How to "return an object" in C++? - c++

Question

I know the title sounds familiar as there are many similar questions, but I'm asking for a different aspect of the problem (I know the difference between having things on the stack and putting them on the heap).
In Java I can always return references to "local" objects
public Thing calculateThing() {
Thing thing = new Thing();
// do calculations and modify thing
return thing;
}
In C++, to do something similar I have 2 options
(1) I can use references whenever I need to "return" an object
void calculateThing(Thing& thing) {
// do calculations and modify thing
}
Then use it like this
Thing thing;
calculateThing(thing);
(2) Or I can return a pointer to a dynamically allocated object
Thing* calculateThing() {
Thing* thing(new Thing());
// do calculations and modify thing
return thing;
}
Then use it like this
Thing* thing = calculateThing();
delete thing;
Using the first approach I won't have to free memory manually, but to me it makes the code difficult to read. The problem with the second approach is, I'll have to remember to delete thing;, which doesn't look quite nice. I don't want to return a copied value because it's inefficient (I think), so here come the questions
Is there a third solution (that doesn't require copying the value)?
Is there any problem if I stick to the first solution?
When and why should I use the second solution?

Answer 1

I don't want to return a copied value because it's inefficient
Prove it.
Look up RVO and NRVO, and in C++0x move-semantics. In most cases in C++03, an out parameter is just a good way to make your code ugly, and in C++0x you'd actually be hurting yourself by using an out parameter.
Just write clean code, return by value. If performance is a problem, profile it (stop guessing), and find what you can do to fix it. It likely won't be returning things from functions.
That said, if you're dead set on writing like that, you'd probably want to do the out parameter. It avoids dynamic memory allocation, which is safer and generally faster. It does require you have some way to construct the object prior to calling the function, which doesn't always make sense for all objects.
If you want to use dynamic allocation, the least that can be done is put it in a smart pointer. (This should be done all the time anyway) Then you don't worry about deleting anything, things are exception-safe, etc. The only problem is it's likely slower than returning by value anyway!

Answer 2

Just create the object and return it
Thing calculateThing() {
Thing thing;
// do calculations and modify thing
return thing;
}
I think you'll do yourself a favor if you forget about optimization and just write readable code (you'll need to run a profiler later - but don't pre-optimize).

Answer 3

Just return a object like this:
Thing calculateThing()
{
Thing thing();
// do calculations and modify thing
return thing;
}
This will invoke the copy constructor on Things, so you might want to do your own implementation of that. Like this:
Thing(const Thing& aThing) {}
This might perform a little slower, but it might not be an issue at all.
Update
The compiler will probably optimize the call to the copy constructor, so there will be no extra overhead. (Like dreamlax pointed out in the comment).

Answer 4

Did you try to use smart pointers (if Thing is really big and heavy object), like shared_ptr:
std::shared_ptr calculateThing()
{
std::shared_ptr<Thing> thing(new Thing);
// .. some calculations
return thing;
}
// ...
{
std::shared_ptr<Thing> thing = calculateThing();
// working with thing
// shared_ptr frees thing
}

Answer 5

One quick way to determine if a copy constructor is being called is to add logging to your class's copy constructor:
MyClass::MyClass(const MyClass &other)
{
std::cout << "Copy constructor was called" << std::endl;
}
MyClass someFunction()
{
MyClass dummy;
return dummy;
}
Call someFunction; the number of "Copy constructor was called" lines that you will get will vary between 0, 1, and 2. If you get none, then your compiler has optimised the return value out (which it is allowed to do). If you get don't get 0, and your copy constructor is ridiculously expensive, then search for alternative ways to return instances from your functions.

Answer 6

Firstly you have an error in the code, you mean to have Thing *thing(new Thing());, and only return thing;.
Use shared_ptr<Thing>. Deref it as tho it was a pointer. It will be deleted for you when the last reference to the Thing contained goes out of scope.
The first solution is very common in naive libraries. It has some performance, and syntactical overhead, avoid it if possible
Use the second solution only if you can guarantee no exceptions will be thrown, or when performance is absolutely critical (you will be interfacing with C or assembly before this even becomes relevant).

Answer 7

I don't want to return a copied value because it's inefficient
This may not be true. Compilers can do optimisation to prevent this copying.
For example, GCC does this optimisation. In the following program, neither move constructor nor copy constructor are called, since no copying or moving is done. Also, notice the address of c. Even though the object c is instantiated inside the function f(), c resides in the stack frame of main().
class C {
public:
int c = 5;
C() {}
C(const C& c) {
cout << "Copy constructor " << endl;
}
C(const C&& c) noexcept {
cout << "Move Constructor" << endl;
}
};
C f() {
int beforeC;
C c;
int afterC;
cout << &beforeC << endl; //0x7ffee02f26ac
cout << &c << endl; //0x7ffee02f2710 (notice: even though c is instantiated inside f(), c resides in the stack frame of main()
cout << &afterC << endl; //0x7ffee02f26a8
return c;
}
C g() {
C c = f(); ///neither copy constructor nor move constructor of C are called, since none is done
cout << &c << endl; //0x7ffee02f2710
return c;
}
int main() {
int beforeC;
C c = g(); ///neither copy constructor nor move constructor of C are called, since none is done
int afterC;
cout << &beforeC << endl; //0x7ffee02f2718
cout << &c << endl; //0x7ffee02f2710 (notice:even though c is returned from f,it resides in the stack frame of main)
cout << &afterC << endl; //0x7ffee02f270c
return 0;
}

Answer 8

I'm sure a C++ expert will come along with a better answer, but personally I like the second approach. Using smart pointers helps with the problem of forgetting to delete and as you say, it looks cleaner than having to create an object before hand (and still having to delete it if you want to allocate it on the heap).