Returning a local object from a function

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". Shou开发者_运维技巧ld I use this way:

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

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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.

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Your problem is:

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

you should use a const reference:

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

This function:

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

is OK as it stands, but all it is doing is what the constructor does, so it is redundant. Passing a value back, rather than a reference, is the right thing to do for this type of function, however, The model parameter should be passed by const reference:

Car getCar( const string & model, int year) {

In general, for class types like string or Car, your default choice for a parameter should always be a const reference.

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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;
}

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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).

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Even better:

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