A pointer to 2d array

int a[2][3];

a is read as an array 2 of array 3 of int which is simply an array of arrays. When you write,

int (*p)[3] = a;

It declares p as a pointer to the first element which is an array. So, p points to the array of 3 ints which is a element of array of arrays.

Consider this example:

        int a[2][3]
+----+----+----+----+----+----+
|    |    |    |    |    |    |
+----+----+----+----+----+----+
\_____________/
       |
       |    
       |
       p    int (*p)[3]

Here, p is your pointer which points to the array of 3 ints which is an element of array of arrays.

Rather than referring to int[2][3] as a '2d array', you should consider it to be an 'array of arrays'. It is an array with two items in it, where each item is itself an array with 3 ints in it.

int (*p)[3] = a;

You can use p to point to either of the two items in a. p points to a three-int array--namely, the first such item. p+1 would point to the second three-int array. To initialize p to point to the second element, use:

int (*p)[3] = &(a[1]);

The following are equivalent ways to point to the first of the two items.

int (*p)[3] = a; // as before
int (*p)[3] = &(a[0]);

Stricly speaking, no, int (*p)[3] = a; is not a pointer to a. It is a pointer to the first element of a. The first element of a is an array of three ints. p is a pointer to an array of three ints.

A pointer to the array a would be declared thus:

int (*q)[2][3] = &a; 

The numeric value of p and q are likely (or maybe even required to be) the same, but they are of different types. This will come into play when you perform arithmetic on p or q. p+1 points to the second element of array a, while q+1 points to the memory just beyond the end of array a.

Remember: cdecl is your friend: int a[2][3], int (*q)[2][3].