What will be the output of the following C code? #include
void main()
{
int arr[5] = {10, 20, 30, 40, 50};
int *p = (int*) (&arr + 1);
printf("%d %d", *(arr + 1), *(p-1));
}

Question

What will be the output of the following C code? #include void main() { int arr[5] = {10, 20, 30, 40, 50}; int *p = (int*) (&arr + 1); printf("%d %d", *(arr + 1), *(p-1)); }

Accepted Answer

Correct answer: B — Answer: 20 50 Explanation: The array arr has elements: 10, 20, 30, 40, 50. *(arr + 1) accesses the element at index 1, which is 20. &arr has type pointer-to-array (int (*)[5]). Adding 1 to &arr moves past the entire array (by 5 ints). Casting (&arr + 1) to int* yields a pointer to one-past-the-last element (arr + 5). So p points to arr[5]. Therefore p-1 points to arr[4], which is 50. Putting it together, the printf prints the values 20 and 50, so the output is "20 50".

Answer

A. 10 50

Answer

B. 20 50

Answer

C. 30 40

Answer

D. 20 40

What will be the output of the following C code? #include void main() { int…