How to “return an object” in 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 t

Answer 1

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 2

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 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

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).

Answer 5

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 6

Did you try to use smart pointers (if Thing is really big and heavy object), like auto_ptr:

std::auto_ptr<Thing> calculateThing()
{
  std::auto_ptr<Thing> thing(new Thing);
  // .. some calculations
  return thing;
}


// ...
{
  std::auto_ptr<Thing> thing = calculateThing();
  // working with thing

  // auto_ptr frees thing 
}