Pointers to arrays in C

Question

I just saw this code snippet Q4 here and was wondering if I understood this correctly.

#include <stdio.h>

int main(void)
{
   int a[5] = { 1, 2, 3, 4, 5 };
   int *ptr = (int*)(&a + 1);

   printf("%d %d\n", *(a + 1), *(ptr - 1));

   return 0;
}

Here's my explanation:

int a[5] = { 1, 2, 3, 4, 5 }; => a points to the first element of the array. In other words: a contains the address of the first element of the array.

int *ptr = (int*)(&a + 1); => Here &a will be a double pointer and point to the whole array. I visualize it like this: int b[1][5] = {1, 2, 3, 4, 5};, here b points to a row of a 2D array. &a + 1 should point to the next array of integers in the memory (non-existent) [kind of like, b + 1 points to the second (non-existent) row of a 2D array with 1 row]. We cast it as int *, so this should probably point to the first element of the next array (non-existent) in memory.

*(a + 1) => This one's easy. It just points to the second element of the array.

*(ptr - 1) => This one's tricky, and my explanation is probably flawed for this one. As ptr is an int *, this should point to int previous to that pointed by ptr. ptr points to the non-existent second array in memory. So, ptr - 1 should probably point to the last element of the first array (a[4]).

Answer 1

Here &a will be a double pointer.

No. It is a pointer to an array. In this example, int (*)[5]. Refer C pointer to array/array of pointers disambiguation

so when you increment pointer to an array, it will crosses the array and points to non-existent place.

In this example, It is assigned to integer pointer. so when int pointer is decremented, it will point to previous sizeof(int) bytes. so 5 is printed.