Dereferencing pointer to entire array

Question

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:

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

Why would you expect *&arr to refer to arr[0][0] instead of arr? That's like expecting *&some_int to refer to the int's first byte, rather than the entire int. — user2357112
– user2357112, Commented Oct 25, 2015 at 5:55
I am confused because int is a primitive type whereas array is a collection of primitives. Printing the array will require looping over all it's elements, unlike an int. — aditya_medhe
– aditya_medhe, Commented Oct 25, 2015 at 7:54

R Sahu · Accepted Answer · 2015-10-25 05:58:41Z

2

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.

edited Oct 25, 2015 at 5:58

answered Oct 25, 2015 at 5:53

R Sahu

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4 Comments

aditya_medhe Over a year ago

Thanks for your explanation! Got it.

tstanisl Over a year ago

%p expects void*. So you trigger UB. It should be printf("%p", (void*)*ptr);

R Sahu Over a year ago

@tstanisl, Any object pointer can be used where void* is expected. I don't understand why leaving out the explicit cast would cause UB.

tstanisl Over a year ago

@RSahu, because C standard does not require pointer types to have the same binary representation as void*. Serializing pointer of type int* and deserializing it as void* is not defined by the standard. Thus one must do a cast to be portable

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