Why is the copy constructor called twice in this code snippet?

  • A+
Category:Languages

I'm playing around with a few things to understand how copy constructors work. But I can't make sense of why the copy constructor is called twice for the creation of x2. I would have assumed it would be called once when the return value of createX() is copied into x2.
I also looked at a few related questions on SO, but as far as I can tell I couldn't find the same simple scenario as I am asking here.

By the way, I'm compiling with -fno-elide-constructors in order to see what's going on without optimizations.

#include <iostream>  struct X {     int i{2};      X() {         std::cout << "default constructor called" << std::endl;     }      X(const X& other) {         std::cout << "copy constructor called" << std::endl;     } };  X createX() {     X x;     std::cout << "created x on the stack" << std::endl;     return x; }  int main() {     X x1;     std::cout << "created x1" << std::endl;     std::cout << "x1: " << x1.i << std::endl << std::endl;          X x2 = createX();     std::cout << "created x2" << std::endl;     std::cout << "x2: " << x2.i << std::endl;          return 0; } 

This is the output:

default constructor called created x1 x1: 2  default constructor called created x on the stack copy constructor called copy constructor called created x2 x2: 2 

Can someone help me what I'm missing or overlooking here?

 


What you have to remember here is that the return value of a function is a distinct object. When you do

return x; 

you copy initialize the return value object with x. This is the first copy constructor call you see. Then

X x2 = createX(); 

uses the returned object to copy initialize x2 so that is the second copy you see.


One thing to note is that

return x; 

will try to move x into the return object if it can. Had you made a move constructor you would have seen this called. The reason for this is that since local objects go out of scope at the end of the function, the compiler treats the object as an rvalue and only if that does not find a valid overload does it fall back to returning it as an lvalue.

Comment

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen: