What is the difference between that ways, when assign an array to pointer

Question

When we need to assign an array to pointer, we doing something like that.

int numbers[] = {7,5,9,3};
int *ptr = NULL;
ptr = &numbers[0];  // <<

Also we can do same thing by doing this.

int numbers[] = {7,5,9,3};
int *ptr = NULL;
ptr = numbers;  // <<

What is the difference between two ways ?
And which one is the Recommended ?

Answer 1

It's the same thing. The expression a[i] is the same as *(a + i), so &a[i] is the same as &*(a + i), which is defined to be identical to a + i (without even evaluating the dereference). For i = 0 this means that a is the same as &a[0]. The name of the array decays to a pointer to the first element of the array.

Answer 2

No difference (except when they are the oprand of sizeof operator). In both case you are assigning address of the first element of array. &numbers[0] assign the address of the first element of the array while numbers will decay to pointer to the first element of the array.

Answer 3

They will both give you the same value, however there is slightly a difference. When you do

ptr = &numbers[0];

you are adding two unnecessary steps, a dereference and a "address of" operation, so using

ptr = numbers;

would be more performant, unless the compiler optimizes it away. It may be useful to think of int numbers[] = ...; as int *numbers = ...