Pointer arithmetic in multidimensional array

Question

How does pointer arithmetic - specifically, addition with numbers - work if the pointer points to a multidimensional array?

When I debugged the sample program below:

int main(void) {
    int a[3][4] = {{1, 2, 3, 4},
                 {5, 6, 7, 8},
                 {9, 10, 11, 12}};
    int *p1 = a;
    // set breakpoint ...
    ...
}

If I use debugger (CLion) to print *(p1 + N), it gives me an array having (N + 1) elements, starting from element a[N / col_size][N % col_size]. This doesn't match my expectation. Listing actual result and expectation as below:

Enter into debugger to print	Debugger result	Expectation
*(p1)	{1}	{1, 2, 3, 4}
*(p1 + 2)	{3, 4, 5}	{9, 10, 11, 12}
*(p1 + 5)	{6, 7, 8, 9, 10, 11}	{an array of random numbers}
*(p1 + 6)	{7, 8, 9, 10, 11, 12, -11}. -11 is random as it already exceeded the allocated size of the array.	{an array of random numbers}

I understand that C stores multidimensional array in row-major order, but what confuses me is why does C provide an array of (N + 1) length for *(p1 + N)?

Answer 1

This is wrong:

int *p1 = a;

a degenerates into a pointer to its first element. It's first element is an int[4] so a degenerates into an int (*)[4]. gcc says as much:

warning: initialization of 'int *' from incompatible pointer type 'int (*)[4]'

You're assigning it to an int *, which strict aliasing violation and therefore undefined behaviour.

If you had used the correct type, the output you would received would have been more in line with your expectations.