Question

I’m messing around with multidimensional arrays and pointers. I’ve been looking at a program that prints out the contents of, and addresses of, a simple array. Here’s my array declaration:

int zippo[4][2] = { {2,4},
            {6,8},
            {1,3},
            {5,7}   };

My current understanding is that zippo is a pointer, and it can hold the address of a couple of other pointers. By default, zippo holds the address of pointer zippo[0], and it can also hold the addresses of pointers zippo[1], zippo[2], and zippo[3].

Now, take the following statement:

printf("zippo[0] = %p\n", zippo[0]);
printf("  *zippo = %p\n", *zippo);
printf("   zippo = %p\n", zippo);

On my machine, that gives the following output:

zippo[0] = 0x7fff170e2230
  *zippo = 0x7fff170e2230
   zippo = 0x7fff170e2230

I perfectly understand why zippo[0] and *zippo have the same value. They’re both pointers, and they both store the address (by default) of the integer 2, or zippo[0][0]. But what is up with zippo also sharing the same memory address? Shouldn’t zippo be storing the address of the pointer zippo[0]? Whaaaat?

Answer 1

When you declare a multidimensional array, the compiler treats it as a single dimensional array. Multidimensional arrays are just an abstraction to make our life easier. You have a misunderstanding: This isn’t one array pointing to 4 arrays, its always just a single contigous block of memory.

In your case, doing:

int zippo[4][2]

Is really the same as doing

int zippo[8]

With the math required for the 2D addressing handled for you by the compiler.

For details, see this tutorial on Arrays in C++.

This is very different than doing:

int** zippo

or

int* zippo[4]

In this case, you’re making an array of four pointers, which could be allocated to other arrays.