* WHAT IS THE OUTPUT OF V(3) -3
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![WHAT IS THE OUTPUT OF V(3) -3
20*4
u = [10
10+20];
V3+4
v = [-1; 20; 3];
x = [u' v] ;
v (3) = x (2, 1)
23.0409 O
23.0940 O
22.940
25.0940 O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bb9001e-7485-498a-b628-c46db5f07ea7%2Fa0b3a927-090e-49c3-ad27-0ef922e646b2%2Fz1epdhn_processed.jpeg&w=3840&q=75)
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- v = (3, 5)print(v)p, q = v[0], v[1]r = p + qif r % 2 == 0:p = p + r - 5q = q + r - pr = r + 1v = (p, q)print(v)else:p = p - 1q = q - 1if r % 3 != 0:v = (p, q)print(v)else:p, q = q, pv = (p, q)print(v)Suppose we number the bytes in a w-bit word from 0 (least significant) to w/8 – 1 (most significant). Write code for the following C function, which will return an unsigned value in which byte i of argument x has been replaced by byte b: unsigned replace_byte (unsigned x, int i, unsigned char b); Here are some examples showing how the function should work: replace_byte(0x12345678, 2, 0xAB) --> 0x12AB5678 replace_byte(0x12345678, 0, 0xAB) --> 0x123456AB Bit-Level Integer Coding Rules In several of the following problems, we will artificially restrict what programming constructs you can use to help you gain a better understanding of the bit-level, logic, and arithmetic operations of C. In answering these problems, your code must follow these rules: Allowed operations All bit-level and logic operations. Left and right shifts, but only with shift amounts between 0 and w – 1. Addition and subtraction. Equality (==) and inequality (!=) tests. (Some of the problems do not allow…Let L={ w in(0 + 1)* 1 | w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expression below represents L? (A) (0* 10* 1)* (B) 0* (10* 10* )* (C) 0* (10* 1* )* 0* (D) 0* 1(10* 1)* 10*
- Suppose we number the bytes in a w-bit word from 0 (least significant)to w/8 −1 (most significant).Write code for the following C function, which will return an unsignedvalue in which byte i of argument x has been replaced by byte b:unsigned replace byte (unsigned x, int i, unsigned char b);Examples:replace byte(0x12345678, 2, 0xAB) → 0x12AB5678replace byte(0x12345678, 0, 0xAB) → 0x123456ABWrite a program that swaps 5th~11th bits in data_a with 25th~31th bits in data_bYour program must work for any data given, not just the example belowIn this question, we assume that the positions of bits count from right to left.That is, the first bit is the least significant bit.data_a DCD 0x77FFD1D1data_b DCD 0x12345678You are writing code that is equivalent to: X = A / (5 + B) Assume signed bytes and the following first 3 lines of code: mov al,A mov bl,B add bl,5 What is the one line of code that will perform the division correctly?
- implement anyEvenBit(x) Return 1 if any even bit in x is set to 1 you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * anyEvenBit - return 1 if any even-numbered bit in word set to 1* Examples anyEvenBit(0xA) = 0, anyEvenBit(0xE) = 1* Legal ops: ! ~ & ^ | + << >>* Max ops: 12*/int anyEvenBit(int x) {return 2;}Find Output of following code. mov ax, 20 mov bx, 0 subtraction: sub ax, [num1+bx] dec ax add bx, 2 cmp bx, 6 jne subtraction num1: dw 2, 1, 4, 1, 2Let A denote the set of bit strings of length 5 that begins with 10 and B denote the set of bit strings of length 5 that terminate at 00. the number of elements in A∩B is ____
- Build a _lengOfSt function in MIPS, that takes an argument in $a1 which is the address of a null-terminated string, returning the length of the given string (number of characters excluding the null-character) in $v0.For the following code fragment.LOOP: LD R1, 0(R2) ;load R1 from address 0+(R2) DADDI R1, R1, #1 ;R1 <– R1 + 1 SD R1, 0, (R2) ;store (R1) at address 0(R2) DADDI R2, R2, #4 ;R2 <– R2 + 4 DSUB R4, R3, R2 ;R4 <– R3 - R2 BNEZ R4, LOOP ;branch to LOOP if (R4) != 0Assume that the initial value of R3 is (R2)+20. Show the timing of this code fragmentPractice 3: Let’s compile following C sequence for MIPS and run on the emulator: int divide(int N, int M) { // map q and i to $s2 and $s3 int q = 0; int i = N; while(i > 0) { q += 1; i = i - M; } return q; } int multiply(int N, int M) { // map sum and i to $s0 and $s1 int sum = 0; for(int i = 0; i < N; ++i) { sum += M; } return sum; } int main() { // map a, b, c, k, and j to $s0, $s1, $s2, $s3, and $s4 int a = 4; int b = 5; int c = 120; int k = multiply(a, b); int j = divide(c, k); } Make sure multiply procedure properly gets backup of $s0 to $s4. Verify that after running this sequence: • value of S0 is (4)10 • value of S1 is (5)10 • value of S2 is (120)10 • value of S3 is (20)10 • value of S4 is (6)10 When ready, copy your MIPS assembly code from emulator and save as a text file.
