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

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14















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 i2;

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?










share|improve this question




























    14















    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 i2;

    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?










    share|improve this question


























      14












      14








      14


      1






      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 i2;

      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?










      share|improve this question
















      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 i2;

      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?







      c++ c++14 copy-constructor






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 28 at 16:08









      NathanOliver

      96.8k16137211




      96.8k16137211










      asked Feb 28 at 15:37









      c_studentc_student

      32311




      32311






















          2 Answers
          2






          active

          oldest

          votes


















          19














          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.






          share|improve this answer




















          • 1





            @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

            – NathanOliver
            Feb 28 at 15:50






          • 1





            I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

            – François Andrieux
            Feb 28 at 15:54






          • 1





            @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

            – NathanOliver
            Feb 28 at 15:56






          • 1





            So we have to assume OP is using a pre-C++17 standard?

            – François Andrieux
            Feb 28 at 15:57






          • 1





            Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

            – NathanOliver
            Feb 28 at 16:00


















          12














          First copy is in return of createX



          X createX() 
          X x;
          std::cout << "created x on the stack" << std::endl;
          return x; // First copy



          Second one is to create x2 from the temporary return by createX.



          X x2 = createX(); // Second copy


          Notice that in C++17, second copy is forced to be elided.






          share|improve this answer






















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            19














            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.






            share|improve this answer




















            • 1





              @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

              – NathanOliver
              Feb 28 at 15:50






            • 1





              I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

              – François Andrieux
              Feb 28 at 15:54






            • 1





              @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

              – NathanOliver
              Feb 28 at 15:56






            • 1





              So we have to assume OP is using a pre-C++17 standard?

              – François Andrieux
              Feb 28 at 15:57






            • 1





              Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

              – NathanOliver
              Feb 28 at 16:00















            19














            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.






            share|improve this answer




















            • 1





              @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

              – NathanOliver
              Feb 28 at 15:50






            • 1





              I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

              – François Andrieux
              Feb 28 at 15:54






            • 1





              @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

              – NathanOliver
              Feb 28 at 15:56






            • 1





              So we have to assume OP is using a pre-C++17 standard?

              – François Andrieux
              Feb 28 at 15:57






            • 1





              Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

              – NathanOliver
              Feb 28 at 16:00













            19












            19








            19







            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.






            share|improve this answer















            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.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Feb 28 at 15:49

























            answered Feb 28 at 15:41









            NathanOliverNathanOliver

            96.8k16137211




            96.8k16137211







            • 1





              @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

              – NathanOliver
              Feb 28 at 15:50






            • 1





              I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

              – François Andrieux
              Feb 28 at 15:54






            • 1





              @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

              – NathanOliver
              Feb 28 at 15:56






            • 1





              So we have to assume OP is using a pre-C++17 standard?

              – François Andrieux
              Feb 28 at 15:57






            • 1





              Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

              – NathanOliver
              Feb 28 at 16:00












            • 1





              @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

              – NathanOliver
              Feb 28 at 15:50






            • 1





              I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

              – François Andrieux
              Feb 28 at 15:54






            • 1





              @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

              – NathanOliver
              Feb 28 at 15:56






            • 1





              So we have to assume OP is using a pre-C++17 standard?

              – François Andrieux
              Feb 28 at 15:57






            • 1





              Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

              – NathanOliver
              Feb 28 at 16:00







            1




            1





            @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

            – NathanOliver
            Feb 28 at 15:50





            @FrançoisAndrieux Yeah. Updating the wording to make that more explicit that isn't something that is just allowed but has to be done.

            – NathanOliver
            Feb 28 at 15:50




            1




            1





            I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

            – François Andrieux
            Feb 28 at 15:54





            I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.

            – François Andrieux
            Feb 28 at 15:54




            1




            1





            @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

            – NathanOliver
            Feb 28 at 15:56





            @FrançoisAndrieux It does not prevent guaranteed RVO if using C++17 since that is mandated that it has to happen, there isn't actually a temporary object created. See here where it only uses a single copy as the second one is mandated not to happen.

            – NathanOliver
            Feb 28 at 15:56




            1




            1





            So we have to assume OP is using a pre-C++17 standard?

            – François Andrieux
            Feb 28 at 15:57





            So we have to assume OP is using a pre-C++17 standard?

            – François Andrieux
            Feb 28 at 15:57




            1




            1





            Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

            – NathanOliver
            Feb 28 at 16:00





            Kind of. The compiler they are using could have a bug, or they are using pre C++17. I'm inclined to believe it is the latter.

            – NathanOliver
            Feb 28 at 16:00













            12














            First copy is in return of createX



            X createX() 
            X x;
            std::cout << "created x on the stack" << std::endl;
            return x; // First copy



            Second one is to create x2 from the temporary return by createX.



            X x2 = createX(); // Second copy


            Notice that in C++17, second copy is forced to be elided.






            share|improve this answer



























              12














              First copy is in return of createX



              X createX() 
              X x;
              std::cout << "created x on the stack" << std::endl;
              return x; // First copy



              Second one is to create x2 from the temporary return by createX.



              X x2 = createX(); // Second copy


              Notice that in C++17, second copy is forced to be elided.






              share|improve this answer

























                12












                12








                12







                First copy is in return of createX



                X createX() 
                X x;
                std::cout << "created x on the stack" << std::endl;
                return x; // First copy



                Second one is to create x2 from the temporary return by createX.



                X x2 = createX(); // Second copy


                Notice that in C++17, second copy is forced to be elided.






                share|improve this answer













                First copy is in return of createX



                X createX() 
                X x;
                std::cout << "created x on the stack" << std::endl;
                return x; // First copy



                Second one is to create x2 from the temporary return by createX.



                X x2 = createX(); // Second copy


                Notice that in C++17, second copy is forced to be elided.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Feb 28 at 15:41









                Jarod42Jarod42

                119k12104189




                119k12104189



























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