copy assignment operator is implicitly deleted c

cppreference.com

Copy assignment operator.

A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . For a type to be CopyAssignable , it must have a public copy assignment operator.

[ edit ] Syntax

[ edit ] explanation.

• Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
• Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
• Forcing a copy assignment operator to be generated by the compiler.
• Avoiding implicit copy assignment.

The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.

[ edit ] Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:

• each direct base B of T has a copy assignment operator whose parameters are B or const B & or const volatile B & ;
• each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M & or const volatile M & .

Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.)

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)

The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification (until C++17) exception specification (since C++17)

Because the copy assignment operator 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 ] Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

• T has a user-declared move constructor;
• T has a user-declared move assignment operator.

Otherwise, it is defined as defaulted.

A defaulted copy assignment operator for class T is defined as deleted if any of the following is true:

• T has a non-static data member of non-class type (or array thereof) that is const ;
• T has a non-static data member of a reference type;
• T has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
• T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

[ edit ] Trivial copy assignment operator

The copy 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) , , and if it is defaulted, its signature is the same as implicitly-defined (until C++14) ;
• T has no virtual member functions;
• T has no virtual base classes;
• the copy assignment operator selected for every direct base of T is trivial;
• 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.

[ edit ] Implicitly-defined copy assignment operator

If the implicitly-declared copy 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 . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.

[ 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 copy assignment operator (same applies to move assignment ).

See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.

[ edit ] Example

[ edit ] defect reports.

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

• Pages with unreviewed CWG DR marker
• Recent changes
• Offline version
• What links here
• Related changes
• Upload file
• Special pages
• Printable version
• Permanent link
• Page information
• In other languages
• This page was last modified on 9 January 2019, at 07:16.
• This page has been accessed 570,566 times.
• Privacy policy
• About cppreference.com
• Disclaimers

Copy assignment operator

A copy assignment operator of class T is a non-template non-static member function with the name operator = that takes exactly one parameter of type T , T & , const T & , volatile T & , or const volatile T & . A type with a public copy assignment operator is CopyAssignable .

[ edit ] Syntax

[ edit ] explanation.

• Typical declaration of a copy assignment operator when copy-and-swap idiom can be used
• Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used
• Forcing a copy assignment operator to be generated by the compiler
• Avoiding implicit copy assignment

The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.

[ edit ] Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T & T :: operator = ( const T & ) if all of the following is true:

• each direct base B of T has a copy assignment operator whose parameters are B or const B& or const volatile B &
• each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M& or const volatile M &

Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument)

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 .

Because the copy assignment operator 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 ] Deleted implicitly-declared copy assignment operator

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 non-static data member that is const
• T has a non-static data member of a reference type.
• T has a non-static data member that cannot be copy-assigned (has deleted, inaccessible, or ambiguous copy assignment operator)
• T has direct or virtual base class that cannot be copy-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
• T has a user-declared move constructor
• T has a user-declared move assignment operator

[ edit ] Trivial copy assignment operator

The implicitly-declared copy assignment operator for class T is trivial if all of the following is true:

• T has no virtual member functions
• T has no virtual base classes
• The copy assignment operator selected for every direct base of T is trivial
• The copy assignment operator selected for every non-static class type (or array of class type) memeber 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.

[ edit ] Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is not deleted or trivial, it is defined (that is, a function body is generated and compiled) by the compiler. For union types, the implicitly-defined copy assignment copies the object representation (as by std:: memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated (since C++11) if T has a user-declared destructor or user-declared copy constructor.

[ edit ] Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either prvalue such as a nameless temporary or xvalue such as the result of std:: move ), and selects the copy assignment if the argument is 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.

[ edit ] Copy and swap

Copy assignment operator can be expressed in terms of copy constructor, destructor, and the swap() member function, if one is provided:

T & T :: operator = ( T arg ) { // copy/move constructor is called to construct arg     swap ( arg ) ;     // resources exchanged between *this and arg     return * this ; }   // destructor is called to release the resources formerly held by *this

For non-throwing swap(), this form provides strong exception guarantee . For rvalue arguments, this form automatically invokes the move constructor, and is sometimes referred to as "unifying assignment operator" (as in, both copy and move).

[ edit ] Example

Copy assignment operator

A copy assignment operator of class T is a non-template non-static member function with the name operator= that takes exactly one parameter (that isn't an explicit object parameter ) of type T , T& , const T& , volatile T& , or const volatile T& . For a type to be CopyAssignable , it must have a public copy assignment operator.

Explanation

The copy assignment operator is called whenever selected by overload resolution , e.g. when an object appears on the left side of an assignment expression.

Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type ( struct , class , or union ), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T& T::operator=(const T&) if all of the following is true:

• each direct base B of T has a copy assignment operator whose parameters are B or const B& or const volatile B& ;
• each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M& or const volatile M& .

Otherwise the implicitly-declared copy assignment operator is declared as T& T::operator=(T&) . (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.).

A class can have multiple copy assignment operators, e.g. both T& T::operator=(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) .

The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17) .

Because the copy assignment operator 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.

Deleted implicitly-declared copy assignment operator

An implicitly-declared copy assignment operator for class T is defined as deleted if any of the following is true:

• T has a user-declared move constructor;
• T has a user-declared move assignment operator.

Otherwise, it is defined as defaulted.

A defaulted copy assignment operator for class T is defined as deleted if any of the following is true:

• T has a non-static data member of a const-qualified non-class type (or array thereof);
• T has a non-static data member of a reference type;
• T has a non-static data member or a direct base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
• T is a union-like class , and has a variant member whose corresponding assignment operator is non-trivial.

Trivial copy assignment operator

The copy 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 copy assignment operator selected for every direct base of T is trivial;
• 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.

Eligible copy assignment operator

Triviality of eligible copy assignment operators determines whether the class is a trivially copyable type .

Implicitly-defined copy assignment operator

If the implicitly-declared copy 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 copy assignment copies the object representation (as by std::memmove ). For non-union class types ( class and struct ), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.

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 copy assignment operator (same applies to move assignment ).

See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

• converting constructor
• copy constructor
• copy elision
• default constructor
• aggregate initialization
• constant initialization
• copy initialization
• default initialization
• direct initialization
• initializer list
• list initialization
• reference initialization
• value initialization
• zero initialization
• move assignment
• move constructor

© cppreference.com Licensed under the Creative Commons Attribution-ShareAlike Unported License v3.0. https://en.cppreference.com/w/cpp/language/as_operator

• Windows Programming
• UNIX/Linux Programming
• General C++ Programming
• Assignment operator implicitly deleted

  Assignment operator implicitly deleted

This browser is no longer supported.

Upgrade to Microsoft Edge to take advantage of the latest features, security updates, and technical support.

Explicitly Defaulted and Deleted Functions

• 8 contributors

In C++11, defaulted and deleted functions give you explicit control over whether the special member functions are automatically generated. Deleted functions also give you simple language to prevent problematic type promotions from occurring in arguments to functions of all types—special member functions, and normal member functions and nonmember functions—which would otherwise cause an unwanted function call.

Benefits of explicitly defaulted and deleted functions

In C++, the compiler automatically generates the default constructor, copy constructor, copy-assignment operator, and destructor for a type if it doesn't declare its own. These functions are known as the special member functions , and they're what make simple user-defined types in C++ behave like structures do in C. That is, you can create, copy, and destroy them without extra coding effort. C++11 brings move semantics to the language and adds the move constructor and move-assignment operator to the list of special member functions that the compiler can automatically generate.

This is convenient for simple types, but complex types often define one or more of the special member functions themselves, and this can prevent other special member functions from being automatically generated. In practice:

If any constructor is explicitly declared, then no default constructor is automatically generated.

If a virtual destructor is explicitly declared, then no default destructor is automatically generated.

If a move constructor or move-assignment operator is explicitly declared, then:

No copy constructor is automatically generated.

No copy-assignment operator is automatically generated.

If a copy constructor, copy-assignment operator, move constructor, move-assignment operator, or destructor is explicitly declared, then:

No move constructor is automatically generated.

No move-assignment operator is automatically generated.

Additionally, the C++11 standard specifies the following additional rules:

• If a copy constructor or destructor is explicitly declared, then automatic generation of the copy-assignment operator is deprecated.
• If a copy-assignment operator or destructor is explicitly declared, then automatic generation of the copy constructor is deprecated.

In both cases, Visual Studio continues to automatically generate the necessary functions implicitly, and doesn't emit a warning by default. Since Visual Studio 2022 version 17.7, C5267 can be enabled to emit a warning.

The consequences of these rules can also leak into object hierarchies. For example, if for any reason a base class fails to have a default constructor that's callable from a deriving class—that is, a public or protected constructor that takes no parameters—then a class that derives from it can't automatically generate its own default constructor.

These rules can complicate the implementation of what should be straight-forward, user-defined types and common C++ idioms—for example, making a user-defined type noncopyable by declaring the copy constructor and copy-assignment operator privately and not defining them.

Before C++11, this code snippet was the idiomatic form of noncopyable types. However, it has several problems:

The copy constructor has to be declared privately to hide it, but because it's declared at all, automatic generation of the default constructor is prevented. You have to explicitly define the default constructor if you want one, even if it does nothing.

Even if the explicitly defined default constructor does nothing, the compiler considers it to be nontrivial. It's less efficient than an automatically generated default constructor and prevents noncopyable from being a true POD type.

Even though the copy constructor and copy-assignment operator are hidden from outside code, the member functions and friends of noncopyable can still see and call them. If they're declared but not defined, calling them causes a linker error.

Although this is a commonly accepted idiom, the intent isn't clear unless you understand all of the rules for automatic generation of the special member functions.

In C++11, the noncopyable idiom can be implemented in a way that is more straightforward.

Notice how the problems with the pre-C++11 idiom are resolved:

Generation of the default constructor is still prevented by declaring the copy constructor, but you can bring it back by explicitly defaulting it.

Explicitly defaulted special member functions are still considered trivial, so there's no performance penalty, and noncopyable isn't prevented from being a true POD type.

The copy constructor and copy-assignment operator are public but deleted. It's a compile-time error to define or call a deleted function.

The intent is clear to anyone who understands =default and =delete . You don't have to understand the rules for automatic generation of special member functions.

Similar idioms exist for making user-defined types that are nonmovable, that can only be dynamically allocated, or that can't be dynamically allocated. Each of these idioms have pre-C++11 implementations that suffer similar problems, and that are similarly resolved in C++11 by implementing them in terms of defaulted and deleted special member functions.

Explicitly defaulted functions

You can default any of the special member functions—to explicitly state that the special member function uses the default implementation, to define the special member function with a nonpublic access qualifier, or to reinstate a special member function whose automatic generation was prevented by other circumstances.

You default a special member function by declaring it as in this example:

Notice that you can default a special member function outside the body of a class as long as it's inlinable.

Because of the performance benefits of trivial special member functions, we recommend that you prefer automatically generated special member functions over empty function bodies when you want the default behavior. You can do this either by explicitly defaulting the special member function, or by not declaring it (and also not declaring other special member functions that would prevent it from being automatically generated.)

Deleted functions

You can delete special member functions and normal member functions and nonmember functions to prevent them from being defined or called. Deleting of special member functions provides a cleaner way of preventing the compiler from generating special member functions that you don't want. The function must be deleted as it's declared; it can't be deleted afterwards in the way that a function can be declared and then later defaulted.

Deleting of normal member function or nonmember functions prevents problematic type promotions from causing an unintended function to be called. This works because deleted functions still participate in overload resolution and provide a better match than the function that could be called after the types are promoted. The function call resolves to the more-specific—but deleted—function and causes a compiler error.

Notice in the preceding sample that calling call_with_true_double_only by using a float argument would cause a compiler error, but calling call_with_true_double_only by using an int argument wouldn't; in the int case, the argument will be promoted from int to double and successfully call the double version of the function, even though that might not be what you intend. To ensure that any call to this function by using a non-double argument causes a compiler error, you can declare a template version of the deleted function.

Was this page helpful?

Coming soon: Throughout 2024 we will be phasing out GitHub Issues as the feedback mechanism for content and replacing it with a new feedback system. For more information see: https://aka.ms/ContentUserFeedback .

Submit and view feedback for

Additional resources

Copy assignment operators (C++ only)

The copy assignment operator lets you create a new object from an existing one by initialization. A copy assignment operator of a class A is a nonstatic non-template member function that has one of the following forms:

• A::operator=(A)
• A::operator=(A&)
• A::operator=(const A&)
• A::operator=(volatile A&)
• A::operator=(const volatile A&)

If you do not declare a copy assignment operator for a class A , the compiler will implicitly declare one for you which will be inline public.

The assignment x = y calls the implicitly defined copy assignment operator of B , which calls the user-defined copy assignment operator A::operator=(const A&) . The assignment w = z calls the user-defined operator A::operator=(A&) . The compiler will not allow the assignment i = j because an operator C::operator=(const C&) has not been defined.

The implicitly declared copy assignment operator of a class A will have the form A& A::operator=(const A&) if the following are true:

• A direct or virtual base B of class A has a copy assignment operator whose parameter is of type const B& , const volatile B& , or B .
• A non-static class type data member of type X that belongs to class A has a copy constructor whose parameter is of type const X& , const volatile X& , or X .

If the above are not true for a class A , the compiler will implicitly declare a copy assignment operator with the form A& A::operator=(A&) .

The implicitly declared copy assignment operator returns a reference to the operator's argument.

The copy assignment operator of a derived class hides the copy assignment operator of its base class.

• Class A has a nonstatic data member of a const type or a reference type
• Class A has a nonstatic data member of a type which has an inaccessible copy assignment operator
• Class A is derived from a base class with an inaccessible copy assignment operator.

An implicitly defined copy assignment operator of a class A will first assign the direct base classes of A in the order that they appear in the definition of A . Next, the implicitly defined copy assignment operator will assign the nonstatic data members of A in the order of their declaration in the definition of A .

• Assignment operators