What does the line "void *ptr = arr;" do?123456789101112#include <stdio.h>int main() { int arr[2] = {10, 20},i=0; void *ptr = arr; while(i<2) { printf("%d ", *((int *)ptr + i)); i++; } return 0;}Marks : 1Negative Marks : 0Answer hereModifies the values of the array arr.Initializes a void pointer to the array arr.Prints the memory addresses of the array arr.Results in a compilation error.ClearPrev

What does the line "void *ptr = arr;" do?123456789101112#include <stdio.h>int main() { int arr[2] = {10, 20},i=0; void *ptr = arr; while(i<2) { printf("%d ", *((int *)ptr + i)); i++; } return 0;}

What is the output of the following code?int arr[5] = {10, 20, 30, 40, 50};int *ptr = arr;cout << *(ptr + 2);10203040

What is the result of the following code?#include <stdio.h>int main() {    int arr[5] = {1, 2, 3, 4, 5};    int *ptr = (int *)(&arr + 1);    printf("%d\n", *(ptr - 1));    return 0;}

What is the output of the following code?int arr[5] = {1, 2, 3, 4, 5};int *p = arr;printf("%d\n", p[2]);

What is the output of the following code?1234567891011#include <stdio.h>int main() { char c = 'K'; char* ptr = &c; (*ptr)++; printf("%c ", *ptr); (*ptr)--; printf("%c", *ptr); return 0;}