Why does this code print addresses?

Question

Why didn't I get a compile time error while accidentally printing only one dimension of a 2D array?

#include <stdio.h>

void main() {
    int i;
    int arr[2][3] = { 1, 2, 3, 4, 5, 6 }; //<- Declared a 2D array

    for (i = 0; i < 6; i++) {
        printf("%d\n", arr[i]);  // <- Accidently forgot a dimension
    }
}

I should have received a compile time error but instead I got a group of addresses! Why? What did arr[0] mean in this context to the compiler?

Answer 1

An expression with array type evaluates to a pointer to the first array element in most contexts (a notable exception, among others, is the sizeof operator).

In your example, arr[i] has array type. So it evaluates to a pointer of type int (*)[] (a pointer to an array). That's what's getting printed. Printing a pointer with %d is undefined behavior, because printf() will read the pointer as if it was an int.