Pointer to an 2D array increment in c

Question

#include<stdio.h>
int main()
{
    int arr[3][4] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
    printf("%u ,%u\n",arr, &arr+1);
    return 0;
}

In the print statement: (Assuming integer size 4 bytes)
Let arr's memory location be 1000

arr+1--gives 1016 (ie.next row memory)

1000 1004 1008 1012 1016

1     2     3    4    5

&arr+1 gives 1048

I understood the logic but I can't understand that how the compiler identifies
arr as 'pointer to an array of 4 integer' and &arr as 'pointer to array of 3 arrays of 4 integer'

Even though both points to the same address (1000) before increment

arr decays to the same type as &arr[0], which is an array of 4 integers. &arr on the other hand is a pointer to array of 3 by 4 integers — Eugene Sh.
– Eugene Sh., Commented Jun 23, 2021 at 16:25
&arr gives you a pointer to the array. arr gives you a pointer to the array's first element. — Steve Summit
– Steve Summit, Commented Jun 23, 2021 at 16:55

MikeCAT · Accepted Answer · 2021-06-23 16:25:35Z

arr is a 3-element array of 4-element array of int.

Arrays in expressions are automatically converted to pointers pointing at the first element of the array (some exception exists). Therefore arr is converted to a pointer pointing at its first element. The type of the first element is "a 4-element array of int, so arr is converted to "a pointer to a 4-element array of int" (int(*)[4]).

Operand of unary & (address) operator is one of the exceptions. arr is a 3-element array of 4-element array of int, so &arr is "a pointer to a 3-element array of 4-element array of int".

Note that the line

printf("%u ,%u\n",arr, &arr+1);

invokes undefined behavior by passind data having wrong type to printf(). %u expects unsigned int. You should cast the pointer to void* and use %p to print pointers like this:

printf("%p ,%p\n",(void*)arr, (void*)(&arr+1));

Alternatively, you can convert the pointers to integers (in an implementation-defined manner) and print them.

#include<stdio.h>
#include<inttypes.h>
int main(void)
{
    int arr[3][4] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
    printf("%" PRIuPTR " ,%" PRIuPTR "\n",(uintptr_t)arr, (uintptr_t)(&arr+1));
    return 0;
}

Vlad from Moscow · Accepted Answer · 2021-06-23 16:36:16Z

1

Arrays designates used in expressions with rare exceptions are converted to pointers to their first elements.

So the array arr declared like

int arr[3][4] = { /*...*/ };

is converted to pointer to its first element when it is used as an argument of a call of printf. The type of the array element is int[4]. So a pointer to an object of this type has the type int ( * )[4].

On the other hand the expression &arr has the type int ( * )[3][4]. Incrementing the pointer like &arr + 1 yields the value that greater than the value of the expression &arr by sizeof( int[3][4] ) (that is equal to 12 * sizeof( int )).

Pay attention to that you have to use the conversion specifier %p to output values of pointers.

printf("%p ,%p\n", (void * )arr, ( void * )( &arr+1 ) );

edited Jun 23, 2021 at 16:36

answered Jun 23, 2021 at 16:30

Vlad from Moscow

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