Question

#include <iostream>

int main()
{
  int anything[] = {5};
  int *something = new int;
  *something = 5;

  std::cout << &anything  << "==" << &anything[0]  << "==" << anything  << std::endl;
  std::cout << &something << "!=" << &something[0] << "==" << something << std::endl;
}

Why is the memory address in &something different from &something[0] and something? Although it is a dynamic allocation, I don’t understand why the memory address is different. I tried it with more than one value; it’s the same thing. Here I used one value for both for simplicity.

Answer 1

&something is the memory address of the pointer itself (hey, it needs to store that value somewhere!), while &something[0] is the address of the actual memory that is storing your stuff.