Q:

How is an array aligned in C++ compared to a type contained? 2

Following the topic How is an array aligned in C++ compared to a type contained? I made an experiment.
Here is the code:
#include<iostream>
using namespace std;

int main()
{
  const int N = 12;
  {
    float p1[N], p2[N];

    cout << p1 << " " << p2 << " " << p2 - p1 << endl;
  }

  {
    float* p1, *p2;

      // allocate memory
    p1 = new float[N];
    p2 = new float[N];

    cout << p1 << " " << p2 << " " << p2 - p1 << endl;
    delete[] p1;
    delete[] p2;
  }
}

According to the cited question and wiki I would expect that p1 and p2 would be sizeof(float) == 4 bytes aligned. But the result is:
0x7fff4fd2b410 0x7fff4fd2b440 12
0x7f8cc9c03bd0 0x7f8cc9c03c00 12

Same 12 distance between arrays for N = 9, 11 and 12. Distance (p2-p1) is 8 for N = 8.
So it looks like float arrays is 16 bytes aligned. Why?
P.S. 
My processor is Intel Core i7
Compiler - g++ 4.6.3

It appears that putting arrays in a structure one can get 10 floats distance:
const int N = 10;

struct PP{
  float p1[N], p2[N];
};

int main() {

  PP pp;
  cout << pp.p1 << " " << pp.p2 << " " << pp.p2 - pp.p1 << endl;
}

0x7fff50139440 0x7fff50139468 10

A:

Memory allocated by operator new always has suitable alignment for any object type, whatever the actual type being created. Additionally, it may also have a wider alignment (for example, alignment with a cache line) to improve performance.
If you replace your dynamic arrays with automatic ones (or better still, make them consecutive members of a class), then you may see narrower alignments. Or you may not; the exact details of how objects are aligned are up to the compiler, as long as it meets the minimum requirement for the type.

