C++ copy constructor called at return
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++
asked Mar 21 at 20:23
SobuchSobuch
886
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++
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 (tmphere) its destructor must be called. Therefore, when youreturn tmpthen you're asking it to make a copy oftmpto be return to whomever calls the function. Once copied,tmpwill 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
5 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++
asked Mar 21 at 20:23
SobuchSobuch
886
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++
asked Mar 21 at 20:23
SobuchSobuch
886
asked Mar 21 at 20:23
SobuchSobuch
886
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
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 (tmphere) its destructor must be called. Therefore, when youreturn tmpthen you're asking it to make a copy oftmpto be return to whomever calls the function. Once copied,tmpwill 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
5 hours ago
add a comment |
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 (tmphere) its destructor must be called. Therefore, when youreturn tmpthen you're asking it to make a copy oftmpto be return to whomever calls the function. Once copied,tmpwill 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
5 hours ago
1
1
see: In which situations is the C++ copy constructor called?
– kmdreko
Mar 21 at 20:29
see: In which situations is the C++ copy constructor called?
– kmdreko
Mar 21 at 20:29
5
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
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
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
@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
5 hours ago
A little surprising as I would have thought that RVO would have been invoked, resulting in no move or copy.
– Adrian
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
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.
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 youdeletean object of classXthrough a pointer to a base class ofXand 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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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.
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 youdeletean object of classXthrough a pointer to a base class ofXand 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.
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 youdeletean object of classXthrough a pointer to a base class ofXand 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.
answered Mar 21 at 20:32
NathanOliverNathanOliver
97.6k16138214
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.
answered Mar 21 at 20:32
NathanOliverNathanOliver
97.6k16138214
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
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 youdeletean object of classXthrough a pointer to a base class ofXand 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
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 youdeletean object of classXthrough a pointer to a base class ofXand 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
10
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
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
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
@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
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
@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
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
@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
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
@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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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 (
tmphere) its destructor must be called. Therefore, when youreturn tmpthen you're asking it to make a copy oftmpto be return to whomever calls the function. Once copied,tmpwill 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
5 hours ago