copy assignment operator c syntax

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)

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
• 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 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.

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 ).

[ 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). However, this approach is not always advisable due to potentially significant overhead: see assignment operator overloading for details.

[ edit ] Example

• 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 30 November 2015, at 07:24.
• This page has been accessed 110,155 times.
• Privacy policy
• About cppreference.com
• Disclaimers

• C++ Data Types
• C++ Input/Output
• C++ Pointers
• C++ Interview Questions
• C++ Programs
• C++ Cheatsheet
• C++ Projects
• C++ Exception Handling
• C++ Memory Management

Copy Constructor vs Assignment Operator in C++

Copy constructor and Assignment operator are similar as they are both used to initialize one object using another object. But, there are some basic differences between them:

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 , same as Test t3(t1);

Must Read: When is a Copy Constructor Called in C++?

Similar Reads

Please login to comment..., improve your coding skills with practice.

What kind of Experience do you want to share?