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C++ copy constructor called at return

17

1

error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~

Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?

struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;

c++

share|improve this question

edited Mar 21 at 21:18

Sobuch

asked Mar 21 at 20:23

SobuchSobuch

886

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  see: In which situations is the C++ copy constructor called?
  
  – kmdreko
  Mar 21 at 20:29
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.
  
  – Everyone
  Mar 21 at 20:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Everyone except that it is usually a move rather than a copy, which is what the question is about.
  
  – Quentin
  Mar 22 at 9:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  A little surprising as I would have thought that RVO would have been invoked, resulting in no move or copy.
  
  – Adrian
  7 hours ago
  

  
  
  
  

add a comment | 

17

1

error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~

Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?

struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;

c++

share|improve this question

edited Mar 21 at 21:18

Sobuch

asked Mar 21 at 20:23

SobuchSobuch

886

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  see: In which situations is the C++ copy constructor called?
  
  – kmdreko
  Mar 21 at 20:29
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.
  
  – Everyone
  Mar 21 at 20:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Everyone except that it is usually a move rather than a copy, which is what the question is about.
  
  – Quentin
  Mar 22 at 9:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  A little surprising as I would have thought that RVO would have been invoked, resulting in no move or copy.
  
  – Adrian
  7 hours ago
  

  
  
  
  

add a comment | 

error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~

Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?

struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;

c++

share|improve this question

edited Mar 21 at 21:18

Sobuch

asked Mar 21 at 20:23

SobuchSobuch

886

error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~

struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;

share|improve this question

edited Mar 21 at 21:18

Sobuch

asked Mar 21 at 20:23

SobuchSobuch

886

share|improve this question

edited Mar 21 at 21:18

Sobuch

asked Mar 21 at 20:23

SobuchSobuch

886

asked Mar 21 at 20:23

SobuchSobuch

886

asked Mar 21 at 20:23

SobuchSobuch

886

1 Answer
1

active

oldest

votes

30

virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.

There are two ways you can fix this. You can add a move constructor like

struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;

or you can create a base class that has the virtual destructor and inherit from that like

struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;

This works because A no longer has a user declared destructor (Yes, C does but we only care about A) so it will still generate a move constructor in A. The important part of this is that C doesn't have a deleted move constructor, it just doesn't have one period, so trying to move it will cause a copy. That means
C's copy constructor is called in A's implicitly generated move constructor since C(std::move(A_obj_to_move_from)) will copy as long as it doesn't have a deleted move constructor.

share|improve this answer

edited Mar 22 at 12:44

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  
  Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.
  
  – 0x5453
  Mar 21 at 20:33
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.
  
  – NathanOliver
  Mar 21 at 20:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.
  
  – eerorika
  Mar 21 at 20:40
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.
  
  – Angew
  Mar 21 at 20:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.
  
  – NathanOliver
  Mar 21 at 20:41
  
  

  
  
  
  

 | 
show 24 more comments

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30

virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.

There are two ways you can fix this. You can add a move constructor like

struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;

or you can create a base class that has the virtual destructor and inherit from that like

struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;

This works because A no longer has a user declared destructor (Yes, C does but we only care about A) so it will still generate a move constructor in A. The important part of this is that C doesn't have a deleted move constructor, it just doesn't have one period, so trying to move it will cause a copy. That means
C's copy constructor is called in A's implicitly generated move constructor since C(std::move(A_obj_to_move_from)) will copy as long as it doesn't have a deleted move constructor.

share|improve this answer

edited Mar 22 at 12:44

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  
  Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.
  
  – 0x5453
  Mar 21 at 20:33
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.
  
  – NathanOliver
  Mar 21 at 20:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.
  
  – eerorika
  Mar 21 at 20:40
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.
  
  – Angew
  Mar 21 at 20:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.
  
  – NathanOliver
  Mar 21 at 20:41
  
  

  
  
  
  

 | 
show 24 more comments

30

virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.

There are two ways you can fix this. You can add a move constructor like

struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;

or you can create a base class that has the virtual destructor and inherit from that like

struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;

This works because A no longer has a user declared destructor (Yes, C does but we only care about A) so it will still generate a move constructor in A. The important part of this is that C doesn't have a deleted move constructor, it just doesn't have one period, so trying to move it will cause a copy. That means
C's copy constructor is called in A's implicitly generated move constructor since C(std::move(A_obj_to_move_from)) will copy as long as it doesn't have a deleted move constructor.

share|improve this answer

edited Mar 22 at 12:44

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214

• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  
  Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.
  
  – 0x5453
  Mar 21 at 20:33
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.
  
  – NathanOliver
  Mar 21 at 20:34
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.
  
  – eerorika
  Mar 21 at 20:40
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.
  
  – Angew
  Mar 21 at 20:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.
  
  – NathanOliver
  Mar 21 at 20:41
  
  

  
  
  
  

 | 
show 24 more comments

virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.

There are two ways you can fix this. You can add a move constructor like

struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;

or you can create a base class that has the virtual destructor and inherit from that like

struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;

This works because A no longer has a user declared destructor (Yes, C does but we only care about A) so it will still generate a move constructor in A. The important part of this is that C doesn't have a deleted move constructor, it just doesn't have one period, so trying to move it will cause a copy. That means
C's copy constructor is called in A's implicitly generated move constructor since C(std::move(A_obj_to_move_from)) will copy as long as it doesn't have a deleted move constructor.

share|improve this answer

edited Mar 22 at 12:44

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214

struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;

struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;

share|improve this answer

edited Mar 22 at 12:44

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214

answered Mar 21 at 20:32

NathanOliverNathanOliver

97.6k16138214