Array of pointers and Pointer to array

An array is known as the contiguous run of elements while a pointer is an address pointing variable.
A pointer could represent the same array.

   int arr[5];
   int *a;
   a = arr;

The compiler reads arr[2] as, get the base address that is 100, next add 2 as the pointer arithmetic got 108 and then dereference it. Hence we got 30.

It’s like *(arr + 2). In the same we can also write *(a + 2) or a[2], as a = arr means a = &arr[0].
Thus an array acts like a pointer but it’s not a pointer. The difference could be seen when both passed to sizeof method. sizeof(arr) gives 20 while sizeof(a) gives 4 (for 32 bit architecture). Therefore, array names in a C program are converted mostly to pointers, leaving cases like involving sizeof operator.

Array of Pointer and Pointer to array:

1. int *x[5]
2. int (*x)[5]

The first one is an array of pointer while the second one is a pointer to an array of 5 blocks. Let’s understand step by step.
So in this post, the table of Precedence and Associativity of operators will be used, so it is better to go through this post to have a better understanding.
Operator [] has higher priority than *.
So in int *x[5], first x[5] is read that says an array of 5 blocks. Next an asterisk * is read, means every block of the array is a pointer. See below,

This could be seen as there are 5 different pointer variables.

In int (*x)[5], as per the table both () and [] has the same precedence, associativity comes into the picture that says left to right reading. (*x)read first means a pointer. Next [5] that says an array of 5 blocks. Therefore, it’s a pointer that points to an array of 5 blocks.

In a 2-d or multidimensional Array

int z[3][5], this is a 2-d array i.e. three 1-d arrays of 5 blocks each. To create a pointer variable pointing to this first set of 5 blocks of this 2-d array, we’ll write x = &z[0] or in other ways x = z (because starting index of 2d array and first 1d array is same).

Doing x++ here, will give z[1] i.e. the starting index of 2nd 1-d array, and for accessing any particular block using x will be like a pointer to pointer accessing. Say z[1][4], it would be *(*(x+1)+4) or simply x[1][4].

Interesting Problems:

Try to find the answer by yourself, then run these problems and use solution hint/output.

Problem1.

int main(void) {
  int a[5];
  printf("%u %u %u %u\n", (void *)a, (void *)(a + 1), (void *)(&a),
         (void *)(&a + 1));
}

Solution Hint-

Here, a+1 and (&a+1), both are very different, a is like a pointer while &a is like pointer to an array of 5 blocks i.e. (*)[5]. Thus &a+1 result a jump of 5*4(for 32 bit).

Problem 2:

int main() {
  int arr[3][3] = {{1, 2, 3}, {4, 5, 6}, {0, 0, 0}};
  int(*p)[3] = arr;

  printf("%d  %d  %d\n", (*p)[0], *(p[1] + 1), (*p)[2]);

  ++p;

  printf("%d  %d  %d\n", (*p)[-1], (*p)[1], (*p)[2]);
  /* (*p)[-1] write it as *(*(p + 0)-1) */

  return 0;
}

Output:-
1  5  3
3  5  6

Problem 3:

int main(void) {
  int a[5] = {0, 1, 2, 3, 4};
  int *p[5] = {a, a + 2, a + 1, a + 4, a + 3};

  printf("%u %u %u %u %u\n", &a[0], &a[2], &a[1], &a[4], &a[3]);
  printf("%u %u %u %u %u", p[0], p[1], p[2], p[3], p[4]);

  return 0;
}

Solution Hint-

Both the printf statement will give the same output, as p is array of pointer.

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