Question

First off, I apologize if this has been asked before. I can’t seem to find the right info.

The following code does not print “300” as I thought it would:

#include <iostream>

int main()
{
  int *array;
  int *arrayCopy = array;

  array = new int[4];

  array[0] = 100;
  array[1] = 200;
  array[2] = 300;
  array[3] = 400;

  std::cout << arrayCopy[2];

  return 0;
}

However, it does, if I move the line

int *arrayCopy = array;

below the line that follows it in the above code. Why is that?

(PS: I know there is a memory leak, and that std::vector is better… I’m just curious).

Answer 1

Maybe you’re thinking of using a reference to a pointer? Here’s what happens with your current code:

int *array; // Currently points to an undefined (invalid) memory location.
int *arrayCopy = array; // Now this points to the same undefined memory location as array.

array = new int[4]; // Now array points to valid memory, but arrayCopy still points to undefined space.

If you did something like this though, it’s different:

int *array; // Points to undefined
int *&arrayCopy = array; // This is a reference to array. That means if you change array, arrayCopy will also reflect the changes.

array = new int[4]; // Now since array points to valid space, arrayCopy does too.

Technically, this isn’t entirely true because different things are happening. A reference is essentially the same level of indirection as a pointer, except the compiler does all the dereferencing for you. But what I described is essentially how it works. If you just stick the extra & in there, your code will do what you were thinking.