Question

In the following bit of code, pointer values and pointer addresses differ as expected.

But array values and addresses don’t!

How can this be?

Output

my_array = 0022FF00
&my_array = 0022FF00
pointer_to_array = 0022FF00
&pointer_to_array = 0022FEFC

#include <stdio.h>

int main()
{
  char my_array[100] = "some cool string";
  printf("my_array = %p\n", my_array);
  printf("&my_array = %p\n", &my_array);

  char *pointer_to_array = my_array;
  printf("pointer_to_array = %p\n", pointer_to_array);
  printf("&pointer_to_array = %p\n", &pointer_to_array);

  printf("Press ENTER to continue...\n");
  getchar();
  return 0;
}

Answer 1

The name of an array usually evaluates to the address of the first element of the array, so array and &array have the same value (but different types, so array+1 and &array+1 will not be equal if the array is more than 1 element long).

There are two exceptions to this: when the array name is an operand of sizeof or unary & (address-of), the name refers to the array object itself. Thus sizeof array gives you the size in bytes of the entire array, not the size of a pointer.

For an array defined as T array[size], it will have type T *. When/if you increment it, you get to the next element in the array.

&array evaluates to the same address, but given the same definition, it creates a pointer of the type T(*)[size] — i.e., it’s a pointer to an array, not to a single element. If you increment this pointer, it’ll add the size of the entire array, not the size of a single element. For example, with code like this:

char array[16];
printf("%p\t%p", (void*)&array, (void*)(&array+1));

We can expect the second pointer to be 16 greater than the first (because it’s an array of 16 char’s). Since %p typically converts pointers in hexadecimal, it might look something like:

0x12341000    0x12341010