Is there a right way to return a new object instance by reference in C++?

18,500

Solution 1

You can't return a reference to a temporary object on the stack. You have three options:

  1. Return it by value
  2. Return by reference via a pointer to something that you created on the heap with the new operator.
  3. Return by reference what you received by reference as an argument. [EDIT: Thanks to @harshath.jr for pointing this out]

Note that when you return by value as in the code below, the compiler should optimize the assignment to avoid the copy - i.e. it will just create a single Rectangle (rect) by optimizing the create+assign+copy into a create. This only works when you create the new object when returning from the function.

Rectangle GetRect( void ) const
{
    return Rectangle( x, y, w, h );
}

Rectangle rect = theBox.GetRect();

Solution 2

No you cannot do this. Essentially what you're trying to do in this sample is return a reference to a temporary variable on the stack. By the time the reference is returned, the variable it's pointing to will be destroyed and hence the reference is invalid.

Solution 3

Returning an object by value (see example below) may actually be less expensive than you think. The compiler often optimizes out the extra copy. This is called the return value optimization.

    Rectangle GetRect( void ) const
    {
            return Rectangle( x, y, w, h );
    }

Solution 4

Is there a right way to return a new object instance by reference in C++?

No, not by reference. There are two ways to create a new object:

On the stack:

Rectangle makeRect()
{
  return Rectangle(x, y, w, h);
}
Rectangle r = makeRect(); // return by value

On the heap:

Rectangle * makeRect()
{
  return new Rectangle(x, y, w, y);
}
Rectangle * r = makeRect(); // returned a pointer, don't forget to delete it later

Why not something like this?

class Box
{
  private:
    Rectangle mRectangle;

  public:
    Box(float x, float y, float w, float h) :
      mRectangle(x, y, w, h) // Forgive me for making assumptions
                             // about the inner workings of your
                             // code here.
    {
    }

    const Rectangle & GetRect() const
    {
      return mRectangle;
    }
};

Rectangle rect = theBox.GetRect();

The 'assignment' should work now. (Technically this is not an assignment operator, but a copy constructor being invoked.)

Hoping to help

Solution 5

This is not possible. A reference is another form of a pointer and you in fact return an address of an object that will have been destroyed (destructor called ) and possibly even overwritten by the time the caller receives control.

You can either

  • call new and return a pointer (maybe you should think of a smart pointer) to a heap-allocated object or
  • return by value or
  • pass the object by reference into a function so it fills it.
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Patrick Hogan
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Patrick Hogan

Patrick has been fiddling with computers since he was a young boy. He has a wide range of experience in programming from game development and design to full-stack web development.

Updated on June 07, 2022

Comments

  • Patrick Hogan
    Patrick Hogan almost 2 years

    So I was writing some code, and I had something like this:

    class Box
    {
        private:
        float x, y, w, h;
    
        public:
        //...
        Rectangle & GetRect( void ) const
        {
            return Rectangle( x, y, w, h );
        }
    };
    

    Then later in some code:

    Rectangle rect = theBox.GetRect();
    

    Which worked in my debug build, but in release there were "issues" returning that Rectangle by reference -- I basically got an uninitialized rectangle. The Rectangle class has an = operator and a copy constructor. Without getting into why this broke, I'm actually more interested in the correct way to return a (new) object by reference for the purpose of assigning copying to a variable. Am I just being silly? Should it not be done? I know I can return a pointer and then dereference on assignment, but I'd rather not. Some part of me feels like returning by value would result in redundant copying of the object -- does the compiler figure that out and optimize it?

    It seems like a trivial question. I feel almost embarrassed I don't know this after many years of C++ coding so hopefully someone can clear this up for me. :)