extern "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,#0 add r2,r3 mov r1, #3 lp: add r0,#4 // add next element ldr r3,[r0] add r2,r3 sub r1,#1 bgt lp mov r0,r2 bx lr Describe what operation is being performed in this code
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extern "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,#0
add r2,r3
mov r1, #3
lp:
add r0,#4 // add next element
ldr r3,[r0]
add r2,r3
sub r1,#1
bgt lp
mov r0,r2
bx lr
Describe what operation is being performed in this code
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- 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 O//Assignment 06 */public static void main[](String[] args) { String pass= "ICS 111"; System.out.printIn(valPassword(pass));} /* public static boolean valPassword(String password){ if(password.length() > 6) { if(checkPass(password) { return true; } else { return false; } }else System.out.print("Too small"); return false;} public static boolean checkPass (String password){ boolean hasNum=false; boolean hasCap = false; boolean hasLow = false; char c; for(int i = 0; i < password.length(); i++) { c = password.charAt(1); if(Character.isDigit(c)); { hasNum = true; } else if(Character.isUpperCase(c)) { hasCap = true; } else if(Character.isLowerCase(c)) { hasLow = true; } } return true; { return false; } }#include #include #include #define SIZE 5 int nums [SIZE] = [5,0,6,1,2); int main() { int i; pid_t pid; pid= fork(); if (pid == 0) { for (i = 0; i 0) { wait (NULL); } for (i = 0; i < SIZE; i++) printf("PARENT: %d\n", nums [i] + 1); /* LINE Y */ return 0; From the above-given program, please state what will be the output at lines X and Y. Explanation in detail required for output at both lines X and Y.
- int func(int a, int b) { return (avoid fact(int num) { int k,f=1; for (k=1;k<=num; k++) f=f*k; cout<< f; int main() { int arri[3]={3,2,1} ; for(int i=0;i<3;i++) fact(arri[i]); } 261 O 621 O 7 462int 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 steps#include for(int i = 0; i { int main() for(int j = 0; j< 2; j++) { { int matrix[2][2] = { printf(" %d", matrix[i][j]); } {2,3,}, //rowo printf("\n"); } {5,7}//row1 getch(); }; } printf("\n Resultant \n"); Try to execute, screenshot the resul and try to explain the sequence of the program. Post it in space-ners provided. FRAMEWORK HTPE ADACHE e OrcE outATION roR TECHO A LEint sum = 0; for (int i 0; i < 5; i++){ sum += i; } cout << sum;const int size =4; int r1[size]; int i; for (i=0;i<2;i++) r1[i]= 2 * i+3; r1[3]=2 * i+ i; r1[2]=r1[3]; for (i=0;iint 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 *p#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…double tab1[5] = {2,3,4,5,6}; double tab2[5] = {6,5,4,3,2}; for(int i = 0;i<5;i++) { tab1[i] = tab1[i]*tab2[i]; cout<<tab1[i]<<" "; } 12 15 16 15 16 15 15 16 15 12 12 15 16 15 12 12 15 12 15 12SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
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