void swap (int arr[], int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] } temp; void selectionSort (int arr[], int n) { for (int i = 0; i < n 1; i++) { int min i; for (int j=i+1; j < n; j++) { if (arr[j] arr[min]) min = j; swap (arr, min, i); } } } int main() { int arr[] = { 3, 5, 8, 4, 1, 9, -2 }; int n = 7; selectionSort (arr, n); return 0; } 1. Tranlaste the above C program to MIPS codes, suppose the starting address of codes is binary number 1000, the starting address of data is binary number 100000. 2. Translate the first MIPS codes of "if (arr[j]
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- sum = 0; for (int i = 1; i< n; i = sum++ || 2*i)int 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 stepsextern "C" int f(int *,int,int); int a[2][2] = {{11,12},{21,22}}; void setup(){ Serial.begin(115200); while(!Serial); delay(500); int *arr = (int*)a; Serial.println(f(arr,2,2)); } void loop(){ } .global f f: ldr r3,[r0] // get first element mov r2,#0add r2,r3 mov r1, #3 lp: add r0,#4 // add next elementldr r3,[r0]add r2,r3sub r1,#1 bgt lp mov r0,r2 bx lr Describe what operation is being performed in this code
- In questions 4-10 estimate the Big O value by analyzing the code. Note the algorithms are written in English. Hint: you are interested in the number of operations for each algorithm.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 Oint c[2][3]; int a[2][3] ={ {5}, {1,2} }; int b[2][3] = { 2,4,6,8} for( int i=0; i<2 ;i++)5. for( int j=0; j<3 ; j++) c[i] [j] = 3 * ( a[i] [j] + b[i] [j] ); program with c++
- for (i = 0; i < 2; ++i) { a = i* 20; ecrire (a); } lire (b);int sum, k, i, j; int x[4] [4]={1,2,3,4}, {5,6,7,8},{9,8,7,3},{2, 1,7,1}; sum=x[0] [0]; for (k=1; k<=3;k++) sum+=x[k] [k]; Give the value in sum after the statements are executed:12.14 Zylab 3 - Single Procedure Call Given an array of at least one integer, write a program to create a new array with elements equal to the power of each element in the original array raised to the index, i.e., P[i] = A[i]^i. For this, write two functions that will be called in main function independently. ● power inputs: element (A[i]) and index (1) • task: returns the value of element raised to index (A[i]^i). } O • newElement inputs: base address of new array P (*P), current size of P (variable k) and the new element (A[i]^i) o task: add the new element at the end. o This function does not return any value (void). O Following is a sample C code to perform the required task. You may modify the code for the functions, but the task performed should not be changed. int main() { // Variable Declaration int *A, *P; int n, k; int pow; // Task of main function P[0] 1; for (int j = k = j; pow // Base addresses of A and P // Lengths of arrays A and B // Return value from power function }…
- #include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…void changeAll(int x[]) { x[0] = x[0] + 5; return; The first element in an array would be increased by 5 The array would be the same because it was passed by value x would be increased by 5 All the elements in the passed array would be incremented by 5int sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;