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

What are the rules for calling the superclass constructor?Can I call a constructor from another constructor (do constructor chaining) in C++?C++ vector::push_back using default copy constructorWhy does r-value reference to object generator call require copy constructor?Why copy constructor is not called here?Copy constructor is not called when return by valueWhy this move constructor is so greedy?object as return value, no constructor called?Copy constructor not called when initializing an object with return value of a functionIs there any special reason why the move constructor is not elided in the snippet shown below?

7

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?

c++ c++14 copy-constructor

share|improve this question

edited 2 hours ago

NathanOliver

93.7k16130198

asked 3 hours ago

c_studentc_student

564

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add a comment | 

7

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?

c++ c++14 copy-constructor

share|improve this question

edited 2 hours ago

NathanOliver

93.7k16130198

asked 3 hours ago

c_studentc_student

564

New contributor

c_student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

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?

c++ c++14 copy-constructor

share|improve this question

edited 2 hours ago

NathanOliver

93.7k16130198

asked 3 hours ago

c_studentc_student

564

New contributor

c_student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

#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;

}

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

share|improve this question

edited 2 hours ago

NathanOliver

93.7k16130198

asked 3 hours ago

c_studentc_student

564

New contributor

c_student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 2 hours ago

NathanOliver

93.7k16130198

asked 3 hours ago

c_studentc_student

564

New contributor

c_student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

NathanOliver

93.7k16130198

edited 2 hours ago

NathanOliver

93.7k16130198

asked 3 hours ago

c_studentc_student

564

New contributor

c_student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

c_studentc_student

564

2 Answers
2

active

oldest

votes

9

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

edited 2 hours ago

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.
  
  – François Andrieux
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So we have to assume OP is using a pre-C++17 standard?
  
  – François Andrieux
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

8

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

answered 2 hours ago

Jarod42Jarod42

117k12103186

add a comment | 

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9

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

edited 2 hours ago

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.
  
  – François Andrieux
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So we have to assume OP is using a pre-C++17 standard?
  
  – François Andrieux
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

9

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

edited 2 hours ago

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not sure of the details of -fno-elide-constructors but I'm assuming it prevents RVO.
  
  – François Andrieux
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So we have to assume OP is using a pre-C++17 standard?
  
  – François Andrieux
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

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

edited 2 hours ago

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

return x;

X x2 = createX();

return x;

share|improve this answer

edited 2 hours ago

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

answered 2 hours ago

NathanOliverNathanOliver

93.7k16130198

8

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

answered 2 hours ago

Jarod42Jarod42

117k12103186

add a comment | 

8

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

answered 2 hours ago

Jarod42Jarod42

117k12103186

add a comment | 

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

answered 2 hours ago

Jarod42Jarod42

117k12103186

X createX() {

    X x;

    std::cout << "created x on the stack" << std::endl;

    return x; // First copy

}

X x2 = createX(); // Second copy

share|improve this answer

answered 2 hours ago

Jarod42Jarod42

117k12103186

answered 2 hours ago

Jarod42Jarod42

117k12103186

answered 2 hours ago

Jarod42Jarod42

117k12103186