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Copy constructors and copy assignment operators (C++)
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Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment . In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see Move Constructors and Move Assignment Operators (C++) .
Both the assignment operation and the initialization operation cause objects to be copied.
Assignment : When one object's value is assigned to another object, the first object is copied to the second object. So, this code copies the value of b into a :
Initialization : Initialization occurs when you declare a new object, when you pass function arguments by value, or when you return by value from a function.
You can define the semantics of "copy" for objects of class type. For example, consider this code:
The preceding code could mean "copy the contents of FILE1.DAT to FILE2.DAT" or it could mean "ignore FILE2.DAT and make b a second handle to FILE1.DAT." You must attach appropriate copying semantics to each class, as follows:
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x); .
Use the copy constructor.
If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you. Similarly, if you don't declare a copy assignment operator, the compiler generates a member-wise copy assignment operator for you. Declaring a copy constructor doesn't suppress the compiler-generated copy assignment operator, and vice-versa. If you implement either one, we recommend that you implement the other one, too. When you implement both, the meaning of the code is clear.
The copy constructor takes an argument of type ClassName& , where ClassName is the name of the class. For example:
Make the type of the copy constructor's argument const ClassName& whenever possible. This prevents the copy constructor from accidentally changing the copied object. It also lets you copy from const objects.
Compiler generated copy constructors
Compiler-generated copy constructors, like user-defined copy constructors, have a single argument of type "reference to class-name ." An exception is when all base classes and member classes have copy constructors declared as taking a single argument of type const class-name & . In such a case, the compiler-generated copy constructor's argument is also const .
When the argument type to the copy constructor isn't const , initialization by copying a const object generates an error. The reverse isn't true: If the argument is const , you can initialize by copying an object that's not const .
Compiler-generated assignment operators follow the same pattern for const . They take a single argument of type ClassName& unless the assignment operators in all base and member classes take arguments of type const ClassName& . In this case, the generated assignment operator for the class takes a const argument.
When virtual base classes are initialized by copy constructors, whether compiler-generated or user-defined, they're initialized only once: at the point when they are constructed.
The implications are similar to the copy constructor. When the argument type isn't const , assignment from a const object generates an error. The reverse isn't true: If a const value is assigned to a value that's not const , the assignment succeeds.
For more information about overloaded assignment operators, see Assignment .
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Copy constructors, assignment operators, and exception safe assignment
| MyClass& other ); MyClass( MyClass& other ); MyClass( MyClass& other ); MyClass( MyClass& other ); |
| MyClass* other ); |
| MyClass { x; c; std::string s; }; |
| MyClass& other ) : x( other.x ), c( other.c ), s( other.s ) {} |
| ); |
| print_me_bad( std::string& s ) { std::cout << s << std::endl; } print_me_good( std::string& s ) { std::cout << s << std::endl; } std::string hello( ); print_me_bad( hello ); print_me_bad( std::string( ) ); print_me_bad( ); print_me_good( hello ); print_me_good( std::string( ) ); print_me_good( ); |
| , ); |
| =( MyClass& other ) { x = other.x; c = other.c; s = other.s; * ; } |
| < T > MyArray { size_t numElements; T* pElements; : size_t count() { numElements; } MyArray& =( MyArray& rhs ); }; |
| <> MyArray<T>:: =( MyArray& rhs ) { ( != &rhs ) { [] pElements; pElements = T[ rhs.numElements ]; ( size_t i = 0; i < rhs.numElements; ++i ) pElements[ i ] = rhs.pElements[ i ]; numElements = rhs.numElements; } * ; } |
| <> MyArray<T>:: =( MyArray& rhs ) { MyArray tmp( rhs ); std::swap( numElements, tmp.numElements ); std::swap( pElements, tmp.pElements ); * ; } |
| < T > swap( T& one, T& two ) { T tmp( one ); one = two; two = tmp; } |
| <> MyArray<T>:: =( MyArray tmp ) { std::swap( numElements, tmp.numElements ); std::swap( pElements, tmp.pElements ); * ; } |
21.12 — Overloading the assignment operator
cppreference.com
Move assignment operator.
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A move assignment operator is a non-template non-static member function with the name operator = that can be called with an argument of the same class type and copies the content of the argument, possibly mutating the argument.
| Syntax Explanation Implicitly-declared move assignment operator Implicitly-defined move assignment operator Deleted move assignment operator Trivial move assignment operator Eligible move assignment operator Notes Example Defect reports See also |
[ edit ] Syntax
For the formal move assignment operator syntax, see function declaration . The syntax list below only demonstrates a subset of all valid move assignment operator syntaxes.
| return-type parameter-list | (1) | ||||||||
| return-type parameter-list function-body | (2) | ||||||||
| return-type parameter-list-no-default | (3) | ||||||||
| return-type parameter-list | (4) | ||||||||
| return-type class-name parameter-list function-body | (5) | ||||||||
| return-type class-name parameter-list-no-default | (6) | ||||||||
| class-name | - | the class whose move assignment operator is being declared, the class type is given as in the descriptions below |
| parameter-list | - | a of only one parameter, which is of type , const T&&, volatile T&& or const volatile T&& |
| parameter-list-no-default | - | a of only one parameter, which is of type , const T&&, volatile T&& or const volatile T&& and does not have a default argument |
| function-body | - | the of the move assignment operator |
| return-type | - | any type, but is favored in order to be consistent with scala types |
[ edit ] Explanation
The move assignment operator is called whenever it is selected by overload resolution , e.g. when an object appears on the left-hand side of an assignment expression, where the right-hand side is an rvalue of the same or implicitly convertible type.
Move assignment operators typically transfer the resources held by the argument (e.g. pointers to dynamically-allocated objects, file descriptors, TCP sockets, thread handles, etc.), rather than make copies of them, and leave the argument in some valid but otherwise indeterminate state. Since move assignment doesn’t change the lifetime of the argument, the destructor will typically be called on the argument at a later point. For example, move-assigning from a std::string or from a std::vector may result in the argument being left empty. A move assignment is less, not more restrictively defined than ordinary assignment; where ordinary assignment must leave two copies of data at completion, move assignment is required to leave only one.
[ edit ] Implicitly-declared move assignment operator
If no user-defined move assignment operators are provided for a class type, and all of the following is true:
- there are no user-declared copy constructors ;
- there are no user-declared move constructors ;
- there are no user-declared copy assignment operators ;
- there is no user-declared destructor ,
then the compiler will declare a move assignment operator as an inline public member of its class with the signature T & T :: operator = ( T && ) .
A class can have multiple move assignment operators, e.g. both T & T :: operator = ( const T && ) and T & T :: operator = ( T && ) . If some user-defined move assignment operators are present, the user may still force the generation of the implicitly declared move assignment operator with the keyword default .
The implicitly-declared move assignment operator has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17) .
Because some assignment operator (move or copy) is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.
[ edit ] Implicitly-defined move assignment operator
If the implicitly-declared move assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used or needed for constant evaluation (since C++14) .
For union types, the implicitly-defined move assignment operator copies the object representation (as by std::memmove ).
For non-union class types, the move assignment operator performs full member-wise move assignment of the object's direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).
| The implicitly-defined move assignment operator for a class is if is a , and that is of class type (or array thereof), the assignment operator selected to move that member is a constexpr function. | (since C++14) (until C++23) |
| The implicitly-defined move assignment operator for a class is . | (since C++23) |
As with copy assignment, it is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined move assignment operator:
[ edit ] Deleted move assignment operator
The implicitly-declared or defaulted move assignment operator for class T is defined as deleted if any of the following conditions is satisfied:
- T has a non-static data member of a const-qualified non-class type (or possibly multi-dimensional array thereof).
- T has a non-static data member of a reference type.
- T has a potentially constructed subobject of class type M (or possibly multi-dimensional array thereof) such that the overload resolution as applied to find M 's move assignment operator
- does not result in a usable candidate, or
- in the case of the subobject being a variant member , selects a non-trivial function.
A deleted implicitly-declared move assignment operator is ignored by overload resolution .
[ edit ] Trivial move assignment operator
The move assignment operator for class T is trivial if all of the following is true:
- It is not user-provided (meaning, it is implicitly-defined or defaulted);
- T has no virtual member functions;
- T has no virtual base classes;
- the move assignment operator selected for every direct base of T is trivial;
- the move assignment operator selected for every non-static class type (or array of class type) member of T is trivial.
A trivial move assignment operator performs the same action as the trivial copy assignment operator, that is, makes a copy of the object representation as if by std::memmove . All data types compatible with the C language are trivially move-assignable.
[ edit ] Eligible move assignment operator
| A move assignment operator is eligible if it is not deleted. | (until C++20) |
| A move assignment operator is eligible if all following conditions are satisfied: (if any) are satisfied. than any other move assignment operator. | (since C++20) |
Triviality of eligible move assignment operators determines whether the class is a trivially copyable type .
[ edit ] Notes
If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move ), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.
It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined move assignment operator (same applies to copy assignment ).
See assignment operator overloading for additional detail on the expected behavior of a user-defined move-assignment operator.
[ edit ] Example
[ edit ] defect reports.
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
| DR | Applied to | Behavior as published | Correct behavior |
|---|---|---|---|
| C++11 | the conditions where defaulted move assignment operators are defined as deleted did not consider multi-dimensional array types | consider these types | |
| C++11 | a defaulted move assignment operator that would call a non-trivial copy assignment operator was deleted; a defaulted move assignment operator that is deleted still participated in overload resolution | allows call to such copy assignment operator; made ignored in overload resolution | |
| C++11 | specification for a defaulted move assignment operator involving a virtual base class was missing | added | |
| C++11 | a volatile subobject made of a defaulted move assignment operator non-trivial ( ) | triviality not affected | |
| C++11 | a defaulted move assignment operator for class was not defined as deleted if is abstract and has non-move-assignable direct virtual base classes | the operator is defined as deleted in this case | |
| C++20 | a move assignment operator was not eligible if there is another move assignment operator which is more constrained but does not satisfy its associated constraints | it can be eligible in this case | |
| C++11 | the implicitly-defined move assignment operator for union types did not copy the object representation | they copy the object representation |
[ edit ] See also
- constructor
- converting constructor
- copy assignment
- copy constructor
- default constructor
- aggregate initialization
- constant initialization
- copy initialization
- default initialization
- direct initialization
- list initialization
- reference initialization
- value initialization
- zero initialization
- move constructor
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C++ Assignment Operator Overloading
Prerequisite: Operator Overloading
The assignment operator,”=”, is the operator used for Assignment. It copies the right value into the left value. Assignment Operators are predefined to operate only on built-in Data types.
- Assignment operator overloading is binary operator overloading.
- Overloading assignment operator in C++ copies all values of one object to another object.
- Only a non-static member function should be used to overload the assignment operator.
In C++, the compiler automatically provides a default assignment operator for classes. This operator performs a shallow copy of each member of the class from one object to another. This means that if we don’t explicitly overload the assignment operator, the compiler will still allow us to assign one object to another using the assignment operator ( = ), and it won’t generate an error.
So, when we should perform assignment operator overloading? when our class involves dynamic memory allocation (e.g., pointers) and we need to perform a deep copy to prevent issues like double deletion or data corruption.
here, a and b are of type integer, which is a built-in data type. Assignment Operator can be used directly on built-in data types.
c1 and c2 are variables of type “class C”.
The above example can be done by implementing methods or functions inside the class, but we choose operator overloading instead. The reason for this is, operator overloading gives the functionality to use the operator directly which makes code easy to understand, and even code size decreases because of it. Also, operator overloading does not affect the normal working of the operator but provides extra functionality to it.
Now, if the user wants to use the assignment operator “=” to assign the value of the class variable to another class variable then the user has to redefine the meaning of the assignment operator “=”. Redefining the meaning of operators really does not change their original meaning, instead, they have been given additional meaning along with their existing ones.
Also, always check if the object is not being assigned to itself (e.g., if (this != &other)), as assigning an object to itself does not make sense and may cause runtime issues.
While managing dynamic resources, the above approach of assignment overloading have few flaws and there is more efficient approach that is recommended. See this article for more info – Copy-and-Swap Idiom in C++
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C++ variant can't be assigned as implicitly deleted [closed]
Im trying to use a variant to be able to store and access two different classes, Node and Nbody, for a barnes Hutt algorithm. However, I get the error:
error: object of type 'std::variant<Nbody, Blank, std::unique_ptr>' cannot be assigned because its copy assignment operator is implicitly deleted
Here is the header file sections using that.. If more code is nessecary I can provide:
- 6 std::unique_ptr can't be copied only moved from, if it could be copied then the object it manages would have 2 owners. – Richard Critten Commented 2 days ago
- 1 Please provide a minimal reproducible example that 1) closes the Node class properly, 2) uses the Node class, and 3) demonstrates the error you are seeing. Because I do not see any error when I try using this code: onlinegdb.com/zpQrFpKSn – Remy Lebeau Commented 2 days ago
- This is too mach irrelevant and invalid/not complete code. – 3CxEZiVlQ Commented 2 days ago
Browse other questions tagged c++ arrays c++11 or ask your own question .
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COMMENTS
the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial. A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD types) are trivially copy-assignable.
I'd say, default operator= is a copy. It copies each member. The distinction between a shallow copy and a deep copy doesn't arise unless the members being copied are some kind of indirection such as a pointer. As far as the default operator= is concerned, it's up to the member being copied what "copy" means, it could be deep or shallow.
Output: can't use default assignment operator. The compiler doesn't create default assignment operator in the following cases: 1. Class has a non-static data member of a const type or a reference type. 2. Class has a non-static data member of a type that has an inaccessible copy assignment operator. 3.
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x);. Use the copy constructor. If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you.
Shallow copying. Because C++ does not know much about your class, the default copy constructor and default assignment operators it provides use a copying method known as a memberwise copy (also known as a shallow copy).This means that C++ copies each member of the class individually (using the assignment operator for overloaded operator=, and direct initialization for the copy constructor).
The implicitly-declared (or defaulted on its first declaration) copy constructor has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17). [] Implicitly-defined copy constructoIf the implicitly-declared copy constructor is not deleted, it is defined (that is, a function body is generated and compiled) by the compiler if ...
C++ compiler implicitly provides a copy constructor, if no copy constructor is defined in the class. A bitwise copy gets created, if the Assignment operator is not overloaded. Consider the following C++ program. Explanation: Here, t2 = t1; calls the assignment operator, same as t2.operator= (t1); and Test t3 = t1; calls the copy constructor ...
A class can have multiple copy assignment operators, e.g. both T & T:: operator = (const T &) and T & T:: operator = (T). If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword default. (since C++11)
Note that none of the following constructors, despite the fact that. they could do the same thing as a copy constructor, are copy. constructors: 1. 2. MyClass( MyClass* other ); MyClass( const MyClass* other ); or my personal favorite way to create an infinite loop in C++: MyClass( MyClass other );
The Copy Assignment Operator in a class is a non-template non-static member function that is declared with the " operator= ". When you create a class or a type that is copy assignable (that you can copy with the = operator symbol), it must have a public copy assignment operator. Here is a simple syntax for the forced (defaulted) copy ...
The Copy Assignment Operator in a class is a non-template non-static member function that is declared with the operator=. When you create a class or a type that is copy assignable (that you can copy with the = operator symbol), it must have a public copy assignment operator. Here is a simple syntax for the typical declaration of a copy ...
In situations like these, you'll need to override C++'s default behavior by providing your own copy constructors and assignment operators. The Rule of Three There's a well-established C++ principle called the "rule of three" that almost always identifies the spots where you'll need to write your own copy constructor and assignment operator.
The rule of three is a well known C++ principle that states that if a class requires a user-defined copy constructor, destructor, or copy assignment operator, then it probably requires all three. In C++11, this was expanded to the rule of five, which adds the move constructor and move assignment operator to the list.
21.12 — Overloading the assignment operator. Alex July 22, 2024. The copy assignment operator (operator=) is used to copy values from one object to another already existing object. As of C++11, C++ also supports "Move assignment". We discuss move assignment in lesson 22.3 -- Move constructors and move assignment.
the move assignment operator selected for every non-static class type (or array of class type) member of T is trivial. A trivial move assignment operator performs the same action as the trivial copy assignment operator, that is, makes a copy of the object representation as if by std::memmove. All data types compatible with the C language are ...
Copy Assignment Operator is a special type of function that takes care of assigning the data of one object to another object. It gets called when you use this assignment operator (=) between objects. ... The default constructors and assignment operators do shallow copy and we create our own constructor and assignment operators when we. 5 min ...
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 ...
Overloading assignment operator in C++ copies all values of one object to another object. Only a non-static member function should be used to overload the assignment operator. In C++, the compiler automatically provides a default assignment operator for classes. This operator performs a shallow copy of each member of the class from one object ...
Through A's field buf, it looks like probably that the default assign operation will call something like memcpy to assign an object X to Y, so what if assign an object to itself and there are no explicit assign operation defined, like a = a; above. memcpy manual page: DESCRIPTION. The memcpy() function copies n bytes from memory area src to ...
error: object of type 'std::variant<Nbody, Blank, std::unique_ptr>' cannot be assigned because its copy assignment operator is implicitly deleted. Here is the header file sections using that.. If more code is nessecary I can provide: