c style pointer arithmetic?

[bluefrog]

Hi

Could somebody explain why the following 2 lines in the code below

cout << 2[arr] << "\n";
cout << 2[p1] << "\n";

are valid?

I understand all the other uses of pointers, but not the above 2.
Full code is:

#include <iostream>

using namespace std;
int main() {

int *p, i;
i = 10;
{
  int i = 5;
  int *p = &i;
  std::cout << "p " << *p << "\n";
}
std::cout << "Before assignment \n";
p = &i;
std::cout << "After assignment " << *p << "\n";
int arr[] = {1, 2, 3, 4, 5};
int *p1 = arr;

cout << p1[2] << "\n";
cout << 2[p1] << "\n";
cout << *(p1 + 2) << "\n";
cout << arr[2] << "\n";
cout << 2[arr] << "\n";
cout << *(arr + 2) << "\n";
return 0;
}

[Peter87]

2[arr] is just the same as arr[2].

arr[2]  =  *(arr + 2)  =  *(2 + arr)  =  2[arr]