C++ object as parameter, but no constructor for it [duplicate] - c++

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?

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).

I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero

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.

Related

C++ rule of 5 and rule of 0 [duplicate]

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?
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).
I've found the diagram below very useful.
from Sticky Bits - Becoming a Rule of Zero Hero
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.

Is a defaulted move constructor considered user-declared?

The question is as my title states.
I'm asking because I have a class with a defaulted move constructor but code trying to perform copy assignment is failing stating that the copy assignment operator is deleted (according to Visual Studio 2015).
So I checked the rules here for implicitly declared copy assignment operators:
The implicitly-declared or defaulted copy assignment operator for class T is defined as deleted in any of the following is true:
...
T has a user-declared move constructor
T has a user-declared move assignment operator
So basically I'm not sure if a defaulted move constructor counts as user-declared. My gut tells me yes but when it comes to standardese I always like to be sure since assumptions can be costly.
The standard says:
12.8 Copying and moving class objects [class.copy]
If the class definition does not explicitly declare a copy constructor, 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 (8.4). The latter case is deprecated if the class has
a user-declared copy assignment operator or a user-declared destructor.
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 (8.4). The latter case is deprecated if the class has a user-declared copy constructor or a user-declared destructor.
Your class has defaulted move constructor, but it is explicitly declared.
So according to standard implicitly declared copy constructor and copy assignment operator is defined as deleted.
8.4.2 Explicitly-defaulted functions [dcl.fct.def.default]
Explicitly-defaulted functions and implicitly-declared functions are collectively called defaulted functions, and the implementation shall provide implicit definitions for them (12.1 12.4, 12.8), which might mean defining them as deleted. A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration. A user-provided explicitly-defaulted function (i.e., explicitly defaulted after its first declaration) is defined at the point where it is explicitly defaulted.
Using this terminology your move constructor is user-declared, but not user-provided.
A defaulted special member function is user-declared but it is also defined by the user as defaulted. The standard doesn't explicitly define the term "user-declared" but it essentially means any special member function that has to be written out by the user. So the following declares a constructor and defines it as defaulted.
struct X {
X() = default; // declaration and definition
};
Defining a member function as defaulted means the definition is equivalent to the implicit definition. It is user-declared by virtue of the fact that it had to be typed out by the user.

Does deleting a copy constructor or copy assignment operator count as "user declared"?

Per this presentation, if either the copy constructor or copy assignment operator is "user declared", then no implicit move operations will be generated. Does deleteing the copy constructor or copy assignment operator count as "user declared"?
struct NoCopy {
NoCopy(NoCopy&) = delete;
NoCopy& operator=(const NoCopy&) = delete;
};
Will implicit move operations be generated for the NoCopy class? Or does deleting the relevant copy operations count as "user declared" and thus inhibit implicit move generation?
If possible, I'd prefer an answer referencing the relevant parts of the standard.
According to slide 14 of your presentation, a deleted copy constructor is "user declared" thus inhibiting the move generation.
The term "user declared" doesn't have a formal definition in the standard. It is meant to be the opposite of "implicitly declared" in the context of special member functions. [dcl.fct.def.default]/4 could be a bit clearer about this fact, but the intention is there:
Explicitly-defaulted functions and implicitly-declared functions are collectively called defaulted functions, and the implementation shall provide implicit definitions for them (12.1 12.4, 12.8), which might mean defining them as deleted. A special member function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration. A user-provided explicitly-defaulted function (i.e., explicitly defaulted after its first declaration) is defined at the point where it is explicitly defaulted; if such a function is implicitly defined as deleted, the program is ill-formed.
Both NoCopy(NoCopy&) = delete; and NoCopy& operator=(const NoCopy&) = delete; are declarations of special member functions. Since you are explicitly declaring them, as opposed to allowing the compiler to declare them implicitly, they are user-declared. Those declarations will therefore suppress the implicit declarations of the move constructor and move assignment operator per [class.copy]/9:
If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if
— X does not have a user-declared copy constructor,
— X does not have a user-declared copy assignment operator,
— X does not have a user-declared move assignment operator,
— X does not have a user-declared destructor, and
— the move constructor would not be implicitly defined as deleted.

Could once_flag be movable?

How does it come, the standard says nothing about the movable'ity of once_flag? I would expect the same arguments to apply as for std::mutex. At least for gcc (version 4.8) moving seems to be disabled. If a certain compiler would allow moving, one could end up with non-portable code.
The synopsis is (§30.4 [thread.mutex])
struct once_flag {
constexpr once_flag() noexcept;
once_flag(const once_flag&) = delete;
once_flag& operator=(const once_flag&) = delete;
};
Since the copy constructor and copy assignment operator are user-declared (and explicitly deleted), the move constructor and move assignment operator are not implicitly declared (§12.8 [class.copy]/p9, 20):
9 If the definition of a class X does not explicitly declare a move
constructor, one will be implicitly declared as defaulted if and only
if
X does not have a user-declared copy constructor,
X does not have a user-declared copy assignment operator,
X does not have a user-declared move assignment operator, and
X does not have a user-declared destructor.
20 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
X does not have a user-declared copy constructor,
X does not have a user-declared move constructor,
X does not have a user-declared copy assignment operator, and
X does not have a user-declared destructor.
Hence, once_flag cannot be moved.

If Move semantics(Move constructor and Move assignment operator) are not defined does compiler optimize by default? [duplicate]

This question already has answers here:
In what scenarios should I expect to explicitly need to implement a move constructor and move assignment operator?
(5 answers)
Closed 8 years ago.
While defining class if we forget to write Move constructor and Move assignment operator does compiler is smart enough to optimize and add automatically.
Maybe, maybe not. It's not a question of whether the compiler is smart enough, but what other special member functions you have remembered, or forgotten, to define. The exact conditions when a move constructor will be implicitly defined by the compiler are listed under §12.8/9 [class.copy]
If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if
— X does not have a user-declared copy constructor,
— X does not have a user-declared copy assignment operator,
— X does not have a user-declared move assignment operator, and
— X does not have a user-declared destructor.
Similarly, the conditions for implicit generation of a move assignment operator are listed under §12.8/20
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
— X does not have a user-declared copy constructor,
— X does not have a user-declared move constructor,
— X does not have a user-declared copy assignment operator, and
— X does not have a user-declared destructor.