Printing the address of the first element of 2D array in C

Question

#include <stdio.h>

int main ()  
{
    int  arr[4][5] = {{1, 2, 3, 4, 5},
                      {6, 7,8, 9, 10},
                      {11, 12, 13, 14, 15},
                      {16, 17,18, 19, 20}
                     };
    printf("%p\n", arr);
    printf("%p\n",*arr);

    return(0);
}

My doubt is that when I print arr and *arr it is printing same value. Why is this happening?

can you please check again, I've updated. – Sanjay Verma Apr 06 '20 at 08:27 — Sanjay Verma, Apr 06 '20 at 08:27

haccks · Accepted Answer · 2020-04-06T08:45:53.280

arr is an array of arrays of int. As per the array to pointer conversion rule, arr in printf(“%p\n”, arr); will decay to pointer to its first element. First element of array arr is of type int [5], i.e an array of 5 int. Therefore, it will print the address of first element of the array arr (address of arr[0]).

Dereferencing, arr will return the first element of arr. This first element is of type int [5] as discussed earlier. So *arr will return an array of 5 int but again as per the array to pointer conversion rule *arr will decay to pointer to it's first element which is arr[0][0].

So, in fact first printf is printing the address of the array arr[0] while later prints the address of the element arr[0][0].

To understand why both these addresses are same I would suggest you to read this post.

Man you're awesome your comment cleared all of my doubt..!!!! Thanks for the Answer. :) — Sanjay Verma, Apr 06 '20 at 08:46

RobertS supports Monica Cellio · Answer 2 · 2020-04-06T09:34:44.010

0

printf("%p\n", arr);
printf("%p\n",*arr);

At the first example, arr decays to a pointer to the first element in the first dimension of the two-dimensional array arr (the first int array of 5 elements - type (int[5])) which also decays to a pointer to its first element which is a[0][0].

At the second example you explicitly address the first element of the first dimension (which is of type int[5] - int array of 5 elements) and this one also decays to a pointer to its first element, which is again a[0][0].

So both have exactly the same result - Access and print the address of the first element of the first array at the first dimension of arr - arr[0][0].

Note that you need to cast the pointer to void, to make the program C-standard-conform as the standard implies that the %p format specifier needs to have an associated argument of type void*:

printf("%p\n", (void*) arr);
printf("%p\n", (void*) *arr);

Quote from ISO:IEC 9899:2018 (C18), Section 7.21.6.1/8:

"p - The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner."

edited Apr 06 '20 at 09:34

answered Apr 06 '20 at 08:30

RobertS supports Monica Cellio

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why do they have same address i mean ? can you share any where i can study this deeply? – Sanjay Verma Apr 06 '20 at 08:32
For example [here](https://en.wikibooks.org/wiki/C_Programming/Pointers_and_arrays) you can find some information. Basically an array is a constant pointer, which is auto-dereferenced each time you access it. – Alex Sveshnikov Apr 06 '20 at 08:35
@SanjayVerma The array `arr´ decays to a pointer to the first elemet of array `arr`. The first element always has the same address, doesn´t matter if you explicitly address or if address it via the array to pointer decay. – RobertS supports Monica Cellio Apr 06 '20 at 08:36
@RobertSsupportsMonicaCellio; `arr` and `*arr` not giving the address of the same element. Though address is same, both giving the address of different objects. Your answer is misleading. – haccks Apr 06 '20 at 08:48
@haccks Hmm, I edited my answer a little bit. You mean `arr[0]` addresses is a different object than `a[0][0]` ? - I think it is the same object even though its representation is different. Do you got any reference to that from the standard? – RobertS supports Monica Cellio Apr 06 '20 at 09:12

score 0 · Answer 3 · answered Apr 06 '20 at 12:01

because you should know that name of any array means the first address of this array say

 int arr[4]; 
printf("%p",arr);// the same as printf("%p",&arr[0]);

and with respect to 2D array also the name of array acts as visuals pointer which points to the address of the first element in array

int arr[4][5]; 
printf("%p",&arr[0][0]);//the same as printf("%p",arr);=printf("%p"*arr);
```````````````

Printing the address of the first element of 2D array in C

3 Answers3