代写CSC 252 Computer Organization Midterm Exam代做回归

Midterm Exam

CSC 252

7 March 2025

Problem 0: Warm-up (2 Points)

What’s the most surprising thing you’ve learned so far?

Problem 1: Fixed-Point Arithmetics (12 points)

Part a) (2 points) Represent decimal number 31 in the hexadecimal form.

Part b) (4 points) Represent octal (base 8) number 73 in the decimal form. and binary form.

Part c) (6 points) Consider two signed binary numbers A = 10101 and B = 11001, both are represented using two’s complement representation. Do the math below.

(2 points) What’s the result of !A || B (! is NOT and || is OR)?

(2 points) What’s the result of A^B (^ is XOR)?

(2 points) How to write A in base 7?

Problem 2: Floating-Point Arithmetics (21 points)

Part a) (4 points)

(2 points) Write 10 8/7 using the normalized scientific notation.

(2 points) Write 32 16/5 using the normalized scientific notation.

Part b) (8 points) The engineering team is designing a new 12 bit floating point standard. The format follows the principles of the IEEE-style. floating point representation we discussed in the class. The engineering team is considering two possible standard:

● Standard A: 6 fraction bits

● Standard B: 9 fraction bits

(4 points) What are the biases for Standard A and Standard B?

(4 points) Given the requirement that the smallest gap between two representable numbers is at most 224/1, which of the two options meets the requirement? Show your work.

Part c) (9 points) Now the engineering team wants to design another 12-bit floating point representation. They want the representation to be able to precisely represent -100 and 100.

(5 points) What is the representation that meets this requirement while having the smallest gap between two consecutive, representable numbers?

(4 points) What is the smallest positive number that can be precisely represented in this format?

Problem 3: Logic Design (27 points)

Part a) (4 points)

(2 points) What is the result of a bitwise XOR operation between 01100 and 00111?

(2 points) What is the result of a bitwise NAND operation between 01000 and 01010?

Part b) (13 points)

(8 points) Given the circuit above, complete the following truth table. Additional columns are provided to show your partial work (for partial credit).

(2 points) Briefly describe the behaviour of the circuit above in words.

(3 points) Using the same circuit, it is possible to implement the equivalent of an AND gate. Complete the following table by assigning each of the three input wires to variables A, B or constants 0, 1 such that UTPUT = A AND B.

Part c) (8 points)

Consider the circuit below, which consists of N+1 XOR gates cascaded with a single NOT gate placed immediately before the (N+1) th XOR gate.

(6 points) What is the output of the circuit for the following values of N? You can write the output as a function of A.

N = 50

N = 63

(4 points) Draw a simpler circuit that will achieve the same output as the circuit above.

Problem 4: Assembly Programming (28 points)

Conventions:

1. For this section, the assembly shown uses the AT&T/GAS syntax opcode src, dst for instructions with two arguments where src is the source argument and dst is the destination argument. For example, this means that mov a, b moves the value a into b.

2. All C code is compiled on a 64-bit machine, where arrays grow toward higher addresses.

3. We use the x86 calling convention. That is, for functions that take two arguments, the first argument is stored in %edi (%rdi) and the second is stored in %esi (%rsi) at the time the function is called; the return value of a function is stored in %eax (%rax) at the time the function returns.

4. We use the Little Endian byte order when storing multi-byte variables in memory.

Part a) (10 points)

Consider the following function `fibonacci` and the assembly code that implements the C function.

int fibonacci(int n) { if (n <= 1) return n; return fibonacci(n - 1) + fibonacci(n - 2); } 0000000000401106 : 401106: push %rbp 401107: mov %rsp,%rbp 40110a: push %rbx 40110b: sub $0x18,%rsp 40110f: mov %_A_, -0x14(%rbp) 401112: _B_ $0x1, -0x14(%rbp) 401116: _C_ 40111d 401118: mov -0x14(%rbp), %eax 40111b: jmp 40113b 40111d: mov -0x14(%rbp), %eax 401120: sub $0x1, %eax 401123: mov %eax, %edi 401125: call 401106 40112a: mov %eax, %ebx 40112c: mov _D_(%rbp), %eax 40112f: ___________E___________ 401132: mov %eax, %edi 401134: call 401106 401139: add %ebx, %eax 40113b: mov -0x8(%rbp), %rbx 40113f: leave 401140: ret

Your Task: Fill in the missing instructions A, B, C, D, and E to complete the assembly code.

(2 points) A:

(2 points) B:

(2 points) C:

(2 points) D:

(2 points) E:

Part b) (18 points)

Consider the following unknown x86_64 assembly function: 000055555555516d : 0x55555555516d: mov $0x1,%eax 0x555555555172: cmp $0x1,%rsi 0x555555555176: jbe 0x5555555551aa 0x555555555178: mov %rdi,%rdx 0x55555555517b: lea -0x4(%rdi,%rsi,4),%r8 0x555555555180: mov $0x1,%r9d 0x555555555186: mov $0x0,%edi 0x55555555518b: jmp 0x555555555199 0x55555555518d: mov %edi,%r9d 0x555555555190: add $0x4,%rdx 0x555555555194: cmp %r8,%rdx 0x555555555197: je 0x5555555551a7 0x555555555199: mov 0x4(%rdx),%esi 0x55555555519c: mov (%rdx),%ecx 0x55555555519e: cmp %ecx,%esi 0x5555555551a0: jg 0x55555555518d 0x5555555551a2: cmovl %edi,%eax 0x5555555551a5: jmp 0x555555555190 0x5555555551a7: or %r9d,%eax 0x5555555551aa: ret

At the start of the function, the first argument is set to a pointer to the following integer array: [1, 1, 3, 2]. The second argument is set to 0x4.

The cmovl instruction conditionally moves a value from the source register to the destination register only if the sign flag is not equal to the overflow flag.

(3 points) Which line of assembly sets the condition codes for the cmovl operation on line 0x555555551a2?

(3 points) How many times is the jump at 0x5555555551a5 taken?:

(3 points) How many times is the jump at 0x5555555551a0 taken?

(3 points) Assume that the following line of assembly was executed immediately after the first execution of the instruction located at 0x55555555517b:

mov (%r8), %r9

What would the value in r9 be after this instruction?

(3 points) What is the meaning of the second argument of the function?

(3 points) Describe the function. What does it do?