] Assignment 3 Trivia - uname1, uname2
for your email. 
  1. Run the program makecookie with your team name as a parameter.  What is the output? 
  2. What is the purpose of the hex2raw program? 
  3. What is the very first instruction executed at the beginning of most x86 functions?  What is its purpose? 
  4. Where is the return address of a calling function stored relative to the callee function’s frame pointer? 
  5. Where is the return value of a function stored when control returns to the calling function? 

Generating Byte Codes

Using gcc as an assembler and objdump as a disassembler makes it convenient to generate the byte codes for instruction sequences.  For example, suppose we write a file example.s containing the following assembly code: 

  # Example of hand-generated assembly code
        pushl $0x89abcdef       # Push value onto stack
        addl $17,%eax           # Add 17 to %eax
        .align 4                # Following will be aligned on multiple of 4
        .long    0xfedcba98     # A 4-byte constant
        .long    0x00000000     # Padding

The code can contain a mixture of instructions and data.  Anything to the right of a ‘#’ character is a comment.  We have added an extra word of all 0s to work around a shortcoming in objdump to be described shortly. 

We can now assemble and disassemble this file: 

    unix> gcc -c example.s
    unix> objdump -d example.o > example.d

The generated file example.d contains the following lines

    0: 68 ef cd ab 89 push $0x89abcdef
    5: 83 c0 11       add  $0x11,%eax
    8: 98             cwtl              Objdump tries to interpret
    9: ba dc fe 00 00 mov  $0xfedc,%edx these as instructions

Each line shows a single instruction.  The number on the left indicates the starting address (starting with 0), while the hex digits after the ':' character indicate the byte codes for the instruction.  Thus, we can see that the instruction pushl $0x89ABCDEF has hex-formatted byte code 68 ef cd ab 89

Starting at address 8, the disassembler gets confused.  It tries to interpret the bytes in the file example.o as instructions, but these bytes actually correspond to data.  Note, however, that if we read off the 4 bytes starting at address 8 we get:  98 ba dc fe.  This is a byte-reversed version of the data word 0xFEDCBA98.  This byte reversal represents the proper way to supply the bytes as a string, since a little endian machine lists the least significant byte first.  Note also that it only generated two of the four bytes at the end with value 00.  Had we not added this padding, objdump gets even more confused and does not emit all of the bytes we want. 

Finally, we can read off the byte sequence for our code (omitting the final 0s) as: 

    68 ef cd ab 89 83 c0 11 98 ba dc fe


There is no explicit turn-in.  The bomb will notify us automatically after you have successfully defused it with the -s option.  You can keep track of how you (and the other groups) are doing by looking at

This web page is updated continuously to show the progress of each group. 


For the “trivia” assignment:  11am, Tuesday, February 18. 

For the main assignment:  11:59pm, Friday, February 28. 

Last Change: 21 February 2014 / Michael Scott's email address