Please explain!

Question

How the output is 3 ?
int arr[]={1,2,3};
int *p=(int *)(&arr + 1);
printf("%d",*(p-1));

https://code.sololearn.com/cYANoAqimJYM/?ref=app

Answer 1

&arr is a pointer to the array (int(*)[3]). arr decays to a pointer to the first element of the array (int*). Both &arr and arr point to the same address, but their types differ so pointer arithmetic is performed differently. If instead you wrote arr + 1, you would have been correct.

When you perform pointer arithmetic p + n, you get a pointer that is moved n times the size of the type pointed to (sizeof(*p)).
The type pointed to by &arr is int[3] which is 12 bytes. &arr + 1 then points to a int[3] after the end of the array.

(int*)(&arr + 1) == arr + 3

(int *)(&arr + 1) converts the pointer to array (12 bytes) to pointer to int (4 bytes). p is then pointing to an int after the end of the array. p - 1 gives you a pointer to the last element of the array which is 3.

int* p = arr + 3;
printf("%d", *(p - 1)); // *(arr + 3 - 1) == *(arr + 2) == arr[2]

Answer 2

int *p=(int *)(arr + 1);
I think this will help you to find you answer.

Answer 3

Try this and u can answer by yourself.
int *p=(int *)(arr + 1);