CODEWITHSUNDEEP
carrayspointers
aditya_medhe
aditya_medhe

Reputation: 372

Dereferencing pointer to entire array

I'm learning pointers to (entire) arrays in C.

Suppose I declare a 2d matrix of ints:

int arr[3][3] = 
    {
        {1, 2, 3},
        {4, 5, 6},
        {7, 8, 9}
    };

Now, I declare a pointer of appropriate type:

int (*ptr)[3][3];

Initialize it:

ptr = &arr;

Now, ptr contains the address of arr.

  ptr --->&(arr[0][0]   arr[0][1]   arr[0][2]
            arr[1][0]   arr[1][1]   arr[1][2]
            arr[2][0]   arr[2][1]   arr[2][2])

So, when we dereference ptr, it should yield the first location, isn't it?

printf("%d", *ptr) gives an error at compile time. To print the first element, I've to use ***ptr.

I have 2 questions:

  1. I can't understand the need for a triple dereferencing.
  2. If *ptr doesn't point to first element of array, what does it point to?

Upvotes: 0

Views: 1124

Answers (1)

R Sahu
R Sahu

Reputation: 206567

printf(*ptr);

is wrong since the first argument to printf needs to be a string that specifies the format to use to print the rest of the arguments. To print the address, use:

printf("%p", *ptr);

I can't understand the need for a triple dereferencing.

The type of p is int (*)[3][3].

The type of *p is int [3][3].

The type of **p is int [3]

The type of ***p is int.

When you have a pointer like that, it's best to use the array syntax.

(*p)[0][0]

The generic form would be:

(*p)[i][j]

That is a lot less confusing than using ***p. Besides, ***p can be used to access the [0][0]-th element of the array only.

If *ptr doesn't point to first element of array, what does it point to?

Hopefully the answer to the previous question explains this.

Upvotes: 2

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