Convert to LEGV8 int main() int set1[5]; int set2[5]; int total=0; set1[0] = 1; for(int i=1; i<5; i++) { set1[i] = set1[i-1] + 2; } set2[0] = 2; for(int j=1; j<5; j+t) { set2[j] = set2[j-1]+ 2; for(int k-0; k<5; k++) { total += set1[k] * set2[
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convert this code and explain each line and what it does/ what it corresponds to. Also do NOT use ADDI. You can use terms such as LSL,MOV, ADD,LDUR,STUR... etc.
![Convert to LEGV8
int main()
{
int set1[5];
int set2[5];
int total=0;
setl[0] = 1;
for(int i=1; i<5; i++)
{
set1[i] = set1[i-1]+ 2;
}
set2[0] = 2;
for(int j=1; j<5; j+t)
{
set2[j] = set2[j-1]+ 2;
}
for(int k-0; k<5; k++)
{
total += set1[k] * set2[k];
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa77d439e-6e4f-4050-a6c7-d8833fccbd3e%2F453f217d-aa64-4a21-80e2-43665b18a5b2%2Fpccpat_processed.jpeg&w=3840&q=75)
Step by step
Solved in 2 steps with 5 images
- int a[10] = {0,1,2,3,4,5,6,7,8,9};int *m = &a[0];int *p = &a[5];int *q = &a[1]; p = (int) m + (int) p - (int) q; what is the value of of *pint main() //default function for call { int a[100],n,i,j; for (int i = 0; i < n; i++) //Loop for ascending ordering { for (int j = 0; j < n; j++) //Loop for comparing other values { if (a[j] > a[i]) //Comparing other array elements { int tmp = a[i]; //Using temporary variable for storing last value a[i] = a[j]; //replacing value a[j] = tmp; //storing last value } } } printf("\n\nAscending : "); //Printing message for (int i = 0; i < n; i++) //Loop for printing array data after sorting { printf(" %d ", a[i]); } } Need to transform this C code to MIPS Assembly Language Code simple stepsВ.width; } { t = B.type; w = { T.type = C.type; T.width = C.width; } T → B C В > int { B.type = integer; B.width 4; } В — foat { B.type = float; B.width = 8; } { C.type = t; C.width = w; } C - [ num] C1 = array(num. value, C1.type); { C.type C.width = num. value x C1. width; }
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- Given the following struct definition; #define MAX 20 typedef struct{ int items[MAX]; int count;}Array;Define the functions:void addItem(Array *a, int item); // adds item to the arrayvoid printArray(Array a); // displays elements of itemsint isFull(Array a); //returns 1 if the items is full; 0 if otherwiseint isEmpty(Array a); //returns 1 if the items is empty; 0 if otherwise#include using namespace std; class Student { public: void setMarks(){ } double getAV(double a[]){ double sum=0; for(int i=0;i<7;i++) sum-sum+a[i]; return sum/7; double a[]={10, 20, 30, 40, 50, 60, 70); cout<Question 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews OSEE MORE QUESTIONS