How does de-referencing work for pointer to an array?

Question

Pointer to array of elements when dereferenced return an address. Since it is holding the address of the first element of the array, dereferencing it should return a value.

Answer 1

Here is a simple, less technical explanation.

How are you setting p? = arr;

How are you setting ptr? = &arr;

Your question really has nothing to do with arrays at all. arr could be literally any type and the answer would be the same: & gets the address of arr, so whatever arr is, ptr is storing its address while p is storing arr itself.

If you do *ptr, you will dereference that address and thus get a value equal to p.

int arr = 3;
// clearly these are different!
int p = arr;
int* ptr = &arr;

Similarly

int x = 3;
int* arr = &x;
// clearly these are different!
int* p = arr;
int** ptr = &arr;
// so of course they dereference differently
printf("*p = %d, *ptr = %p\n", *p, *ptr);