ECE9609 - assignment 2, PicoCTF buffer overflow challenges

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Assignmnet 2

Starting from assignment 2, I'll using a kali linux VM to do the homework.


Question 1 - File Permissions

Following the steps:

Daddy told me about cool MD5 hash collision today. I wanna do something like that too!

ssh col@pwnable.kr -p2222 (pw:guest)

Login pwnable's server through port 2222:

image-20210202172730531

Checking the file permissions:

image-20210202172846806

The answers to the questions:

  1. Which user owns the file col?

    col_pwn owns the file col.

  2. Which files in this directory can the users in the group col read from?

    col and col.c can be read by the users in group col. Current user is in the group col , so the current user can r-x, which are reading and executing to the file. col.c has been set that everyone is readable(r--).

  3. What does the SUID flag do?

    SUID is a type of special permissions, files with SUID permission could be excuted as if the owner is excuting it. s indicates the file has SUID permission.

    In my opinion, SUID is normally used by the command, for example: while a none root user change its password, he need to use command passwd, which has SUID permission to modify the password file(/etc/passwd).

    image-20210202175109643

  4. What exactly does -r-sr-x--- tell us about the file col? Be sure to explain who is allowed to do what.

    in -r-sr-x---, the first - tells us it is a file.

    For the owner of the file: r-s is the permissions the owner has, means read and SUID, allow other user execute it with the owner's permission.

    For the users in group col: they have r-x permission, which is read and excute.

    For current logged in user col: The current user is in the group col, so he can read and excute the file.

    For other users(root not included): no permission allowed

The following pictures I found at http://www.csit.parkland.edu/~smauney/csc128/permissions_and_links.html and https://pamirwebhost.com/check-linux-file-permissions-with-ls/ are very helpful with leaning the linux file permissions:

image-20210202182208298

image-20210202182304797

Question 2 - Basics of C

Steps:

Create the source file by using vi flag.c and paste the code into the file:

image-20210202181036605

Then save the file (:wq):

image-20210202181124322

Compile the source code by gcc flag.c -O ex_flag, compile and save the binary to ex_flag and finally, execute it:

image-20210202181518796

So for the questions:

  1. What is the flag?

    The program outputs Here's your HINT, and the flag = HINT.

  2. What command(s) did you use to compile and run this program?

    I used gcc to compile the source code. After compiling it, the output file will automatically has the execute permission, so I execute using ./theFileName.

Question 3 - Basics of Computer Memory

What I understand about memory:

  1. in a X86 32-bit system, every process has a 32-bit long virtual address and it has an offset compared to the physical address.

  2. in a X86 32-bit system, virtual address space is allocated to two part, kernel space and user space. Kernel space normally takes up to 1G space at high address, the rest is user space which is 3G.

  3. Each process has the following memory space:

    image-20210203012733285

    Source: https://www.cnblogs.com/chaozhu/p/6377485.html

    On linux system, we can take a look at the memory map:

    image-20210203013259642

And at the bottum of the memory is stack:

image-20210203013354359

What I understand about endianness:

Endianness defines how a operating system read memory, from high address to low address or the oppsite order.

Answers to the questions:

  1. What is the number 3735928559 in hexadecimal form?

    0xDEADBEEF (I used an online converter)

  2. Suppose this number was stored as an integer (i.e., int type) in little-endian format at memory address 0x12345678. Fill in the following memory map showing where each byte is stored. If the value is unknown/not relevant, leave it as 0x??.

    First of all, little-endian means the last byte of the data is stored at the lowest address. In terms of 0xDEADBEEF, the least significant is EF (the last). Because the number is stored at memory address 0x12345678, the start of the memory is 0x12345678 (lowest address). So, EF should be placed at address 0x12345678 .

    Address     | Value
    -------------------------
    ...         |
    0x12345674  | 0x??
    0x12345675  | 0x??
    0x12345676  | 0x??
    0x12345677  | 0x??
    0x12345678  | 0xEF
    0x12345679  | 0xBE
    0x1234567a  | 0xAD
    0x1234567b  | 0xDE
    

Question 4 - Collision Challenge

First login the server:

ssh col@pwnable.kr -p2222

After login, I checked the directory using ls, I find that there's a file called flag, it's possible where the flag is. My mission is to access the file. Then I checked flag file's permission using ls -l .

image-20210203022031664

So, flag is owned by col_pwn. Before I tried other stuff, I checked if my current user is on the sudoer list using sudo cat flag.

image-20210203022229551

But I'm not a sudoer. I have a new idea while dealing with sudo, on January 26, 2021, someone found sudo has a buffer overflow bug, which is exploitable by any user on the machine. So, I look this up on Google and find this article https://www.sudo.ws/alerts/unescape_overflow.html. Here's a test to check if the machine has the bug:

To test whether your version of sudo is vulnerable, the following command can be used:

sudoedit -s /

A vulnerable version of sudo will either prompt for a password or display an error similar to:

sudoedit: /: not a regular file

A patched version of sudo will simply display a usage statement, for example:

usage: sudoedit [-AknS] [-a type] [-C num] [-c class] [-D directory] [-g group]
                [-h host] [-p prompt] [-R directory] [-T timeout] [-u user]
                file ...

If the sudoers plugin has been patched but the sudo front-end has not, the following error will be displayed:

sudoedit: invalid mode flags from sudo front end: 0x20002

Here's what i find:

image-20210203030309869

And the OS has been patched.

Next, I opened col.c to see the code using vi col.c :

image-20210203024316057

So, the question became how to make 0x21DD09EC equal to check_password(INPUT).

After reviewing the check_password() function, it takes the char input, force convert to int, copy it to the location of pointer ip, then add them together like numbers. From the notice, I know the input is 20-bytes long. And one int takes up 4 bytes, so it's 20/4=5 numbers I need know.

Now I convert 0x21DD09EC to decimal(568134124) to figure out the 5 numbers. However, 568134124 cannot be divided by 5.

image-20210203031433211

So, I just take the integer part 113626824 and mutiply it with 4 (I also could set the first 4 number as 113626825, the rest is the same) , 113626824 * 4 = 454507296. Now I have 4 numbers, the rest value would be the fifth number. 568134124 - 454507296 = 113626828. Check if this is correct, 113626824 * 4 + 113626828 = 568134124.

Now I have the 5 numbers: 113626824 - 113626824 - 113626824 - 113626824 - 113626828

In hex format: 0x06C5CEC8 0x06C5CEC8 0x06C5CEC8 0x06C5CEC8 0x06C5CECC

Finally, we input the numbers like the course website(https://whisperlab.org/introduction-to-hacking/lectures/collision):

./col $'\x06\xc5\xce\xc8\x06\xc5\xce\xc8\x06\xc5\xce\xc8\x06\xc5\xce\xc8\x06\xc5\xce\xcc'

Surprisingly, the answer is wrong:

image-20210203033338471

I stucked here, so I googled. In this article https://0xrick.github.io/pwn/collision/#Exploitation, it says that the system is little-endian, I'm not sure where this is come from. I did not find anything related to this. So, I use a python program to check if the system is little-endian.

python -c "import sys;print(sys.byteorder=='little')"

And it is little-endian:

image-20210203034220293

So, I need to reverse the input and try again:

./col $'\xc8\xce\xc5\x06\xc8\xce\xc5\x06\xc8\xce\xc5\x06\xc8\xce\xc5\x06\xcc\xce\xc5\x06'

Eventually, I have the flag:

image-20210203034418244

Flag: daddy! I just managed to create a hash collision :)

Question 5 - bof Challenge

The challenge asks me to download 2 files, so I did. And it seems bof is a executable file, so I give it excute permission and run it:

image-20210203034841156

Ok, so nc pwnable.kr 9000 is running the same program:

image-20210203035007785

The next step is to analysis the source code, I use vim to see the code, vi bof.c.

image-20210203143421150

so, main() calls func() then exit, the logic is in function func(). It has an integer input key, then compare the key with 0xcafebabe. Before comparison, the program requires a std input from keyboard and save it to a 32 bytes char variable overflowme, so it's possible to overflow variable overflowme and write the data (0xcafebabe) to the address where variable key is. This will make the varibale key equal to 0xcafebabe and calls system("/bin/sh") then enter a new bash. (gets() function in C language does not validate the memory, it just wirte everything it gets to the memory address)

I'll try to get the exploit on my local kali linux and then use it on pawnable.kr's server. This time, I'll only use gdb to debug the program, but I could also use a gun program on kali linux to debug it. In order to debug c program with GDB, I need to re-compile bof.c with flag gcc -g3 -m32 bof.c -o new_bof . -g3 means compile full debug information to the binary(It's ok to use g3, because it's a small program, while debug complex program, -g is much preferred). -m32 means compile it to 32-bit binary.

But it did not work:

image-20210203150530376

It seems that I don't have stdio.h header, and this is impossible on a linux machine with gcc. So, I googled the error:

image-20210203150731575

https://stackoverflow.com/questions/54082459/fatal-error-bits-libc-header-start-h-no-such-file-or-directory-while-compili/54082790

ok, I need to install 32-bit library manully. Kali linux use apt to manage the packages, so I only need to apt-get install gcc-multilib to install mulilib.

image-20210203151015221

Then run gcc to compile it again, and it worked just fine:

image-20210203151052880

Great! GDB is not installed too:

image-20210203151346415

After install GDB, I can finally debug the program:

image-20210203151452780

Then I use run to start the program and try something random just to play with it.

image-20210203151914488

I take a look at the source code, I understand I need to set a breakpoint at the comparison to check the value in key. (it's just partial code in the picture)

image-20210203152416042

Now run the program, I put a long string to overflow it:

image-20210203153153702

So, the key's value is replaced with 0xffffd180, I tried again without overflow:

image-20210203153222389

Now I know the buffer overflow could work on this program. So, I use an online tool genrate a string(https://wiremask.eu/tools/buffer-overflow-pattern-generator/) which is very easy to use on buffer overflow. What I used here is Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2Ad3Ad4Ad5Ad6Ad7Ad8Ad9Ae0Ae1Ae2Ae3Ae4Ae5Ae6Ae7Ae8Ae9Af0Af1Af2Af3Af4Af5Af6Af7Af8Af9Ag0Ag1Ag2Ag3Ag4Ag5Ag

I could use the string to find where is target place.

image-20210203154029685

I know it overflows the varibable at the start of Ab . Next is to find the content of variable key, then I know the place to overflow key's data.

image-20210203154931846

I learned to print the key as a sequence of 4 characters on the class tutorial notes.

Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6A KEY b7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2Ad3Ad4Ad5Ad6Ad7Ad8Ad9Ae0Ae1Ae2Ae3Ae4Ae5Ae6Ae7Ae8Ae9Af0Af1Af2Af3Af4Af5Af6Af7Af8Af9Ag0Ag1Ag2Ag3Ag4Ag5Ag

The bold words are where key is. Now I can set any value I want to key.

So, the expoit string is Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6A + \xbe\xba\xfe\xca.

Since it doesn't matter what content is in the overflow variable, so I replace the long string with one char q.

Now I have the overflow string, so I'll use a python script to help me input it:

(python -c "print 'q'*52 +'\xbe\xba\xfe\xca'"; cat) | nc pwnable.kr 9000

This pythin script will input the data to pwnable.kr's 9000 port. It's like running a local program ./bof. The difference is piped the data to a remote server.

image-20210203162452643

Finally, I have the user col's shell and use it to get the flag.

And the flag is daddy, I just pwned a buFFer :)