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In c, what is int(*)[]?

In some c code I inherited, I have seen the following

int (*b)[] = (int(*)[])a;

What is int(*)[] and how is this different than int**?

As per The ``Clockwise/Spiral Rule'' ,

int(*)[] is a pointer to an array of int.

int(*)[]       int[]
+---------+    +---------+
|      ------->|         |
+---------+    +---------+
               :         :

int** is a pointer to a pointer to an int.

int**          int*           int
+---------+    +---------+    +---------+
|      ------->|      ------->|         |
+---------+    +---------+    +---------+
               :?        :?   :?        :?

As you can see, int(*)[] is closer to int* than to int** .

int*           int
+---------+    +---------+
|      ------->|         |
+---------+    +---------+
               :?        :?

All of these have distinctively different meanings:

  • int* = pointer to int
  • int[] = array of int, with incomplete type since the size is missing.
  • int** = pointer to pointer to int .
  • int(*)[] = pointer to (incomplete) array of int

Notably an int** , pointer to pointer to int , cannot be assigned to point at an array, nor to an array of arrays. It has nothing to do with arrays: With one special exception: it can be assigned to point at the first element in an array of int* , that is an array of type int* arr[] . We may then write int** ptr = arr; in that special case and only then.

Most of the misconception regarding int** originates from the misconception that int** together with malloc could be used to allocate a 2D array, which was never the case, see Correctly allocating multi-dimensional arrays .

Regarding int(*)[] it's a handy type in some cases, since an array with no size (incomplete type) is compatible with an array of fixed size (through a special rule called "composite type"). So we can use a int(*)[] to point at any int array type no matter size. Or use it to create a macro to check if something is an array: #define IS_INT_ARRAY(x) _Generic(&(x), int(*)[]: puts(#x " is an int array"))

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