Behavior of 2D arrays

Question

I have created a 2D array, and tried to print certain values as shown below:

int a[2][2] = { {1, 2}, 
                {3, 4}};

printf("%d %d\n", *(a+1)[0], ((int *)a+1)[0]);

The output is:

3 2

I understand why 3 is the first output (a+1 points to the second row, and we print its 0th element.

My question is regarding the second output, i.e., 2. My guess is that due to typecasting a as int *, the 2D array is treated like a 1D array, and thus a+1 acts as pointer to the 2nd element, and so we get the output as 2.

Are my assumptions correct or is there some other logic behind this?
Also, originally what is the type of a when treated as pointer int (*)[2] or int **?

Answer 1

When you wrote expression

(int *)a

then logically the original array can be considered as a one-dimensional array the following way

int a[4] = { 1, 2, 3, 4 };

So expression a points to the first element equal to 1 of this imaginary array. Expression ( a + 1 ) points to the second element of the imaginary array equal to 2 and expression ( a + 1 )[0] returns reference to this element that is you get 2.