How does sizeof work with this dereferencing of a pointer to array?

Question

Here I have a pointer ptr to array arr of 4 integers. ptr points to the whole array. ptr[0] or *ptr points to the first element of the array, so adding 1 to ptr[0] gives the address of the second element of the array.

I can't understand why using sizeof(ptr[0]) gives the size of the whole array, 16 bytes, not the size of only the first element, 4 bytes, (as ptr[0] points to the first element in the array).

int arr[4] = {0, 1, 2, 3};
int (*ptr)[4] = &arr;
printf("%zd", sizeof(ptr[0])); //output is 16

Answer 1

OP: ptr[0] points to the first element in the array.

Type confusion. ptr[0] is an array.

ptr is a pointer to array 4 of int.
ptr[0], like *ptr deferences the pointer to an array.
sizeof(ptr[0]) is the size of an array.

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With sizeof(ptr[0]), ptr[0] does not incur "an expression with type ‘‘pointer to type’’ that points to the initial element of the array object" conversion. (c11dr §6.3.2.1 3). With sizeof, ptr[0] is an array.

Answer 2

ptr here is of type pointer to an array of 4 int elements and the array type has size 16 on your platform (sizeof(int) * (number of elemetns)).

I can't understand why using sizeof(ptr[0]) gives the size of the whole array 16 bytes not the size of only the first element 4 bytes

because C type system has array types. Here both arr and *ptr has it. What you declare that you have. To get sizeof int here you should sizeof(ptr[0][0]) - where ptr[0] evaluates to array.