[SOLVED] CS6035 binary exploitation steps/scripts 2025 100%

Step 0:Welcome to the Binary Exploitation (BinExp) project for CS6035! We’re excited to have you with us for this effort.Binary exploitation is a really interesting and challenging domain within cybersecurity. It rests at the intersection of many sub-disciplines, including reverse engineering, low-level programming, operating systems, code review, etc. As such, you’ll be expected to draw upon a variety of subjects matter in approaching and working through the challenges of this project.Step 1:Before diving in, let’s ensure that our project work environment is appropriately configured.Below is a brief summary of the project’s contents:https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/00_intro.html      1/2 1/27/25, 12:54 PM             Flag 00 – Intro | CS 6035In all of the projects’ binaries, your goal is to have the binary read from /proc/flag! Most of the time, this is via a system() call like: system(“cat /proc/flag”);However, sometimes it’s not that simple – carefully analyze your source code within each challenge to figure out how the binary is meant to read from /proc/flag.Step 2:Enter into /home/binexp/binexp/00_intro and run the flag binary located there. Note the GTID it outputs to verify that the binary is reading the correct value from step 1, above. If so, copy the supplied hash and write it to the appropriate value in project_binexp.json.Finally, try submitting the project_binexp.json file to Gradescope to verify that the hash is correct. Remember, you have unlimited submission attempts for this project, so be sure to submit early and often.NOTE: Still confused? Try heading over to the corresponding Ed Discussion megathread and see if someone else asked a similar question. If not, pose your question there and a TA will respond as soon as they are able.(watch the intro video first please)This is a test on using GDB to read a few things that would be difficult to calculate (you can if you want to), but the main point of this task is to try running a few things in gdb and get comfortable with it first.cd ~/binexp/01_bb_steps/ gdb flagyou will be met with a pwndbg prompt which waits for you to run the program (or set breakpoints, etc) which we’ll do now: b mainsets a breakpoint at the first executable statement of the main() function. rruns the program and you will be met with some text/blocks showing various things like stack traces, register printouts, code prints, etc. Now next you will want to run the STEP command to go into the functionsAfter that, you will notice we are now in the bb_steps() function and can traverse all the code with the NEXT commandnWhich will execute each following instruction. This is useful if you don’t want to go into every single function call, and rather want to just execute those calls/instructions. https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/01_bb_steps.html              1/2 Pay attention to the register window which will show each register value as the instructions are executed.At the end of the ASM instructions, you will be prompted to enter in the answers for the two registers RBX and R15. You can enter these into GDB if you want to, however to get the official/valid flag for submission you will have to save your answers for the two registers, and then enter them into the non-debug binary run, e.g.:binexp@cs6035:~/binexp/01_bb_steps$ ./flag What value is currently in RBX?: 0xdeadbeefUpon correctly answering the questions, you will see your flag printed out, which you can copy into the json file! https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/01_bb_steps.html                                                                      2/2 (watch the intro video first please)This task is a very simple buffer overflow that, upon inspection, will check if a variable is non-zero. Using the information you have gathered from reading and the videos, it is your task to get this program to get to the call_me() function, and get the flag printed.Here are some steps I suggest you follow for the remainder of the project, that will set yourself up for success when debugging/analyzing/reverse engineering the binaries.Note: you are free to use GDB if you need to for this project but you need to run the program on the command line (i.e. ./e.py) in order to get the real flag for submission and submit it! https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/01_buffer_overflow_1.html                                                        1/1 In this task you will learn details about binaries compiled from C code (with gcc) in a Linux environment, and how some basic things can be exploited such as process redirection or control flow hijacking. The steps in this flag are discussed in-depth in the intro video.In this directory you have an executable binary named ‘flag’ which is vulnerable to a buffer overflow in one of its functions. We will be using an exploitation library called pwntools to automate some of the overflow techniques and get the binary to call a function it otherwise wouldn’t have. This function called ‘call_me’ generates a key using your Gradescope User ID to get a valid flag that will pass the autograder.Now we will run the binary just to see what the program is doing by running the executableWe see the binary is asking for a string, input any text you want or just press enter, and you’ll see that the program does nothing and just exits. That’s just to simplify the code so we can focus on the exploit.The binary is statically linked to a shared object which has a lot of methods that construct the key and has a simple function called ‘call_me()’ which will print out your key.This is where we will start learning about binary file formats. Without going into a deep dive about program structure, operating systems, compilers, assembly language, machine code, etc. you will still be able to understand that there are two aspects that are key in binary exploitationThese are logical locations the computer understands.A buffer overflow occurs when too much data is fed into an unprotected (or poorly protected) data buffer. The way that 64-bit C programs work is, a few bytes before the start of the stack frame (this is the function entry point), there is an address stored in memory called the Return Address. This Return Address is read into a register upon function return (when the function ends and intends to return to its caller), andthen process redirection to that address happens. If we override this location with another valid address, we can manipulate the control flow of the program and have it execute arbitrary (or otherwise unintended) code, with a well-formed attack. Starting off easy, we are going to modify e.py and learn a few basics of the pwntools library, which will build up into a successful attack at the end.Open e.py with your favorite text editor and analyze the content and comments.Once you understand what they do, proceed to fill in the cyclic size (this number is up to you, based on your understanding of the program and what would break it) to get a segmentation fault message by running$ ./e.py dbgThis will open up a gdb terminal with a breakpoint set at main()Type ‘c’ to continue from the breakpointWe see the program received an interrupt signal for a SEGMENTATION FAULT (SIGSEV, or an invalid access to memory). This happens when the program tries to access memory at a certain location that it either isn’t allowed to access, or doesn’t exist. In this case the return address for the function was overwritten by cyclic()’s data in the form of long strings of characters. Pay attention to the bottom of the screenshot where the instruction pointer is currently trying to ‘ret’ (return) to 0x6561……616b which is just a string of ascii characters in hexadecimal form. Now we know how to break the binary, let’s figure out how to purposefully break it. Using a pwntools method called‘cyclic_find()’ we enter in the bottom 32 bits (4 bytes) of the return string (in this example is 0x6561616b) which will give the number of characters before reaching that value. This is important because we are now going to reach our first step of control flow hijacking by overflowing enough data that we can place a value and change the course of the program’s normal path.In e.py, on the commented line below your cyclic command, we are now going to use cyclic_find() which will automate our buffer length calculation, and feed that number into cyclic(). Place in your 4 character bytes (preceded by a 0x, like 0x6561616b). Uncomment either of the lines beneath our original cyclic() call (one uses hex value and the other uses the ASCII values), and fill in the hex or ascii value described above This will fill the buffer with our calculated buffer length, appended by the ASCII byte equivalent of the variable by using another pwntools method p64(number). After you have done that, rerun./e.py dbgAnd hit ‘c’If done correctly, you should see something like this screenshot, where if you check the ‘ret’ instruction, we are now failing on an invalid access to our dummy address.Stepping away from the pwntools library for a moment, we now need to find something usable within the binary that will allow us to actually call a function or do something other than just crashing the program.Now we will use a linux command ‘objdump’ which takes a binary file and will output a dump of the binary which will give some key information about the binary. The -D flag will output binary addresses, machine code, and assembly code of the binary into a file. objdump -D flag > flag.asmThen open flag.asmYou will see a bunch of (likely) confusing information that at a high level translates to the code that you can see in the ‘flag.c’ file. You aren’t going to have to go through this file in any extreme expanse (unless you want to?) we are just going to focus on finding an address within the binary file that holds the machine code responsible for making a function call to ‘call_me()’.Search for the string ‘call_me’ in flag.asm and keep looking until you find the assembly instruction:call <some address> <call_me>Note down the highlighted address showing the call (it will be different in your binary):With the hexadecimal value of the address above (prepend 0x to the value highlighted)Now run ./e.py again from the command line (without dbg) and check the terminal output.Did you get it? Awesome! Submit your first flag to gradescope (follow APPENDIX for more details) If not, retrace your steps in this task and also make sure you used the call call_me address in the earlier step and not the address of the actual function call_me() This flag shows a communication between a server and a client. The client binary (flag) will send data to the server, and the server appends some (very conveniently structured) data to that message and sends it back to the client. Your goal for this task is to have the server return the ideal data to overwrite the instruction pointer with the data that is returned from the server.Follow the same steps in previous tasks (buffer_overflow_2, more specifically) to break the program in gdb, and then figure out your buffer size, and try to fill in the response to correctly hit this function call!If you use the pwntools e.py file, it will start the server for you so there is no need to explicitly start the server.If you are running the program on the command line to experiment, then you must start the server each time you run the binary. You can either open a new terminal, and run _./server Or in the same terminal, each time you run the binary, run./server &Your task is to figure out the breaking point, and heavily inspect the last bytes that are returned from the server in order to get the right return and get the flag! This task will get you to determine which series of assembly instructions will direct the code flow into constructing a call to call_me(). Analyze the different instructions and look up the usage/behavior of them to figure out which combination will get you there.There is a small twist, however:In your earlier buffer overflow exercises, you made a concerted effort to jump precisely into the address of a particular instruction. This is for good reason; generally speaking, jumping into an instruction mid-way is not advised as it can result in unexpected behaviors. However – as may often be the case in exploit development – unexpected is not necessarily undesirable (and may even potentially work out in our favor!). While we should try and avoid “splitting” instructions where possible, we shouldn’t rule out testing such outcomes.You can use objdump or gdb to find the address of call_me() and figure out how you calculate it.For debugging, we highly recommend using gdb, setting a breakpoint on the gadget function, and stepping through the options once you think you know the correct path to get to the function call.(FYI:: you don’t have to use pwntools for this one)https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/02_assemble_the_assembly.html      1/2 1/27/25, 12:54 PM Flag 02 – Assemble | CS 6035https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/02_assemble_the_assembly.html                                              2/2 This Program (very conveniently) leaks out part of the libc base address this address is randomized via ASLR so it will change a little bit every time the program is launched run the program a few times and notice what bytes are different and which ones aren’tNext step will be analyzing the C file and see what we are comparing against in order to get to call_me – system() is a libc function, use GDB to get the address of system using command p system –run ./flag multiple times, it will ask you for input and your goal is to guess an address. Put in any random guess and try it a few times to see if you can notice a pattern versus what is leaked and what is being expected.Fortunately there’s only one byte that is missing from our formula, so we can do some scripting in python to try out the remaining values.Your task is going to be: – get the value leaked from the program – modify it with the offset of the system() function – fill in the remaining byte with a random value – send to the process – (repeat until you get a flag) note: i recommend using recvall() after you send in each payload, and write your loop logic around the output (see other flags for what kind of string output you can expect) to see if you got the right value! https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/02_bad_rand0.html 1/2https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/02_bad_rand0.html                                                                   2/2Task 02_p4s5w0rdSTRINGS!Now it’s time to learn a really useful technique to find all the available strings in a program.And by strings, we mean any collection of printable characters that exist in the binary. So things like variable names, hardcoded paths, debug messages, or eeeeevenn…. passwords? Hopefully not in a real program but you would be surprised.This binary has zero debugging information and you do not have the source code available, but guess what? The program is written terribly and is very unsafe, with passwords stored in plain text that can easily be dumped/searched in the binary!I would recommend running the program once or twice to see what it’s doing (checking a series of responses to questions) and if you get every question right, then you will get the flag!To get the strings for the program, run the command:$ strings flagThis will output it all to the terminal which isn’t super helpful, so would suggest redirecting output to a file like:$ strings flag > flag_strNow you will be able to grep/search/navigate the file in a new terminal and will (hopefully) be able to figure out what the correct responses would be for the given questions.(hint, strings are stored in the binary in the order that they’re written in the C code, might be a good idea to search for the questions they’re asking and it should be pretty easy to determine the answer from there!) Good luck!Disclaimer: You are responsible for the information on this website. The content is subject to change at any time.https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/02_p4s5w0rd.html                                                                     2/2Time to rev up those Reverse Engineering motors, because you need to unravel the logic that this program is checking against in order to get to the call_me() function!No buffer overflow this time, you just ‘simply’ need to input the right values that will correctly decode the logic and pass the checks.If you’re unfamiliar with C operators, this TUTORIAL has all the necessary operations detailed.Suggest pen and paper for this one to work through the logic by hand, or do a ton of experimentation to get the right value!XOR TRUTH TABLEXOR Operations are REVERSIBLE, meaning that performing the same XOR operation on a number twice will end up with the original number!Example:0b11110000 ^ 0b00001111 = 0b11111111=>0b11111111 ^ 0b00001111 = 0b11110000https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/03_XORbius.html              1/2 1/27/25, 12:55 PM             Flag 03 – XORbius | CS 6035 in Hex notation0xF0 ^ 0x0F = 0xFF=>0xFF ^ 0x0F = 0xF0Some students have likewise found it helpful to try and break down some of the elements of the binary within their own toy code; below is an example of such an effort using Programmiz.com:Disclaimer: You are responsible for the information on this website. The content is subject to change at any time.https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/03_XORbius.html                                                                      2/2 We see there is an unsafe() function which has some checks for different local variables. The positioning of these variables is important because they are declared before the input buffer which means that a buffer overflow will cause data to be overwritten.This program is a Buffer Overflow, however you will not be changing the control flow to a specific binary address (i.e. overwriting ret), rather you will need to enter in the right values to trick the pointer arithmetic logic and get to the call_me() function.Some students have likewise found it helpful to try and break down some of the elements of the binary within their own toy code; below is an example of such an effort using Programmiz.com:Disclaimer: You are responsible for the information on this website. The content is subject to change at any time.https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/03_pointy_pointy_point.html                                                      1/1You’ve made it! You are now on your final task. In this directory is the entire contents of /usr/bin, a collection of binary files that make up a lot of common linux uses. One of these files has been overwritten by our vulnerable flag program. It is your task to figure out which one.NOTE: just to make it explicitly clear, you will not find a binary named “flag” in this directory. There IS a binary compiled from the flag.c source code present, but it is renamed to something else, overwriting/replacing that binary.To help sort out what’s going on, we’ve supplied you with a checksums file. Inside the checksums file is – as you might guess – a list of known good checksums generated by the linux shasum command. You should leverage this checksums file to figure out which binary present is the vulnerable one; you are free to do this however you would like.NOTE: we suggest skipping over the ‘checksums’ file itself (as well as any additional files you might introduce during your hunt).Once you find the file it is time to begin our exploit. This is a bit more complex than the other flags and will require a full ROP (return oriented programming) exploit to chain calls together, and we will also need a new tool called Ropper to find a ‘gadget’ in order to supply a function argument and pass a specific check.In 64-bit programs, the function gets arguments through registers, in the case of intel architecture the RDI register supplies the first function argument.So we need to find a gadget (a piece of code that we can override the instruction pointer with, that will perform a certain action and then continue with the control flow hijack) that will pop a value from the stack into the RDI register.Let’s use ropper like this$ ropper –file flag | grep ‘pop’https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/03_hunt_then_rop.html      1/2 1/27/25, 12:55 PM             Flag 03 – Hunt/ROP | CS 6035This will give you all gadgets within the binary that have a keyword ‘pop’ (spoiler, there’s a LOT of them). An objective for this task is to figure out what gadget will likely work best to get the required argument passed into the function you are trying to call. This Writeup is a helpful reference to understand how calling convention works for x86_64 cpu’sNote the addresses that are output for each gadget. Once you find a gadget you think will work, we will need that as our first override value in pwntoolsPictorially, this is what our crafted exploit needs to look like (remember stack grows down)Now we will need to supply the argument, which will be on the stack immediately after our pop gadget, figure out what that value needs to be, and add it as p64() after the pop gadgetThen we need to put the address of the function as the next call, use objdump or gdb to find the addresses (you should probably get the second function address while you’re at it). The call to our pop gadget will ‘ret’ and then hit this second function call to enter one of the unsafe functionsFinally, we need to finish our execution chain by calling the second function which will allow for exploitation. Append that address to your chain and see if you get a flag!https://github.gatech.edu/pages/cs6035-tools/cs6035-tools.github.io/Projects/BinExp/03_hunt_then_rop.html                                                              2/2