Returning a local object from a function

Question

Is this the right way to return an object from a function?

Car getCar(string model, int year) {
   Car c(model, year);
   return c;
}

void displayCar(Car &car) {
   cout << car.getModel() << ", " << car.getYear() << endl;
}

displayCar(getCar("Honda", 1999));

I'm getting an error, "taking address of temporary". Should I use this way:

Car &getCar(string model, int year) {
   Car c(model, year);
   return c;
}

Solution 1

getCar returns a Car by value, which is correct.

You cannot pass that return value, which is a temporary object, into displayCar, because displayCar takes a Car&. You cannot bind a temporary to a non-const reference. You should change displayCar to take a const reference:

void displayCar(const Car& car) { }

Or, you can store the temporary in a local variable:

Car c = getCar("Honda", 1999);
displayCar(c);

But, it's better to have displayCar take a const reference since it doesn't modify the object.

Do not return a reference to the local Car variable.

Solution 2

It is not safe to return a local variable's reference from a function.

So yes this is correct:

Car getCar(string model, int year) {
   Car c(model, year);
   return c;
}

Solution 3

Yes, it is definitely not safe to return a reference or a pointer to a temporary object. When it expires (ie, when the function getCar exits), you'll left with what is technical known as a "dangling pointer".

However, if you're keen on reducing copy operations on an object, you should check out C++0x's "Move semantics". It is a relatively new concept, but I'm sure it'll become main-stream soon enough. GCC 4.4 upwards supports C++0x (use the compiler option -std=c++0x to enable).