What is the appropriate MIPS code equivalent to given c-style code? int doAdd (int t0, int t1) { return t0+tl*2; } а) sll $al,$al,1 add $v0,$a0 ,$al jr $ra doAdd: b) doAdd: mul $r1,$rl,2 add $v0,$r0,$rl jr $ra c) sll $t1,$t1,1 add $v0,$t0,$t1 jr $ra doAdd: d) mul $s1,$s1,2 add $v0,$s0,$s1 jr $ra doAdd:
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- Please complete the C code based on your understanding of the following assembly code. long rfun(unsigned long x) { if ( _______) return ______; unsigned long nx = _______; long rv = rfun(nx); return _______; } Options: x > 1 0 x>>4 x + rv x>>2 x > 0 x rx nx 2 Information # long rfun(unsigned long x) # x in %rdi rfun: cmpq $1, %rdi jbe .L8 movl $0, %eax ret .L8: pushq %rbx movq %rdi, %rbx shrq $4, %rdi call rfun addq %rbx, %rax popq %rbx ret Question 17 What value does rfun store in the callee-saved register %rbx? a. rv b. x c. nx d. 0python3 quadratic.py 15 19 4 the answer should be The solutions are -0.27 and -1.00 its only showing -0.27 and -1.0 when ran , it should be -0.27 and -1.00 not -1.0 import sys def main(): if len(sys.argv) == 4: a = float(sys.argv[1]) b = float(sys.argv[2]) c = float(sys.argv[3]) d = (b ** 2) - (4 * a * c) if d < 0 or a == 0: print("There are no roots") else: root1 = (-b + d ** 0.5) / (2 * a) root2 = (-b - d ** 0.5) / (2 * a) print("The solutions are", round(root1,2), "and", round(root2,2)) else: print("Please provide 3 numbers as command line arguments") main()Command line arguments are passed to int main(int argc, char** argv) as arguments argc and argv. You should assume that argc is at ebp+8 and argv is at ebp+12. 00000000 <what>: 0: push %ebp 1: mov %esp,%ebp 3: sub $0x10,%esp 6: mov 0x8(%ebp),%eax 9: add $0x4,%eax c: mov %eax,-0x4(%ebp) f: mov 0x8(%ebp),%eax 12: imul 0xc(%ebp),%eax 16: mov %eax,-0x8(%ebp) 19: mov 0x8(%ebp),%eax 1c: sub 0xc(%ebp),%eax 1f: mov %eax,-0xc(%ebp) 22: mov -0x4(%ebp),%edx 25: mov -0x8(%ebp),%eax 28: add %eax,%edx 2a: mov -0xc(%ebp),%eax 2d: add %edx,%eax 2f: leave 30: ret00000031 <main>: 31: lea 0x4(%esp),%ecx 35: and $0xfffffff0,%esp 38: pushl -0x4(%ecx) 3b: push %ebp 3c: mov %esp,%ebp 3e: push %ebx 3f: push %ecx 40: sub $0x10,%esp 43: mov %ecx,%ebx 45: mov 0x4(%ebx),%eax 48: add $0x4,%eax 4b: mov (%eax),%eax 4d: sub…
- Starting with the following C++ program: #include <iostream> using namespace std; void main () { unsigned char c1; unsigned char c2; unsigned char c3; unsigned char c4; unsigned long i1 (0xaabbccee); _asm { } cout.flags (ios::hex); cout << "results are " << (unsigned int) c1 << ", " << (unsigned int) c2 << ", " << (unsigned int) c3 << ", " << (unsigned int) c4 << endl; } Inside the block denoted by the _asm keyword, add code to move each byte of i1 into the unsigned chars c1 through c4 with the high order byte going into c1 and the low order into c4. Without the use of pointers.suppose r0 = 0x20000000 and r1 = 0x01234567. what is the content of r0 and r1 after the following line of code is executed: LDRH r1, [r0], #4// Your task is to write C code to do the following: // // 1) Include stdio.h and qutyio.h so that you can access the // functions required to write to the serial interface. // 2) Initialise the qutyio serial inteface by calling serial_init(). // 3) Create a variable "state" to store your student number. You // should interpret your student number as a decimal number. Use // the smallest standard unsigned integer type in which your student // number will fit. (you will need to include the stdint header). // e.g. the student number 10000012 would represent the number // ten million and twelve. // 4) Iterate through all the numbers from 0 to 255 in sequence. // For each number in the sequence perform the following steps: // a) Take the bitwise xor of the number with the variable "state", // storing the result back into "state". // b) Rotate right the bits in "state" at least one time, and until // the LSB of "state" is a zero. If there are no…
- We are given an unsigned integer number represented by 16 bits. This number can be kept in a variable defined in C language as "uint16_t". Let's represent this variabe as "uint16_t x;" in our C code. a) keep the least significant 8 bits of this number in an unsigned character variable "x_low". b) Keep the most significant 8 bits of this number in an unsigned character variable "x_high". c) Form a new unsigned integer 16-bit variable in C language called "x_swap". This variable will keep the swapped version of the variable "x" such that the least significant 8 bits of the variable x will become the most significant 8 bits of the variable "x_swap". Also, the most significant 8 bits of the variable x will become the least significant 8 bits of the variable "x_swap".Please write in C program to print out a list of bit strings that can be created by using combination How many bit strings of length 12 containa) exactly three 1s?b) at most three 1s?c) at least three 1s?d) an equal number of 0s and 1s?What is the output of below C++ code:- int x=0, i= 1: do { if(i % 5 == 0) { cout<<x; X++; ++i; }while(i<0): cout<<x<<" "<<i;
- The below program is: A MIPS program that executes the statement: s = (a + b) – (c + 101), where a, b, and c are user provided integer inputs, and s is computed and printed as an output. .data askA: .asciiz "Enter value of a: " askB: .asciiz "Enter value of b: " askC: .asciiz "Enter value of c: " msgS: .asciiz "Result s: ".text # Prompt user to enter value of a li $v0, 4 la $a0, askA syscall # Get user input li $v0, 5 syscall #store the result in t0 move $t0, $v0 # Prompt user to enter value of b li $v0, 4 la $a0, askB syscall # Get user input li $v0, 5 syscall #store the result in t1 move $t1, $v0 add $t2, $t0, $t1 # Prompt user to enter value of c li $v0, 4 la $a0, askC syscall # Get user input li $v0, 5 syscall #store the result in t0 move $t0, $v0 add $t1,$t0,101 sub $t0,$t2,$t1 li $v0, 4 la $a0, msgS syscall #print or show the age li $v0, 1 move $a0, $t0 syscall My question is, I want to…Using C Language, what is the output of this code? #include<stdio.h> #include<conio.h> #define r 3 #define c 4 int z[r][c] = {1,2,3,4,5,6,7,8,9,10,11,12}; main() { int a, b, c = 999; clrscr(); for(a=0;a<r;++a) for(b=0;b<c;++b if(z[a][b]<c) c=z[a][b]; printf(“%d”,c); getch(); return 1; )Please review my code here in C... there are errors occurring and I'm not sure why. InputInput will begin with a line containing 1 integer, n (1 ≤ n ≤ 500,000), representing the number of groupsto process. The following n lines will each contain 3 integers: s, a, and p (1 ≤ s ≤ 1,000,000; 1 ≤ a ≤1,000,000,000; 1 ≤ p ≤ 1,000,000) representing the group size, arrival time, and processing timerespectively.Output should contain 1 integer representing the sum time taken by all people that waited in line. Example: Sample Input:41 4 101 2 101 1 101 3 10Output: 5632 10 103 9 11 1 20 Output: 65 Code below: #include<iostream>#include<string.h>#include<sstream>#include<vector>#include <stdio.h> usingnamespace std; int main() { while(true){ cout<<"\nEnter visitor's data or input xx to stop\n"; string s; getline(cin,s);// Enter the new data line if(s=="xx")//If entered value is xx then stop break; stringstream ss(s);//Conersion…