Plz explain this c code

Question

#include <stdio.h>
int main()
{
int arr[3]={40,41,42};
int *ptr=(int*)(&arr+1);
printf("%d %d",*(arr+1),*(ptr-1));
return 0;
}

Answer 1

"&arr" is of type int (*)[3] "a pointer to an array of 3 ints".

pointer arithmetic says adding an integer N to a pointer to an element of an array with index K produces a pointer to an element of the same array, with index K+N

since “&arr” is pointer to array of 3 ints, addition of 1 resulted in an address with increment of (assuming sizeof int is 4 bytes) 4 x 3 = 12 (we end up to the addr of the next array of 3 ints if there was one ) which is decayed to int* (a pointer to an integer), subtraction of 1 resulted in an address with decrement of 4.

+-----+-----+-----+-----+-----+-----+
| 40 | 41 | 42 | | |
+-----+-----+-----+-----+-----+-----+
^ ^ ^ ^
| | | |
| | | |
| | | |
&arr | ptr-1 &arr + 1 == ptr
^ |
| |
arr arr + 1