Conditions for automatic generation of default/copy/move ctor and copy/move assignment operator?
I want to refresh my memory on the conditions under which a compiler typically auto generates a default constructor, copy constructor and assignment operator.
I recollect there were some rules, but I don't remember, and also can't find a reputable resource online. Can anyone help?
c++ copy-constructor default-constructor move-constructor move-assignment-operator
add a comment |
I want to refresh my memory on the conditions under which a compiler typically auto generates a default constructor, copy constructor and assignment operator.
I recollect there were some rules, but I don't remember, and also can't find a reputable resource online. Can anyone help?
c++ copy-constructor default-constructor move-constructor move-assignment-operator
add a comment |
I want to refresh my memory on the conditions under which a compiler typically auto generates a default constructor, copy constructor and assignment operator.
I recollect there were some rules, but I don't remember, and also can't find a reputable resource online. Can anyone help?
c++ copy-constructor default-constructor move-constructor move-assignment-operator
I want to refresh my memory on the conditions under which a compiler typically auto generates a default constructor, copy constructor and assignment operator.
I recollect there were some rules, but I don't remember, and also can't find a reputable resource online. Can anyone help?
c++ copy-constructor default-constructor move-constructor move-assignment-operator
c++ copy-constructor default-constructor move-constructor move-assignment-operator
edited Jan 19 '18 at 10:39
Community♦
11
11
asked Feb 9 '11 at 11:01
oompahloompahoompahloompah
3,623165386
3,623165386
add a comment |
add a comment |
3 Answers
3
active
oldest
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In the following, "auto-generated" means "implicitly declared as defaulted, but not defined as deleted". There are situations where the special member functions are declared, but defined as deleted.
- The default constructor is auto-generated if there is no user-declared constructor (§12.1/5).
- The copy constructor is auto-generated if there is no user-declared move constructor or move assignment operator (because there are no move constructors or move assignment operators in C++03, this simplifies to "always" in C++03) (§12.8/8).
- The copy assignment operator is auto-generated if there is no user-declared move constructor or move assignment operator (§12.8/19).
- The destructor is auto-generated if there is no user-declared destructor (§12.4/4).
C++11 and later only:
- The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (§12.8/10).
- The move assignment operator is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move assignment operator is valid (e.g. if it wouldn't need to assign constant members) (§12.8/21).
7
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
8
I think that you should mention perhaps that havingconst
members in the class will prevent the constructor from being auto-generated...
– nonsensickle
Jul 31 '14 at 23:08
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
add a comment |
I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
4
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
add a comment |
C++17 N4659 standard draft
For a quick cross standard reference, have a look at the "Implicitly-declared" sections of the following cppreference entries:
- https://en.cppreference.com/w/cpp/language/copy_constructor
- https://en.cppreference.com/w/cpp/language/move_constructor
- https://en.cppreference.com/w/cpp/language/copy_assignment
- https://en.cppreference.com/w/cpp/language/move_assignment
The same information can of course be obtained from the standard. E.g. on C++17 N4659 standard draft:
15.8.1 "Copy/move constructors" says for for copy constructor:
6 If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared copy
constructor is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter case is deprecated if
the class has a user-declared copy assignment operator or a user-declared destructor.
and for move constructor:
8 If the definition of a class X does not explicitly declare a move constructor, a non-explicit one will be implicitly
declared as defaulted if and only if
(8.1)
— X does not have a user-declared copy constructor,
(8.2)
— X does not have a user-declared copy assignment operator,
(8.3)
— X does not have a user-declared move assignment operator, and
(8.4)
— X does not have a user-declared destructor.
15.8.2 "Copy/move assignment operator" says for copy assignment:
2 If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared
copy assignment operator is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter
case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.
and for move assignment:
4 If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly
declared as defaulted if and only if
- (4.1) — X does not have a user-declared copy constructor,
- (4.2) — X does not have a user-declared move constructor,
- (4.3) — X does not have a user-declared copy assignment operator, and
- (4.4) — X does not have a user-declared destructor.
15.4 "Destructors" says it for destructors:
4 If a class has no user-declared destructor, a destructor is implicitly declared as defaulted (11.4). An
implicitly-declared destructor is an inline public member of its class.
add a comment |
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In the following, "auto-generated" means "implicitly declared as defaulted, but not defined as deleted". There are situations where the special member functions are declared, but defined as deleted.
- The default constructor is auto-generated if there is no user-declared constructor (§12.1/5).
- The copy constructor is auto-generated if there is no user-declared move constructor or move assignment operator (because there are no move constructors or move assignment operators in C++03, this simplifies to "always" in C++03) (§12.8/8).
- The copy assignment operator is auto-generated if there is no user-declared move constructor or move assignment operator (§12.8/19).
- The destructor is auto-generated if there is no user-declared destructor (§12.4/4).
C++11 and later only:
- The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (§12.8/10).
- The move assignment operator is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move assignment operator is valid (e.g. if it wouldn't need to assign constant members) (§12.8/21).
7
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
8
I think that you should mention perhaps that havingconst
members in the class will prevent the constructor from being auto-generated...
– nonsensickle
Jul 31 '14 at 23:08
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
add a comment |
In the following, "auto-generated" means "implicitly declared as defaulted, but not defined as deleted". There are situations where the special member functions are declared, but defined as deleted.
- The default constructor is auto-generated if there is no user-declared constructor (§12.1/5).
- The copy constructor is auto-generated if there is no user-declared move constructor or move assignment operator (because there are no move constructors or move assignment operators in C++03, this simplifies to "always" in C++03) (§12.8/8).
- The copy assignment operator is auto-generated if there is no user-declared move constructor or move assignment operator (§12.8/19).
- The destructor is auto-generated if there is no user-declared destructor (§12.4/4).
C++11 and later only:
- The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (§12.8/10).
- The move assignment operator is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move assignment operator is valid (e.g. if it wouldn't need to assign constant members) (§12.8/21).
7
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
8
I think that you should mention perhaps that havingconst
members in the class will prevent the constructor from being auto-generated...
– nonsensickle
Jul 31 '14 at 23:08
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
add a comment |
In the following, "auto-generated" means "implicitly declared as defaulted, but not defined as deleted". There are situations where the special member functions are declared, but defined as deleted.
- The default constructor is auto-generated if there is no user-declared constructor (§12.1/5).
- The copy constructor is auto-generated if there is no user-declared move constructor or move assignment operator (because there are no move constructors or move assignment operators in C++03, this simplifies to "always" in C++03) (§12.8/8).
- The copy assignment operator is auto-generated if there is no user-declared move constructor or move assignment operator (§12.8/19).
- The destructor is auto-generated if there is no user-declared destructor (§12.4/4).
C++11 and later only:
- The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (§12.8/10).
- The move assignment operator is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move assignment operator is valid (e.g. if it wouldn't need to assign constant members) (§12.8/21).
In the following, "auto-generated" means "implicitly declared as defaulted, but not defined as deleted". There are situations where the special member functions are declared, but defined as deleted.
- The default constructor is auto-generated if there is no user-declared constructor (§12.1/5).
- The copy constructor is auto-generated if there is no user-declared move constructor or move assignment operator (because there are no move constructors or move assignment operators in C++03, this simplifies to "always" in C++03) (§12.8/8).
- The copy assignment operator is auto-generated if there is no user-declared move constructor or move assignment operator (§12.8/19).
- The destructor is auto-generated if there is no user-declared destructor (§12.4/4).
C++11 and later only:
- The move constructor is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move constructor is valid (§12.8/10).
- The move assignment operator is auto-generated if there is no user-declared copy constructor, copy assignment operator or destructor, and if the generated move assignment operator is valid (e.g. if it wouldn't need to assign constant members) (§12.8/21).
edited Dec 24 '17 at 21:35
milleniumbug
12.7k33462
12.7k33462
answered Feb 9 '11 at 11:17
PhilippPhilipp
37.6k107097
37.6k107097
7
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
8
I think that you should mention perhaps that havingconst
members in the class will prevent the constructor from being auto-generated...
– nonsensickle
Jul 31 '14 at 23:08
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
add a comment |
7
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
8
I think that you should mention perhaps that havingconst
members in the class will prevent the constructor from being auto-generated...
– nonsensickle
Jul 31 '14 at 23:08
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
7
7
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
Does an inherited destructor count? I mean, say I've got a base class with an empty virtual destructor. Does it prevent creation of move constructors in subclasses? If the answer is yes, will it help if I define a move constructor in the base class?
– kamilk
Jul 6 '14 at 13:29
8
8
I think that you should mention perhaps that having
const
members in the class will prevent the constructor from being auto-generated...– nonsensickle
Jul 31 '14 at 23:08
I think that you should mention perhaps that having
const
members in the class will prevent the constructor from being auto-generated...– nonsensickle
Jul 31 '14 at 23:08
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
Does "There are situations where the special member functions are declared, but defined as deleted." refer to where you for example have const or reference members where move will be impossible? No, that can't be, because there copy will be applied.
– towi
Sep 27 '16 at 7:37
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
I know that it's restricted to send hyperlinks in this forum. But it's also good article - cplusplus.com/articles/y8hv0pDG
– bruziuz
Oct 12 '16 at 0:54
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
Note, that as of the standard an implicitly defaulted copy constructor "is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor" (12.8 Copying and moving class objects [class.copy]).
– sigy
Mar 29 '17 at 10:59
add a comment |
I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
4
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
add a comment |
I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
4
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
add a comment |
I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero
I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero
edited Apr 20 '18 at 11:57
O'Neil
3,35721126
3,35721126
answered Jul 8 '16 at 1:02
Marco M.Marco M.
1,64011920
1,64011920
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
4
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
add a comment |
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
4
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
Beautiful. What does "independent" refer to? Independent from what?
– towi
Sep 27 '16 at 7:29
4
4
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Copy ctor/assignment are 'independent' from each other. If you write just one, the compiler will provide the other. In contrast, if you provide either a move ctor or a move assignment, the compiler won't supply the other.
– Marco M.
Oct 3 '16 at 20:14
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
Wonder what's the reason behind copy operations are being independent. Historic reasons may be? or the fact that copy won't modify it's target but move does?
– Explorer_N
Jul 5 '17 at 7:47
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
@Explorer_N Yes, backward compatibility, so historic reasons. It was a bad design choice long time ago, so now there's a need for good practices like the "rule of three" (define all 3 or none: copy constructor, copy assignment operator, and often destructor) to avoid hard to find bugs.
– atablash
Mar 31 '18 at 18:41
add a comment |
C++17 N4659 standard draft
For a quick cross standard reference, have a look at the "Implicitly-declared" sections of the following cppreference entries:
- https://en.cppreference.com/w/cpp/language/copy_constructor
- https://en.cppreference.com/w/cpp/language/move_constructor
- https://en.cppreference.com/w/cpp/language/copy_assignment
- https://en.cppreference.com/w/cpp/language/move_assignment
The same information can of course be obtained from the standard. E.g. on C++17 N4659 standard draft:
15.8.1 "Copy/move constructors" says for for copy constructor:
6 If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared copy
constructor is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter case is deprecated if
the class has a user-declared copy assignment operator or a user-declared destructor.
and for move constructor:
8 If the definition of a class X does not explicitly declare a move constructor, a non-explicit one will be implicitly
declared as defaulted if and only if
(8.1)
— X does not have a user-declared copy constructor,
(8.2)
— X does not have a user-declared copy assignment operator,
(8.3)
— X does not have a user-declared move assignment operator, and
(8.4)
— X does not have a user-declared destructor.
15.8.2 "Copy/move assignment operator" says for copy assignment:
2 If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared
copy assignment operator is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter
case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.
and for move assignment:
4 If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly
declared as defaulted if and only if
- (4.1) — X does not have a user-declared copy constructor,
- (4.2) — X does not have a user-declared move constructor,
- (4.3) — X does not have a user-declared copy assignment operator, and
- (4.4) — X does not have a user-declared destructor.
15.4 "Destructors" says it for destructors:
4 If a class has no user-declared destructor, a destructor is implicitly declared as defaulted (11.4). An
implicitly-declared destructor is an inline public member of its class.
add a comment |
C++17 N4659 standard draft
For a quick cross standard reference, have a look at the "Implicitly-declared" sections of the following cppreference entries:
- https://en.cppreference.com/w/cpp/language/copy_constructor
- https://en.cppreference.com/w/cpp/language/move_constructor
- https://en.cppreference.com/w/cpp/language/copy_assignment
- https://en.cppreference.com/w/cpp/language/move_assignment
The same information can of course be obtained from the standard. E.g. on C++17 N4659 standard draft:
15.8.1 "Copy/move constructors" says for for copy constructor:
6 If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared copy
constructor is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter case is deprecated if
the class has a user-declared copy assignment operator or a user-declared destructor.
and for move constructor:
8 If the definition of a class X does not explicitly declare a move constructor, a non-explicit one will be implicitly
declared as defaulted if and only if
(8.1)
— X does not have a user-declared copy constructor,
(8.2)
— X does not have a user-declared copy assignment operator,
(8.3)
— X does not have a user-declared move assignment operator, and
(8.4)
— X does not have a user-declared destructor.
15.8.2 "Copy/move assignment operator" says for copy assignment:
2 If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared
copy assignment operator is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter
case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.
and for move assignment:
4 If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly
declared as defaulted if and only if
- (4.1) — X does not have a user-declared copy constructor,
- (4.2) — X does not have a user-declared move constructor,
- (4.3) — X does not have a user-declared copy assignment operator, and
- (4.4) — X does not have a user-declared destructor.
15.4 "Destructors" says it for destructors:
4 If a class has no user-declared destructor, a destructor is implicitly declared as defaulted (11.4). An
implicitly-declared destructor is an inline public member of its class.
add a comment |
C++17 N4659 standard draft
For a quick cross standard reference, have a look at the "Implicitly-declared" sections of the following cppreference entries:
- https://en.cppreference.com/w/cpp/language/copy_constructor
- https://en.cppreference.com/w/cpp/language/move_constructor
- https://en.cppreference.com/w/cpp/language/copy_assignment
- https://en.cppreference.com/w/cpp/language/move_assignment
The same information can of course be obtained from the standard. E.g. on C++17 N4659 standard draft:
15.8.1 "Copy/move constructors" says for for copy constructor:
6 If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared copy
constructor is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter case is deprecated if
the class has a user-declared copy assignment operator or a user-declared destructor.
and for move constructor:
8 If the definition of a class X does not explicitly declare a move constructor, a non-explicit one will be implicitly
declared as defaulted if and only if
(8.1)
— X does not have a user-declared copy constructor,
(8.2)
— X does not have a user-declared copy assignment operator,
(8.3)
— X does not have a user-declared move assignment operator, and
(8.4)
— X does not have a user-declared destructor.
15.8.2 "Copy/move assignment operator" says for copy assignment:
2 If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared
copy assignment operator is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter
case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.
and for move assignment:
4 If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly
declared as defaulted if and only if
- (4.1) — X does not have a user-declared copy constructor,
- (4.2) — X does not have a user-declared move constructor,
- (4.3) — X does not have a user-declared copy assignment operator, and
- (4.4) — X does not have a user-declared destructor.
15.4 "Destructors" says it for destructors:
4 If a class has no user-declared destructor, a destructor is implicitly declared as defaulted (11.4). An
implicitly-declared destructor is an inline public member of its class.
C++17 N4659 standard draft
For a quick cross standard reference, have a look at the "Implicitly-declared" sections of the following cppreference entries:
- https://en.cppreference.com/w/cpp/language/copy_constructor
- https://en.cppreference.com/w/cpp/language/move_constructor
- https://en.cppreference.com/w/cpp/language/copy_assignment
- https://en.cppreference.com/w/cpp/language/move_assignment
The same information can of course be obtained from the standard. E.g. on C++17 N4659 standard draft:
15.8.1 "Copy/move constructors" says for for copy constructor:
6 If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared copy
constructor is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter case is deprecated if
the class has a user-declared copy assignment operator or a user-declared destructor.
and for move constructor:
8 If the definition of a class X does not explicitly declare a move constructor, a non-explicit one will be implicitly
declared as defaulted if and only if
(8.1)
— X does not have a user-declared copy constructor,
(8.2)
— X does not have a user-declared copy assignment operator,
(8.3)
— X does not have a user-declared move assignment operator, and
(8.4)
— X does not have a user-declared destructor.
15.8.2 "Copy/move assignment operator" says for copy assignment:
2 If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.
If the class definition declares a move constructor or move assignment operator, the implicitly declared
copy assignment operator is defined as deleted; otherwise, it is defined as defaulted (11.4). The latter
case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.
and for move assignment:
4 If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly
declared as defaulted if and only if
- (4.1) — X does not have a user-declared copy constructor,
- (4.2) — X does not have a user-declared move constructor,
- (4.3) — X does not have a user-declared copy assignment operator, and
- (4.4) — X does not have a user-declared destructor.
15.4 "Destructors" says it for destructors:
4 If a class has no user-declared destructor, a destructor is implicitly declared as defaulted (11.4). An
implicitly-declared destructor is an inline public member of its class.
answered Nov 15 '18 at 9:09
Ciro Santilli 新疆改造中心 六四事件 法轮功Ciro Santilli 新疆改造中心 六四事件 法轮功
145k34555471
145k34555471
add a comment |
add a comment |
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