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}; ```

Accepted Answer

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 } };
```

Initializing "a pointer to an array of integers"

Solution

Problem