Getting Address of Constructed Item/Member in C++

Question

I'm writing part of a customer database that handles making new customers and selling them items. In "Customer.h", I have the struct:

struct Customer {
    String name;
    int bottles;
    Customer(String);
    Customer(void) { name = "__invalid__"; }
};

I have a class in the file "CustomerDB.h"

class CustomerDB {
private:
    Customer* data; // an array of Customer objects (allocated on the heap)
    int capacity; // size of the array on the heap
    int length; // number of valid customers actually in the array

public:
    CustomerDB(void);

There's a constructor, in "CustomerDB.cpp"

Customer::Customer(string name) {
    this->bottles = 0;
    this->name = name;
}

I create *an object in another function (which is in "CustomerDB.cpp")

Customer& CustomerDB::operator[](string name) {
     Customer Customer(name); 
     return Customer;

And there's an object of CustomerDB:

CustomerDB database; (Which is in another cpp file which handles the purchases).

The string input works. The object is created. There's no issues there. So I have 2 problems, one of which is because of the other.

1. I need to make a new non-Local Customer object in this function (stored into the non-Local Database), and
2. Return the reference to it. But the constructor I just called doesn't give me a reference. It just makes it, as it should. I'm unintentionally making a local object instead of one that's added to the "main" database. As a result, the reference is nothing useful.

If I try return Customer; it says that "Customer does not refer to a value."

Any help/advice is appreciated.

Answer 1

I need to return the address of this freshly made Customer

No, you need to return a reference to it.

but the constructor I just called

You never call constructors. You declared an object.

doesn't give me an address. How do I find/obtain this new Customer and return its address?

You take the address of an object like this:

return &Customer;

But to return a reference, it's simply:

return Customer;

---

But this isn't going to work!

You'd be returning a reference to a local variable. It's about to go out of scope, leaving the reference dangling.

So, you need to rethink your choices with this design.