Q4. pts] Program Analysis - Please analyze the value of following registers after executing the instruction in the same line. # Calculate the value of myEg = 1+x+ x2+ x3 (x =5) .data word var x: . Word 3 var n: word 0 myEg: .text 19lobl main: main $s0,var x Sslavar n addi $s2, $zero, 1 addi $t0,$zero, 1 # $s0 = lw # $s1 # $s2 = # $t0 = $s2s2,$s0 Sto Sto, $s2 # $s2 = # St0 = mul add $s2 = # $t0 = Ss2Ss2, $s0 Sto, $to,$s2 mul add # $s2 $s2s2, $s0 $t0, $to, $s2 mul $t0 add $t0myEa $v0,10 SW li main
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- Assignment for Computer Architecture: N Factual by Recusion *please have comments in the code* You are to write a program in MIPS that computes N! using recursion. Remember N! is the product of all the numbers from 1 to N inclusive, that is 1 x 2 x 3 x (N – 1) x N. It is defined as 1 for N = 0 and is undefined for values less than 0. The programs first requests the user to input the value of N (display a prompt first so the user knows what to do). If the input value is less than 0, the program is to display “N! undefined for values less than 0” and then requests the user to input the value of N again. If the value input is non-negative, it is to compute N! using a recursive function, that is one that calls itself. You are to have your name, the assignment number, and a brief description of the program in comments at the top of your program. Since this is an assembly language program, I expect to see comments on almost every line of code in the program. Also make the…Segmentation: Select all of the following statements that are true. In segmentation, a logical address always has a length of 32 bit. In order to translate logical into physical addresses, the memory management unit uses the segment part of the logical address to determine the start address in the segment table and adds the offset to this to get the physical address. In segmentation, the logical address consists of a segment part and an offset. The segment length is limited by the maximum possible segment number. When applying segmentation, processes are only allowed to access the memory within their segments. Segments can be assigned access rights and privilege levels.2._____'ccccc6 sloc) 497 Bytes Write a recursive function that returns the sum of the digits of a given integer.Input format :Integer NOutput format :Sum of digits of NConstraints :0 <= N <= 10^9Sample Input 1 :12345Sample Output 1 :15Sample Input 2 :9Sample Output 2 :9 Solution:///////////// public class solution { public static int sumOfDigits(int input){ int sum; if(input<10){ return input; } sum = (input % 10) + sumOfDigits(input / 10); return sum; }}..
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- Python programing / must execute more than once with different numbers Write a program whose input is two integers and whose output is the two integers swapped. Ex: If the input is: 3 8 the output is: 8 3 Your program must define and call the following function. swap_values() returns the two values in swapped order.def swap_values(user_val1, user_val2)Instructions: You are strictly not allowed to use anything other than pointers and dynamic memory. One function should perform one functionality only. Task 1 Write a program in C++ that reads data from a file. Create dynamic memory according to the data. Now your task is to perform the following task. Row wise Sum Column wise Sum Diagonal wise Sum Example data.txt 4 5 1.6 10.2 33.7 99 20.5 3 44 50 96.1 2 8 9 4 74 50 99 19.1 Output: Sum row wise: 165, 191, 17, 242.1 Sum col wise: 127.6, 120.1, 228.8, 118.1, 20.5 Sum diagonal wise: Not Possible Note: You are restricted to use pointers and your function should be generic. Avoid memory wastage, memory leakage, dangling pointer. Use regrow or shrink concepts if required.// add.ll define void @add(i32* %ptr1, i32* %ptr2, i32* %val) {ret void} Fill add.ll function to do the following operation: void add(int *ptr1, int *ptr2, int *val) { *ptr1 += *val; *ptr2 += *val; } This is the full question. It is related to LLVM. If it's going to help there is one more code given which is: #include <stdio.h> void add(int *ptr1, int *ptr2, int *val); int main(int argc, char **argv) {FILE *f = fopen(argv[1], "r");int a, b, c;fscanf(f, "%d %d %d", &a, &b, &c);add(&a, &b, &c);printf("%d %d\n", a, b);fclose(f); return 0;}
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