In C, why is dereferenced array pointer identical to the pointer itself?

Question

In the code below, I declare an int array "a" and then a pointer to the entire array "ap":

int main(){
        int a[3] = {10, 100, 1000};
        int (*ap)[] = &a;

        printf("%p\n", ap);
        printf("%p\n", *ap);
        printf("%d\n", **ap);

        return 0;
}

The first print of ap shows that it is actually a pointer to the first value of the array. OK.
But in the second print, ap is dereferenced and, yet, the value is the same: pointer to the first value of the array.
It's only after a "double dereferencing" that I get the value of the first array element (10 in my example):

0x7fff193310cc
0x7fff193310cc
10

Why is that so?
Thanks a lot!

Answer 1

When you do *ap it's basically the same as doing *(&a), which is in essence plain a.

And as arrays decays to pointers to their first elements, plain a is the same as &a[0].

Now here's the interesting thing: The location of the array a and the location of the first element of a is the exact same location.

It's easy to see if we draw it out:

+------+------+------+
| a[0] | a[1] | a[2] |
+------+------+------+
^
|
&a[0]
|
&a

See how both &a[0] and &a are pointing to the same location?

But also remember that while both &a[0] and &a are pointing to the same location, they have very different types:

• &a[0] will have the type int *
• &a will have the type int (*)[3]