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Copy assignment operator.

A copy 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 without mutating the argument.

[ edit ] Syntax

For the formal copy assignment operator syntax, see function declaration . The syntax list below only demonstrates a subset of all valid copy assignment operator syntaxes.

[ edit ] 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.

[ edit ] Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type, 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 & ) .

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.

[ 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 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, the operator performs member-wise copy assignment of the object's direct bases and non-static data members, in their initialization order, using built-in assignment for the scalars, memberwise copy-assignment for arrays, and copy assignment operator for class types (called non-virtually).

[ edit ] Deleted copy assignment operator

An implicitly-declared or explicitly-defaulted (since C++11) copy assignment operator for class T is undefined (until C++11) defined as deleted (since C++11) 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 copy 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.

[ 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);
  • 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 ] Eligible copy assignment operator

Triviality of eligible copy 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 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.

[ edit ] See also

  • 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
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IMAGES

  1. C++ : Deleted implicitly-declared copy assignment operator

    its copy assignment operator is implicitly deleted

  2. C++ : Why do reference type members cause implicitly-declared copy

    its copy assignment operator is implicitly deleted

  3. 2 Wrong Way to Learn Copy Assignment Operator in C++ With Example

    its copy assignment operator is implicitly deleted

  4. What Is Deleted Implicitly-declared Copy Assignment Operator In C++

    its copy assignment operator is implicitly deleted

  5. [100% Working Code]

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  6. C++ : Why defining a destructor deletes the implicitly defined move

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COMMENTS

  1. c++

    I'm trying to make a move operator for a class (which we will call A) that contains another reference to another class (which we will call B ), whose copy constructor has been implicitly deleted because it contains another reference. A simple example has been shown below. public: int & num; B(int & _num) : num(_num) {} public:

  2. c++

    error: object of type 'Socket' cannot be assigned because its copy assignment operator is implicitly deleted connection->second.socketObj = std::move(socketObj); ^ ./socketwrapper.h:44:46: note: copy assignment operator of 'Socket' is implicitly deleted because field 'socket' has a deleted copy assignment operator seastar::output_stream<char ...