C++ copy constructor called at return












9















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++







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  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    5 hours ago






  • 3





    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
    5 hours ago
















9















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









New contributor




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





















  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    5 hours ago






  • 3





    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
    5 hours ago














9












9








9


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









New contributor




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












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++




c++






share|improve this question









New contributor




Sobuch 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









New contributor




Sobuch 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




share|improve this question








edited 4 hours ago







Sobuch













New contributor




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Check out our Code of Conduct.









asked 5 hours ago









SobuchSobuch

485




485




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






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













  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    5 hours ago






  • 3





    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
    5 hours ago



















  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    5 hours ago






  • 3





    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
    5 hours ago

















see: In which situations is the C++ copy constructor called?

– kmdreko
5 hours ago





see: In which situations is the C++ copy constructor called?

– kmdreko
5 hours ago




3




3





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
5 hours ago





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
5 hours ago












1 Answer
1






active

oldest

votes


















20














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;

};





share|improve this answer





















  • 6





    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
    5 hours ago






  • 2





    @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
    5 hours ago






  • 2





    @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
    5 hours ago








  • 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
    5 hours ago






  • 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
    5 hours ago













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









20














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;

};





share|improve this answer





















  • 6





    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
    5 hours ago






  • 2





    @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
    5 hours ago






  • 2





    @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
    5 hours ago








  • 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
    5 hours ago






  • 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
    5 hours ago


















20














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;

};





share|improve this answer





















  • 6





    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
    5 hours ago






  • 2





    @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
    5 hours ago






  • 2





    @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
    5 hours ago








  • 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
    5 hours ago






  • 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
    5 hours ago
















20












20








20







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;

};





share|improve this answer















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;

};






share|improve this answer














share|improve this answer



share|improve this answer








edited 5 hours ago

























answered 5 hours ago









NathanOliverNathanOliver

96k16136209




96k16136209








  • 6





    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
    5 hours ago






  • 2





    @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
    5 hours ago






  • 2





    @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
    5 hours ago








  • 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
    5 hours ago






  • 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
    5 hours ago
















  • 6





    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
    5 hours ago






  • 2





    @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
    5 hours ago






  • 2





    @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
    5 hours ago








  • 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
    5 hours ago






  • 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
    5 hours ago










6




6





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
5 hours ago





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
5 hours ago




2




2





@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
5 hours ago





@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
5 hours ago




2




2





@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
5 hours ago







@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
5 hours ago






3




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
5 hours ago





@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
5 hours ago




3




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
5 hours ago







@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
5 hours ago














Sobuch is a new contributor. Be nice, and check out our Code of Conduct.










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