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?

Answer 1

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] , ie 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 .

Answer 2

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."

Answer 3

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);
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