Pointers with multi dimensional arrays in C

Question

int array[2][3] = {{2,3,6},{4,5,8}};

printf("%d\n",*array);

What will be the output of this and please explain how?

Regards,

Winston

Answer 1

Educate yourself on multidimensional arrays.

Array array is a 2D array:

int array[2][3] = {{2,3,6},{4,5,8}};

*array is the first element of array array because

*array -> * (array + 0) -> array[0]

The first element of array array is array[0], which is {2,3,6}. The type of array[0] is int [3]. When you access an array, it is converted to a pointer to first element (there are few exceptions to this rule).

So, in this statement

printf("%d\n",*array);

*array will be converted to type int *. The format specifier %d expect the argument of type int but you are passing argument of type int *. The compiler must be throwing warning message for this. Moreover, wrong format specifier lead to undefined behaviour.

If you want to print a pointer, use %p format specifier. Remember, format specifier %p expect that the argument shall be a pointer to void, so you should type cast pointer argument to void *.