Are the parentheses required when declaring a pointer to an array?

Question

I was just looking at this question:

How to assign a multi-dimensional array to a temporary variable?

The solution ended up using the lines:

int a[3][2] = {{1, 2}, {11, 12}, {21, 22}};
...
int (*b)[2] = a;

to "assign a statically allocated, multi-dimensional array to a temporary variable."

I'm a little confused about the syntax of the line:

int (*b)[2] = a;

In this instance, are the parentheses required to get the right effect, and if so, why? Is there a way to get the same result without using them?

Answer 1

This:

int (*b)[2]

declares b as a pointer to an array of two ints. This is not the same as:

int *b[2]

which declares b as an array of two pointers-to-int.

You need the first form in order to correctly perform pointer arithmetic.