c++ - Why is arr and &arr the same?

Question

I have been programming c/c++ for many years, but todays accidental discovery made me somewhat curious... Why does both outputs produce the same result in the code below? (arr is of course the address of arr[0], i.e. a pointer to arr[0]. I would have expected &arr to be the adress of that pointer, but it has the same value as arr)

  int arr[3];
  cout << arr << endl;
  cout << &arr << endl;

Remark: This question was closed, but now it is opened again. (Thanks ?)

I know that &arr[0] and arr evaluates to the same number, but that is not my question! The question is why &arr and arr evaluates to the same number. If arr is a literal (not stored anyware), then the compiler should complain and say that arr is not an lvalue. If the address of the arr is stored somewhere then &arr should give me the address of that location. (but this is not the case)

if I write

const int* arr2 = arr;

then arr2[i]==arr[i] for any integer i, but &arr2 != arr.

Answer 1

#include <cassert>

struct foo {
    int x;
    int y;
};

int main() {    
    foo f;
    void* a = &f.x;
    void* b = &f;
    assert(a == b);
}

For the same reason the two addresses a and b above are the same. The address of an object is the same as the address of its first member (Their types however, are different).

                            arr
                      _______^_______
                     /               
                    | [0]   [1]   [2] |
--------------------+-----+-----+-----+--------------------------
      some memory   |     |     |     |        more memory
--------------------+-----+-----+-----+--------------------------
                    ^
                    |
           the pointers point here

As you can see in this diagram, the first element of the array is at the same address as the array itself.