Skip to main content

ARM buffer overflow: chapter 5 ಠ-ಠ (day 71)


Back again with another ARM buffer overflow, this post will pick up what we left in the previous post. Last time, we talk about how to bypass NX protection that changes stack region permission into non-executable by constructing ROP in our exploit and we can see that using one anti-exploit mechanism may not be really effective and we need to add a couple of additional protection to make it more reliable.

One of the additional protection that usually emphasizes in binary along with NX is ASLR that stands for Address Space Layout Randomization.

ASLR is responsible to make the address of used by the program to be randomized this could bring trouble for us since in the previous chapter we always hardcoded our ROP into the payload.

Although ASLR and NX is a pretty good combination in circumventing exploit we can still bypass this two protection together, by:

  • Information leak, this is basically taking advantage of another vulnerability that let us have information about the current state of the memory. We are going to cover this approach in the next post
  • Brute Force, we basically just guess the address until we got the right location. In this post, we are going to use brute force to bypass ASLR protection  
So continuing the explanation of brute force, this approach basically takes advantage of the low entropy of the randomized address. For the sake of POC, I used this following piece of code, this code was based on billie ellis challenge 4


#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>

char find_me[] = "/bin/bash";

void gadget1(){
    __asm__("pop {r0,pc}");
}

void crash_me(){
    char name[16];

    printf("Enter your name:\n");
        scanf("%s",name);
        printf("Welcome, %s!\n",name);
   
}

int main(){
    printf("\x1B[35m================================================\n");
    printf("Welcome to ROPLevel4 by williams\n");
    printf("================================================\x1B[0m\n\n");
   
    crash_me();
   
    return 0;



if you follow the previous post, you can guess we can just basically chain the gadget in function gadget1() with the string "/bin/sh" in global variable named find_me as the r0 and for pc we can use the address of system in libc. But since the ASLR is on we cannot determine where exactly is the system function reside in the memory.

compile the code with this parameter:

~# gcc roplevel4.c -o roplevel4 -fno-stack-protector

and load into gef, run checksec to make sure is compiled the way we expected



the last thing we need to do is to turn on the ASLR module in the machine



to turn on the ASLR you need to have root privilege. Check if aslr is on by using this command:



take a look at the base address showed in libc.so.6 we can see that every time we execute the command the value is changed

we are set to go :)

First, we need to found out how many offsets we need to crash the program and to take control of the PC register.







attach gef to the program and put a long pattern of string and once it is crash we can see that it need 20 bytes of offset to crash the program

Next, we need to find the location of "/bin/bash" string and gadget address in the code. We don't need to worry about ASLR since it will not randomize the code from the program



From this information, we can continue to create python script to craft our exploit



Another important note, we don't need to worry about the \x00 character used by the address since scanf only stop to receive string if it encounters whitespace or newline character

so the last piece of the puzzle that we need to find out is what value we need to put in the system_function variable that will point to system address in the binary. As I mentioned earlier ASLR produce random memory that has low entropy we can analyze this by "set disable-randomization off" in gef and run it couple of time











from this result, we can see that ASLR only randomizes two bytes of the function this leads us to 1/256 possibility and that's not really big number.

we can just try to use one of these addresses and run a loop until we can execute a shell command.





run the script by ~# bash ex.sh and it may take a couple of seconds until it finds the right address



cool :) so we got the right after a couple of iteration.

That's all folks and I hope you enjoy this post wee you next time at the next arm buffer overflow series





Comments

Popular posts from this blog

Having fun analyzing nginx log to find malicious attacker in the net (ง'̀-'́)ง (day 37)

  What makes you sleepless at night? is it because of a ghost or scary stories? is it because you have an important meeting tomorrow? or is it because you have an exam? For me, what keeps me up all night is that I keep thinking about what happens to a website that I just created, is it safe from an attacker (certainly not) or did I missing some security adjustments that lead to vulnerability? well I'm not the best secure programmer in the world, I'm still learning and there is a big possibility that I can make a mistake but for me, a mistake can be a valuable investment to myself or yourself to be better so from this idea, I want to know more about what attackers casually do when attacking a website. Here in this post, I'm going to show you how I analyzed attack to the website that I have permission to design and also some interesting findings that I could get from the analysis Background: All of this analysis comes from the traffic that is targeted to th

Utilize Pwntools for crafting ROP chain :') (day 69)

who doesn't like pwntools? it is a very versatile tool and can be customized according to our need using the python script but did you need to know that pwntools itself can help us to automatically craft a rop chain for us? so in this post, I will show you how to make rop chain less painful and make pwntools do all the heavy lifting. To demonstrate this I will use the binary challenge callme 64 bit from ropemporium link: https://ropemporium.com/challenge/callme.html Crashing the app: Like any other exploitation process, we need to crash the program by generating a long string pattern to determine the offset. based on the information from the above figure we can see that we required to provide 40 bytes of offset Fun stuff: now this where the fun stuff began write the following python script: as in the guideline of the challenged said we need to chain the function call by first to call the callme_one function, callme_two function and then callme_three funct

WriteUp PWN tarzan ROP UNICTF ಠ_ಠ (day 61)

So in this post, I'm going to talk about how to solve the Tarzan pwn challenge from UNICTF 2019. Back in the day when the competition is still going I couldn't finish it and don't have any clue to solve this but this time I was able to finish it :) Also in this post, we will be going to be heavily focused on how to utilize pwntools to construct a ROP chain. If you kinda confused about my explanation in this post you can refer to this following youtube video, link: https://www.youtube.com/watch?v=gWU2yOu0COk I build the python script based on this video Ok, let's get started! In this challenge, you will get two binary first go with tarzan and libc-2.29.so by providing .so file it tell us what version library that the target machine is using this could help us to do ROP chain. first, we run the Tarzan binary to get the basic idea of the program work and as you can see it just show you some text, newline and when you try to input something it doesn't gi