why does this pointer arithmetic on a 2D array work?

Question

I wrote the following code:

#include <iostream>
using namespace std;

int main()
{
    int a[10][10];
    for (int i = 0; i < 10; i++)
        for (int j = 0; j < 10; j++)
            a[i][j] = i * j;
    cout << *(*(a + 3) + 4) << endl;
    return 0;
}

I was expecting it to print some garbage data, or a segmentation fault. What I got was 12. I tested it both in c and c++ (using gcc and g++ respectively), and I herd this works the same on VS although I haven't tested this. Why does this work, and is there a official documentation of this behavior?

Answer 1

When you want a simple answer build the complex type with typedefs

it means:

int a[10][10];

will be:

typedef int a10[10]; //a10 is array of 10 ints 
a10 a[10]; //a is array of 10 a10s

now to understand the sizes and position:

sizeof(a) = 10 * 10 * sizeof int
sizeof(a[0]) = is  10 * sizeof int
a+1 is equal to &a[1]

When you increment a pointer by 3 it is mean increment by sizeof of the type.

address of a[1] == address of a[0] + sizeof(a[0])

Therefore:

*(*(a + 3) + 4) == a[3][4]