c - Initializing "a pointer to an array of integers"

Question

 int (*a)[5];

How can we Initialize a pointer to an array of 5 integers shown above.

Is the below expression correct ?

int (*a)[3]={11,2,3,5,6}; 

Answer 1

Suppose you have an array of int of length 5 e.g.

int x[5];

Then you can do a = &x;

 int x[5] = {1};
 int (*a)[5] = &x;

To access elements of array you: (*a)[i] (== (*(&x))[i]== (*&x)[i] == x[i]) parenthesis needed because precedence of [] operator is higher then *. (one common mistake can be doing *a[i] to access elements of array).

Understand what you asked in question is an compilation time error:

int (*a)[3] = {11, 2, 3, 5, 6}; 

It is not correct and a type mismatch too, because {11,2,3,5,6} can be assigned to int a[5]; and you are assigning to int (*a)[3].

Additionally,

You can do something like for one dimensional:

int *why = (int p[2]) {1,2};

Similarly, for two dimensional try this(thanks @caf):

int (*a)[5] = (int p[][5]){ { 1, 2, 3, 4, 5 } , { 6, 7, 8, 9, 10 } };