程序代写CS代考 scheme python x86 database chain compiler assembly 2020 – cscodehelp代写

2020
CS 161 Computer Security
Midterm
Only one selected option
For questions with square checkboxes, you may select one or more choices on Gradescope.
You can select
multiple squares
For questions with a large box, you need write and label your answer in the corresponding text box on Gradescope.
You have 110 minutes, plus a 10 minute buffer, for a total of 120 minutes. There are 7 questions of varying credit (120 points total).
The exam is open note. You can use an unlimited number of handwritten cheat sheets, but you must work alone.
Clarifications will be posted at https://cs161.org/clarifications.
Q1 MANDATORY – Honor Code (2 points) Read the following honor code and type your name on Gradescope. Failure to do so will result in a grade of 0 for this exam.
For questions with circular bubbles, you may select exactly one choice on Gradescope. Unselected option
I understand that I may not collaborate with anyone else on this exam, or cheat in any way. I am aware of the Berkeley Campus Code of Student Conduct and acknowledge that academic misconduct will be reported to the Center for Student Conduct and may further result in, at minimum, negative points on the exam.
Solution: Don’tworryifyouforgottofillinyourname.Everyonegets2freepointsforembracing the suck this semester.
We also won’t take any points off if you entered something in a text box for a multiple-choice question, or if you bubbled in some options for a free-response question, or if you filled something in for a question that doesn’t exist on your randomized form. To be consistent, we will not consider any unnecessary writing/bubbling on your exam during grading (pretend it’s scratch work).
This is the end of Q1. Proceed to Q2 on your answer sheet.
Page 1 of 23

Grade distribution (out of 120 points):
Midterm Page 2 of 23 CS 161 – Fall 2020

Q2 True/false (30 points) Each true/false is worth 2 points.
Q2.1 True or False: Pointer authentication prevents all buffer overflow attacks. True False
Q2.2 TrueorFalse:TheRET2ESP(ReturntoESP)exploitfromProject1,Question6doesnotrequire knowing the absolute address of the shellcode when crafting the exploit.
True False
Q2.3 True or False: The function f(x) = 1 is a one-way function, since we can’t go from 1 to our original value of x.
True False
Q2.4 TrueorFalse:EvanBotdesignscustombufferoverflowprotectionthatblocksallwritestoRIP’s and SFP’s. This is a successful defense against all buffer overflow attacks.
True False
Q2.5 True or False: Using fgets(buf, size, …) instead of gets(buf) always prevents an attacker from overflowing buf.
Solution: False. If you never overwrite a pointer (e.g. rip, sfp, program-allocated pointer on the stack), your exploit won’t be detected. Remember, the only way to prevent all buffer overflow attacks is to use a memory-safe language.
Solution: True. The RET2ESP technique works even when ASLR is enabled, a case in which we don’t know what the exact address of the shellcode is.
Solution: False. One-way functions are defined as “given f(x) = y, it is hard to find any x′ such that f(x′) = y′. This is false for f(x) = 1, since you can pick any value and satisfy f(x) = f(x′).
Solution: False. We can still overwrite other values on our stack (e.g. an “authenticated” flag).
True
False
Solution: False. It’s still possible for us to have an integer conversion vulnerability, where the value of size is greater than the actual value of the buffer (we saw this in the project!).
Many people asked in clarifications whether size is guaranteed to match the size of buf, which caused Nick to say…“Do you always use fgets correctly?”
Midterm
Page 3 of 23
CS 161 – Fall 2020

Q2.6 True or False: Diffie-Hellman is a protocol for sending messages confidentially between two people who don’t share a key.
True False
Q2.7 True or False: The protocol from lecture guarantees integrity. True False
Q2.8 TrueorFalse:WhenusingCBCmode,weneedtopadmessagesbecausetheblockciphertakes a fixed-length input.
True False
Q2.9 True or False: Kerckhoffs’s principle assumes that everything about a cryptographic system, including the key, is public knowledge.
True False
Q2.10 True or False: Slower hashes are useful for password hashing. True False
Q2.11 True or False: In a digital signature scheme, the verifying key is private, and the signing key is public.
Solution: False. Diffie-Hellman is a key exchange protocol that allows two people to agree on a shared secret key. There is no message being sent in the Diffie-Hellman protocol.
Solution: False. As seen in Homework 2, an attacker could change the ciphertext (c1, c2) to be (c1, 2c2), causing the recipient to see the message 2x instead of x. There is no way for the recipient to detect this, so does not guarantee integrity.
Solution: True. As seen in Homework 2 and Lab 1, CBC mode breaks the plaintext up into blocks and passes each block through block cipher encryption, and since the block cipher takes a fixed-length input, we need the plaintext length to be a multiple of the block size.
Solution: False. Everything except the key is public knowledge.
Solution: True. A slower hash only costs a legitimate user an additional fraction of a second to check their legitimate password, but it increases the cost for an attacker performing a dictionary/brute-force attack by a huge constant factor.
Midterm
Page 4 of 23 CS 161 – Fall 2020

True False
Q2.12 True or False: A 64-bit stack canary on a 64-bit processor provides more protection than a 32-bit stack canary on a 32-bit processor.
True False
Q2.13 True or False: Security is economics, so you should generally not use a $100 lock to secure a $10 product.
True False
Q2.14 True or False: The confidentiality of is compromised if r, the random value chosen for each message sent, is public.
True False
Q2.15 True or False: RSA encryption without padding is IND-CPA secure. True False
This is the end of Q2. Proceed to Q3 on your answer sheet.
Solution: False.Thesigningkeyisprivate,soonlytheownerofthesigningkeycangenerate valid signatures. The verifying key is public, so everyone can verify signatures.
Solution: True.A64-bitrandomcanaryishardertoguessbybrute-forcethana32-bitrandom canary. There is no security disadvantage from having a longer stack canary (although there might be negligible performance impact).
Solution: True. As seen in Homework 1, a rational consumer should not spend more on the lock than the product.
Solution: True. The attacker can recover the original message by multiplying c2 by A−r.
Solution: False. RSA encryption without padding, as seen in lecture (and CS70), is not CPA secure because it is deterministic.
Midterm Page 5 of 23 CS 161 – Fall 2020

Q3 MAC Madness (18 points) Evan wants to store a list of every CS161 student’s firstname and lastname, but he is afraid Mallory will tamper with his list.
Evan is considering adding a cryptographic value to each record to ensure its integrity. For each scheme, determine what Mallory can do without being detected.
Assume MAC is a secure MAC, H is a cryptographic hash, and Mallory does not know Evan’s secret key k. Assume that firstname and lastname are all lowercase and alphabetic (no numbers or special characters), and concatenation does not add any delimiter (e.g. a space or tab), so nick∥weaver = nickweaver.
Clarifications during the exam: Bob is storing the names with the cryptographic value in the database. Duplicate records are not allowed. Mallory can change anything in the database. “A value of her choosing” means any arbitrary value.
Q3.1 (3 points) H(firstname∥lastname)
(A) Mallory can modify a record to be a value of her choosing
(B) Mallory can modify a record to be a specific value (not necessarily of her choosing) (C) Mallory cannot modify a record without being detected
(D)
(E)
(F)
Q3.2 (3points) MAC(k,firstname∥lastname)
Hint: Can you think of two different records that would have the same MAC?
(G) Mallory can modify a record to be a value of her choosing
(H) Mallory can modify a record to be a specific value (not necessarily of her choosing) (I) Mallory cannot modify a record without being detected
(J)
(K)
(L)
Midterm
Page 6 of 23 CS 161 – Fall 2020
Solution: Anybody can hash a value, so Mallory could change a record to be whatever she wants and compute the hash of her new record.

Solution: Because the concatenation doesn’t have any indicator of where the first name ends and the last name begins, Mallory could shift some letters between the first name and last name. For example, she could change the name to Ni Ckweaver, , , etc. Since the MAC would remain unchanged, this edit would be undetectable.
Q3.3 (3 points) MAC(k, firstname∥”-“∥lastname), where “-” is a hyphen character. (A) Mallory can modify a record to be a value of her choosing
(B) Mallory can modify a record to be a specific value (not necessarily of her choosing) (C) Mallory cannot modify a record without being detected
(D)
(E)
(F)
Q3.4 (3points) MAC(k,H(firstname)∥H(lastname))
(G) Mallory can modify a record to be a value of her choosing
(H) Mallory can modify a record to be a specific value (not necessarily of her choosing) (I) Mallory cannot modify a record without being detected
(J)
(K)
(L)
Q3.5 (3points) MAC(k,firstname)∥MAC(k,lastname)
(A) Mallory can modify a record to be a value of her choosing
Solution: Now, the concatenation includes a separator between first name and last name, so the attack from the previous part is no longer possible. Note that names are alphabetical, so they would never include a dash in them.
Solution: Hashes have fixed-length output, so the attack from the previous part (shifting letters between the first and last name) is not possible here either. It will always be unambiguous where the first hash ends and the second hash begins.
Also, since both hashes are used as input to a single MAC, there is no way for an attacker without the key to generate a valid MAC for any different name.
Midterm Page 7 of 23 CS 161 – Fall 2020

(B) Mallory can modify a record to be a specific value (not necessarily of her choosing) (C) Mallory cannot modify a record without being detected
(D)
(E)
(F)
Q3.6 (3 points) Which of Evan’s schemes guarantee confidentiality on his records? (G) All 5 schemes (J) None of the schemes
Solution: Because the first name and last name have separate MACs, Mallory could swap the first name and last name, and swap the two halves of the MAC.
In other words, Mallory could change the name to , and change the MAC from MAC(k, nick)∥MAC(k, weaver) to MAC(k, weaver)∥MAC(k, nick).
(H) Only the schemes with a MAC (I) Only the schemes with a hash
(K) (L)
Solution: MACs and hashes do not have any confidentiality guarantees.
This is the end of Q3. Proceed to Q4 on your answer sheet.
Midterm Page 8 of 23 CS 161 – Fall 2020

Q4 Socially Distanced Cipher (18 points) Bob and Alice want to plan a social distancing picnic, but don’t want to invite Eve because she hasn’t been wearing a mask in public. They decide to send messages using a new block cipher chaining mode, AES-SDC (Socially Distanced Cipher). Note that AES-SDC requires a different key for each block of the message.
C0 = IV
Ci =Eki(Ci−1 ⊕i)⊕Pi
Q4.1 (3 points) Which of the following is the correct decryption expression/diagram for AES-SDC? (A)Pi =Eki(Pi−1 ⊕i)⊕Ci
(B)Pi =Dki(Pi−1 ⊕i)⊕Ci
Midterm
Page 9 of 23 CS 161 – Fall 2020

(C)Pi =Eki(Ci−1 ⊕i)⊕Ci
(D)Pi =Dki(Ci−1 ⊕i)⊕Ci
(E) (F)
Solution: In equations: To solve for Pi, XOR both sides of the encryption expression: Ci ⊕ Eki (Ci−1 ⊕ i) = Pi.
In pictures: Observing the three-way XOR junctions, we see that to retrieve the plaintext, we need to XOR the ciphertext with the block cipher encryption (not decryption) output. This rules out options (B) and (D). The input to the block cipher encryption is the ciphertext, not the plaintext, which rules out options (A) and (B). Thus option (C) is the correct answer.
Q4.2 (3 points) Select all true statements about this encryption scheme.
Hint: The cipher mode you saw in Homework 2, Ci = Ek(Ci−1) ⊕ Pi, is IND-CPA secure.
Midterm
Page 10 of 23
CS 161 – Fall 2020
(G) Encryption can be parallelized
(H) Decryption can be parallelized
(I) It is IND-CPA secure
(J) None of the above

(K) (L)
Solution: Encryptioncannotbeparallelized,becausecalculatingaciphertextblockCirequires the previous ciphertext block Ci−1 to be calculated first.
Decryption can be parallelized, because calculating a plaintext block Pi only requires ciphertext blocks Ci and Ci−1, which are already known before decryption starts.
The scheme is IND-CPA secure. Intuitively, AES-SDC is the same as the cipher mode from Homework 2, with two differences. First, a different key is used for each block cipher. This doesn’t affect IND-CPA security because the attacker still doesn’t know any of the secret keys. Second, a counter is added before encryption. This also doesn’t affect IND-CPA security, since the output of a block cipher looks random to an attacker without the key, regardless of whether the input is XOR’d with a counter.
Suppose Alice loses some of her shared keys with Bob. Alice wants to encrypt an n-block message using AES-SDC. For each scenario below, determine which blocks Alice can still encrypt.
Q4.3 (3 points) Alice has all the keys except k4 and k5.
(A) Alice can encrypt all parts of her message except P4 and P5
(B) Alice can encrypt P1, P2 and P3 only.
(C) Alice can encrypt the entire message
(D) Alice cannot encrypt any block of the message (E) None of the above
(F)
Solution: The intended answer was (B). Without k4 and k5, Alice cannot run the block cipher encryptions needed to output C4 and C5. Since C5 is used as an input to further block cipher encryptions, all blocks after C5 cannot be encrypted either.
Someone in clarifications found an alternate solution to this part. If Alice doesn’t know k4 and k5 and substitutes random/garbage values for the missing keys, then the corresponding ciphertexts C4 and C5 end up being random garbage as well. However, since C5 is random garbage and is fed to the next encryption block, it can be used as an IV for future encryptions, which causes the rest of the encryption to be correct, even if Bob doesn’t know what garbage values Alice used for k4 and k5!
We accepted both (A) and (B) as correct answers for full credit.
Now, suppose Alice now has all the keys, and Alice sends a n-block message to Bob. Eve learns some keys and some blocks of ciphertext. For each scenario below, determine which blocks Eve can decrypt.
Clarification during exam: Eve knows the index of all keys and ciphertext blocks that she learns. Midterm Page 11 of 23 CS 161 – Fall 2020

Q4.4 (3 points) Eve learns the IV, ciphertext blocks C5 and C6, and key k5. (G) Eve can decrypt C5 only
(H) Eve can decrypt C5 and C6 only
(I) Eve can decrypt all messages intercepted
(J) Eve cannot decrypt any intercepted messages (K) None of the above
(L)
Solution: In order to decrypt a ciphertext Ci, Eve needs to gain access to both Ci−1 as well as ki.
To decrypt C5, Eve would need k6, which she doesn’t have, and to decrypt C6, Eve would need k7, which she also doesn’t have. Thus Eve can’t decrypt any intercepted messages.
Someone in clarifications noted that options H and I are technically equivalent, since C5 and C6 are the only messages Eve intercepts. However, neither is the correct answer choice, so this didn’t affect our grading. Sorry if this caused any confusion.
Q4.5 (3 points) Eve learns the IV, ciphertext blocks C2, C3, and C5, and keys k2, k3, and k5. (A) Eve can decrypt C3 and C5 only
(B) Eve can decrypt C2, C3, C5 only
(C) Eve can decrypt C2, C3, C4, C5 only
(D) Eve can decrypt C3 only
(E) Eve cannot decrypt any intercepted messages (F) None of the above
Q4.6 (3 points) Bob receives all the keys and ciphertext blocks C1 through Cn, but C3 is corrupted. Which plaintext blocks can Bob successfully decrypt?
Clarification during exam: “Bob receives all the keys and ciphertext blocks C1 through Cn” should be “ciphertext blocks C0 through Cn.”
Solution: Using the same reasoning as the previous part, Eve has C2, C3, and k3, so she can decrypt C3.
To decrypt C2, Eve would need C1, which she doesn’t have, and to decrypt C5, Eve would need C4, which she also doesn’t have.
Midterm
Page 12 of 23 CS 161 – Fall 2020

(G) Bob can successfully decrypt all blocks except C3
(H) Bob can successfully decrypt all blocks except C4
(I) Bob can successfully decrypt all blocks except C1, C2, C3 (J) Bob can successfully decrypt all blocks except C3 and C4 (K) Bob cannot successfully decrypt any of the blocks
(L) None of the above
This is the end of Q4. Proceed to Q5 on your answer sheet.
Solution: The same reasoning from the previous parts applies here as well, where Bob has all the keys and all the ciphertexts, except C3.
C3 is needed in the decryption of C3 and C4, so Bob can’t decrypt these two blocks. Bob can decrypt all other blocks.
Midterm Page 13 of 23 CS 161 – Fall 2020

Q5
Hacked EvanBot (16 points) Hacked EvanBot is running code to violate students’ privacy, and it’s up to you to disable it before it’s too late!
1 2 3 4 5 6 7 8 9
10 11
#include
void spy_on_students ( void ) { char buffer [16];
fread(buffer , 1, 24, stdin);
}
int main() { spy_on_students () ; return 0;
}
The shutdown code for Hacked EvanBot is located at address 0xdeadbeef, but there’s just one problem— Bot has learned a new memory safety defense. Before returning from a function, it will check that its saved return address (rip) is not 0xdeadbeef, and throw an error if the rip is 0xdeadbeef.
Clarification during exam: Assume little-endian x86 for all questions.
Assume all x86 instructions are 8 bytes long. 1Assume all compiler optimizations and buffer overflow
defenses are disabled.
The address of buffer is 0xbffff110.
Q5.1 (3 points) In the next 3 subparts, you’ll supply a malicious input to the fread call at line 5 that causes the program to execute instructions at 0xdeadbeef, without overwriting the rip with the value 0xdeadbeef.
The first part of your input should be a single assembly instruction. What is the instruction? x86 pseudocode or a brief description of what the instruction should do (5 words max) is fine.
Solution: jmp *0xdeadbeef
You can’t overwrite the rip with 0xdeadbeef, but you can still overwrite the rip to point at arbitrary instructions located somewhere else. The idea here is to overwrite the rip to execute instructions in the buffer, and write a single jump instruction that starts executing code at 0xdeadbeef.
Grading: most likely all or nothing, with some leniency as long as you mention something about jumping to address 0xdeadbeef. We will consider alternate solutions, though.
Q5.2 (3 points) The second part of your input should be some garbage bytes. How many garbage bytes do you need to write?
1In practice, x86 instructions are variable-length.
Midterm Page 14 of 23 CS 161 – Fall 2020

(G) 0 (H) 4 (I) 8 (J) 12 (K) 16 (L)
Q5.3 (3 points) What are the last 4 bytes of your input? Write your answer in Project 1 Python syntax, e.g. x12x34x56x78.
Q5.4 (3 points) When does your exploit start executing instructions at 0xdeadbeef? (G) Immediately when the program starts
(H) When the main function returns
(I) When the spy_on_students function returns (J) When the fread function returns
(K)
(L)
Q5.5 (4 points) Which of the following defenses would stop your exploit from the previous parts? (A) Non-executable pages (also called DEP, WˆX, and the NX bit)
(B) Stack canaries
(C) ASLR
(D) Rewrite the code in a memory-safe language (E) None of the above
(F)
Solution: After the 8-byte instruction from the previous part, we need another 8 bytes to fill buffer, and then another 4 bytes to overwrite the sfp, for a total of 12 garbage bytes.
Solution: x10xf1xffxbf
This is the address of the jump instruction at the beginning of buffer. (The address may be
slightly different on randomized versions of this exam.) Partial credit for writing the address backwards.
Solution: Theexploitoverwritestheripofspy_on_students,sowhenthespy_on_students function returns, the program will jump to the overwritten rip and start executing arbitrary instructions.
Midterm
Page 15 of 23
CS 161 – Fall 2020

Solution: Non-executable pages prevents the exploit because the exploit requires executing the jmp instruction that was written on the stack.
Stack canaries prevent the exploit because the exploit will overwrite the canary between buffer and the rip.
ASLR prevents the exploit because the exploit requires overwriting the rip with a known address on the stack.
Many people asked in clarifications if ASLR would change the address of the shutdown code at 0xdeadbeef. We didn’t answer this clarification because it doesn’t affect the correct answer choice here: even if you knew the absolute address of the shutdown code, you couldn’t overwrite the rip with the address of the buffer on the stack, because ASLR would randomize addresses on the stack.
Using a memory-safe language always prevents buffer overflow attacks.
This is the end of Q5. Proceed to Q6 on your answer sheet.
Midterm Page 16 of 23 CS 161 – Fall 2020

Q6 Chegg (17 points) Engineers at Chegg are analyzing different password management techniques. Unfortunately, the engineers at Chegg used Chegg to make it through CS 161, so they don’t remember anything they learned!
Q6.1 (3 points) Suppose there is an offline attacker (with access to the hashed passwords file) and an online attacker (without access to the hashed passwords file). Chegg implements a CAPTCHA on its login page. Which attacker(s) does the CAPTCHA prevent from performing a dictionary attack?
Clarification during exam: “prevent from performing a dictionary attack” means make the attack significantly more expensive.
(A) The offline attacker only (B) The online attacker only (C) Both attackers
(D) Neither attacker
(E) (F)
Solution: TheCAPTCHAmakestheattacksignificantlymoredifficultfortheonlineattacker, who would have to fill out a CAPTCHA every time they try a guess at the password.
However, it would not stop the offline attacker, since they already have access to all the hashed passwords.
Q6.2 (3 points) Instead of salting each password hash, Chegg engineers XOR the hashed password with the account creation timestamp and store the XOR’d password hash with the timestamp in their database.
True or false: This successfully prevents an offline attacker from performing a dictionary attack.
Clarification during exam: The timestamp and the XOR’d password hash are both stored in the database. The offline attacker has the entire database.
(G) True (H) False (I) (J) (K) (L)
Q6.3 (4 points) One of Chegg’s competitors, Course Hero, has been compromised, and all of their user accounts and passwords have been leaked in plaintext. Select all defenses that Chegg could use to protect students who use the same password for Chegg and Course Hero.
(A) Use a slow hash function
(B) Include a salt in the password hash e.g. store a tuple of (salt, H(salt∥password))
(C) Require every login attempt to also provide a random code sent by a secure SMS to the registered user’s phone (a secure second factor)
Midterm
Page 17 of 23 CS 161 – Fall 2020
Solution: False.AnattackercouldsimplyXOReachhashwiththeaccountcreationtimestamp to retrieve the original unsalted hash, which would be susceptible to dictionary attacks.

(D) During the account creation phase, require every password to end with -CHEGG (E) None of the above
(F)
Solution: Aslowhashfunction,saltedhashes,andpasswordsendingin-CHEGGalldonothing to stop an attacker who knows the plaintext passwords.
Two-factor authentication is the only valid defense here. Now, an attacker can’t use the leaked passwords to log into the user’s Chegg account, because they don’t have the random code sent to the user’s phone.
Some people asked in clarifications whether the attacker knows which password corresponds to which user. This doesn’t change the answer, because even if the attacker doesn’t have a mapping of users to passwords, the list of passwords still significantly reduces the search space for an attacker to try logins to Chegg accounts.
Chegg uses a certificate chain in order to verify tutors. When tutors post responses, they attach a digital signature of their response along with their certificate. Students can verify the authenticity of a response by verifying the certificate and using the public key in the certificate to verify the signature.
The certificate chain is below. Assume that the Chegg Root Certificate Authority (CA) is hardcoded into students’ browsers.
1. Identity: Director of Chegg Recruiting (Verified by Chegg Root CA)
2. Identity: Campus Chegg Recruiter (Verified by Director of Chegg Recruiting) 3. Identity: Authorized Tutor (Verified by Campus Chegg Recruiter)
Q6.4 (4 points) EvanBot is not a valid tutor, but wants to create a fake tutor response with a valid signature. Which of these attacks would allow Bot to accomplish this?
(G) Steal the public key of the Campus Chegg Recruiter
(H) Steal the private key of the Director of Chegg Recruiting (I) Steal the private key of the Chegg Root CA
(J) Steal the certificate of an authorized tutor
(K) None of the above
(L)
Solution: Public keys and certificates are public and well-known, so an attacker who steals them can’t do anything to forge a message.
Stealing the private key of any individual in the certificate chain above the Authorized Tutor level allows EvanBot to create a forged certificate with a valid signature.
Midterm
Page 18 of 23
CS 161 – Fall 2020

For example, EvanBot can use the private key of the Director of Chegg Recruiting to sign a certificate for a fake Campus Chegg Recruiter made up by EvanBot. Then, EvanBot uses the private key of the made-up Campus Chegg Recruiter to sign a certificate of a fake tutor. Finally, EvanBot can use the private key of the fake tutor to sign a tutor response.
Q6.5 (3 points) EvanBot gains access to the private key of Dave, who is an authorized tutor. Which of the following can EvanBot do?
(A) Post a valid response as Nick, an existing tutor
(B) Post a valid response as Dave
(C) Create and sign a certificate for Raluca, a new tutor (D) None of the above
(E)
(F)
Solution: Dave’s private key cannot be used to sign messages as Nick, who would have a different private key.
Dave’s private key can be used to sign messages as Dave.
Dave’s private key is at the tutor level (the lowest level of the certificate chain), so it cannot be used to sign certificates.
This is the end of Q6. Proceed to Q7 on your answer sheet.
Midterm Page 19 of 23 CS 161 – Fall 2020

Q7
1 2 3 4 5 6 7 8 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
no exception handlers, saved registers, or compiler padding.
Q7.1 (3 points) Fill in the numbered blanks on the following stack diagram for prepare_input.
1 (0xbffff494) 2 (0xbffff490) 3 (0xbffff450) 4 (0xbffff44c)
Stack Exchange
Consider the following vulnerable C code:
(19 points)
#include
#include #include
void prepare_input ( void ) { char buffer [64];
int64_t ∗ptr;
printf ( “What is the buffer ?
” ) ; fread(buffer , 1, 68, stdin);
printf ( “What is the pointer ?
” ) ;
fread(&ptr , 1, sizeof(uint64_t ∗) , stdin);
if (ptr < buffer || ptr >= buffer + 68) { printf(“Pointer is outside buffer!”); return ;
}
/∗ Reverse 8 bytes of memory at the address ptr ∗/
∗ ptr = bswap_64 (∗ ptr ) ; }
int main ( void ) { prepare_input () ;
return 0; }
The bswap_64 function 2takes in 8 bytes and returns the 8 bytes in reverse order.
Assume that the code is run on a 32-bit system, no memory safety defenses are enabled, and there are
(A) 1 = sfp, 2 = rip, 3 = buffer, 4 = ptr (B) 1 = sfp, 2 = rip, 3 = ptr, 4 = buffer (C) 1 = rip, 2 = sfp, 3 = buffer, 4 = ptr
2Technically, this is a macro, not a function.
Midterm Page 20 of 23
(D) 1 = rip, 2 = sfp, 3 = ptr, 4 = buffer (E)
(F)
CS 161 – Fall 2020

Solution: Theripispushedontothestackfirst,followedbythesfp,followedbythefirstlocal variable buffer, followed by the second local variable ptr. (Remember that local variables are placed on the stack, highest-to-lowest address, in the order they are defined in the code.)
Q7.2 (4 points) Which of these values on the stack can the attacker write to at lines 10 and 13? Select all that apply.
(G) buffer (J) rip
(H) ptr (K) None of the above
(I) sfp
(L)
Solution: At line 10, the attacker can write 68 bytes starting at buffer. This overwrites all 64 bytes buffer and the 4 bytes directly above it, which is the sfp.
At line 13, the attacker can write exactly 1 uint64_t * into ptr. This overwrites ptr, and nothing else.
Notice that the rip cannot be directly overwritten.
Q7.3 (3 points) Give an input that would cause this program to execute shellcode. At line 10, first input these bytes:
(A) 64-byte shellcode (B) xbfxffxf4x4c (C) x4cxf4xffxbf
Q7.4 (3 points) Then input these bytes: (G) 64-byte shellcode
(H) xbfxffxf4x4c (I) x4cxf4xffxbf
Q7.5 (3 points) At line 13, input these bytes: (A) xbfxffxf4x50
(B) x50xf4xffxbf (C) xbfxffxf4x90
(D) xbfxffxf4x50 (E) x50xf4xffxbf (F)
(J) xbfxffxf4x50 (K) x50xf4xffxbf (L)
(D) x90xf4xffxbf (E) xbfxffxf4x94 (F) x94xf4xffxbf
Midterm
Page 21 of 23
CS 161 – Fall 2020

Solution: Line 10 writes 68 bytes into the 64-byte buffer, which lets us overwrite the sfp, but not the rip.
Line 13 lets us write a value into ptr, which is then dereferenced in a call to bswap_64. This lets us reverse any 8 bytes in memory we want, as long as they are between buffer and buffer + 68, i.e. in the buffer or sfp.
The overarching idea here is to write the address of shellcode in the sfp, and then use the call to bswap_64 to swap the sfp and the rip.
First, we write the 64 bytes of shellcode into the buffer. Then, we overwrite the sfp with xbfxffxf4x50. These bytes are written backwards because bswap_64 will reverse all 8 bytes of the sfp and the rip. Finally, we write the address of the sfp, x90xf4xffxbf, into ptr. These bytes are written normally because bswap_64 never affects ptr.
Suppose the current rip is 0xdeadbeef. Our input causes the 8 bytes starting at the sfp to be xbfxffxf4x50xefxbexadxde. When we call bswap_64 at the location of sfp, the 8 bytes starting at sfp are reversed, so they are now xdexadxbexefx50xf4xffxbf. Notice that the rip is now pointing to the address of shellcode in the correct little-endian order. Also note that the original rip has been swapped into the sfp and is now backwards, although we don’t care because the rip has already been overwritten.
Note: Because you can overwrite the sfp, you might be tempted to use the off-by-one exploit from Q4 of Project 1. However, this does not work here because you need enough space to write the shellcode and the address of shellcode in the buffer, but the buffer only has space for the shellcode.
Partial credit for Q7.4 option (K) and Q7.5 option (C) (correct address, but backwards).
Q7.6 (3points)Supposeyoureplace68with64atline10andline15.Isthismodifiedcodememory-safe? (G) Yes (H) No (I) (J) (K) (L)
Solution: This is still not memory-safe. If you make ptr point at one of the last 4 bytes of buffer, the check at line 15 will pass, but it will cause part of the sfp to be overwritten. For example, if ptr is located 4 bytes before the end of buffer, the last 4 bytes of buffer will be swapped into the sfp.
Because you can overwrite the sfp, you could still exploit this modified code using the technique from Project 1, Question 4 (although as mentioned above, you would need shorter shellcode). Regardless of what shellcode you use, since this code lets you write to the sfp (outside the bounds of buffer), it is not memory-safe.
This is the end of Q7. You have reached the end of the exam.
Midterm Page 22 of 23 CS 161 – Fall 2020

C Function Definitions
bswap_64(x);
Returns a value in which the order of the bytes in its 8-byte argument
is reversed.
char *fgets(char *s, int size, FILE *stream);
fgets() reads in at most one less than size characters from stream and
stores them into the buffer pointed to by s. Reading stops after an
EOF or a newline. If a newline is read, it is stored into the buffer.
A terminating null byte (‘’) is stored after the last character in
the buffer.
size_t fread(void *ptr, size_t size, size_t nmemb, FILE *stream);
The function fread() reads nmemb items of data, each size bytes long,
from the stream pointed to by stream, storing them at the location
given by ptr.
Note that fread() does not add a null byte after input.
Midterm Page 23 of 23 CS 161 – Fall 2020