Pointer Arithmetic in C using Array Variables

Question

I am in the process of learning C, and have begun exploring the world of pointers and pointer arithmetic. For example, in the following code snippet:

int nums[] = {1, 2, 3};

nums is an Array variable and acts like a pointer that points to the first memory location of the array. I wrote the following sample code and am trying to understand why I am getting the results that I am getting:

#include <stdio.h>
#include <stdlib.h> 

int main() 
{
  int nums[] = {1, 2, 3};

  if(nums == &nums)
    puts("nums == &nums");
  else
    puts("nums != &nums");

  if((nums + 1) == (&nums + 1))
    puts("(nums + 1) == (&nums + 1)");
  else
    puts("(nums + 1) != (&nums + 1)");    

  printf("nums: %i\n", nums);
  printf("&nums: %i\n", &nums);
  printf("nums + 1: %i\n", nums + 1);
  printf("&nums + 1: %i\n", &nums + 1);

  return 0;    
}

I am getting that nums == &nums is true as expected; however, when I apply pointer arithmetic and add 1 to nums this result does not equal &nums + 1. In other words (nums + 1) != (&nums + 1) even though nums == &nums.

This is the output of the program that I get:

nums == &nums
(nums + 1) != (&nums + 1)
nums: 2345600
&nums: 2345600
nums + 1: 2345604
&nums + 1: 2345612

It appears that nums and nums + 1 are off set by 4 bytes; however, &nums and &nums + 1 are offset by 12. Why is it that this offset is by 12 bytes and not by 4?

Answer 1

The confusion is related to how in C, arrays implicitly decay into pointers in certain contexts.

The easier one to explain, nums + 1 effectively means &nums[0] + 1. nums[0] is type int, which is 4 bytes per element. Thus &nums[0] + 1 is 4 bytes after &nums.

As for &nums + 1, &nums is of type int(*)[3], which is 12 bytes per element. Thus &nums + 1 is 12 bytes after &nums.