Why an array pointer with index returns the value in that index instead of returning the specific index adress?

Question

i was practicing pointers and i saw that when i define a pointer and assign an array to that pointer, i was able to get the value in a specific index when i use the pointer name with that index instead of array name, like,

  int arr[5] = {1,2,3,4,5};
  int *ptr;

  ptr = arr;
  
  printf("%d", ptr[1]); // prints out 2 
 

I was expecting to see the value when i use *ptr[1] and the specific index adress when i use ptr[1], but when i use *ptr[1] i get compiler error. I thought pointer name keeps the adress and using the name with * gives the value in that adress.

Am i missing something here? Why pointer with array works that way?

Answer 1

The misunderstanding here is that pointers and arrays have similar behaviours in C, as in you can treat a pointer like an array, and an array like a pointer.

In effect x[n] is the same as *(x + n) and vice-versa. x[0] is just *x .

As such, ptr[1] will return a de-referenced int* or in other words an int .

If you want the actual address you need to do either ptr + n or &ptr[n] , both of which are equivalent, they're int* .