The majority of attackers do not focus much on crypto-type attacks, as it is time consuming and requires significant computing power to crack the cipher text to extract meaningful information. But in some cases, the logic of the cryptography implemented can be understood easily.

In this section, we will explore bit-flipping attacks, which use Cipher Block Chain (CBC) to encrypt the given plaintext. In CBC, before you encrypt a block, the plaintext will be XOR'ed with the encrypted output of the previous block by creating a logical chain of blocks, as shown in the following screenshot:

In a nutshell, XOR compares two values, and returns true if they are different.

What is the potential attack scenario here? If anyone can XOR the plaintext block with the encrypted message from the previous block, what would be the XOR input for the first block? All you need is an initialization vector. Access mutillidae by navigating to OWASP 2017 > A1 - Injection (Other) > CBC bit flipping:

http://192.168.0.101/mutillidae/index.php?page=view-user-privilege-level.php&iv=6bc24fc1ab650b25b4114e93a98f1eba

Testers should be able to land on the following page:

As we can see, the current app user is running with User ID 100 and Group ID 100. You need to be user 000 in group 000 to become the highly privileged root user.

The only thing we need to manipulate is the IV value, 6bc24fc1ab650b25b4114e93a98f1eba. As it is hexadecimal and 32 characters long, the length is 128 bits. We start assessing the initialization vector by splitting the value into two characters as a block and change the value in the URL by accessing them one by one:

• http://192.168.0.101/mutillidae/index.php?page=view-user-privilege-level.php&iv=00c24fc1ab650b25b4114e93a98f1eba: No change to the User or Group ID
• http://192.168.0.101/mutillidae/index.php?page=view-user-privilege-level.php&iv=6b004fc1ab650b25b4114e93a98f1eba: No change to the User or Group ID

When we get to the fifth block, 6bc24fc100650b25b4114e93a98f1eba, we see a change in the User ID, as shown in the following screenshot:

Testers can utilize Python to generate the hex value for us, as shown here. We will XOR the value to give us the result, 000:

>>> print hex(0XAB ^ 0X31)
0x9a
>>> print hex(0X9A ^ 0X31)
0xab
>>> print hex(0X9A ^ 0X30)
0xaa

To become root user, both Group ID and User ID need to be 000, so we repeat the same on all the blocks until the value changes. Finally, we get the eighth block, 6bc24fc1ab650b14b4114e93a98f1eba, which changed the group ID; now, we do the same as we did for the User ID:

root@kali:~# python
Type "help", "copyright", "credits" or "license" for more information
>>> print hex(0X25 ^ 0X31)
0x14
>>> print hex(0X14 ^ 0X30)
0x24
>>> exit()

This gives us the following key: 6bc24fc1aa650b24b4114e93a98f1eba. When you pass the IV with the new value, you should now gain access to the application with enhanced privileges, as shown in the following screenshot: