Copy Assignment Operator Derived Class And Base

A copy assignment operator of class 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 constvolatile T&. For a type to be , it must have a public copy assignment operator.

[edit]Syntax

class_nameclass_name ( class_name ) (1)
class_nameclass_name ( const class_name ) (2)
class_nameclass_name ( const class_name ) = default; (3) (since C++11)
class_nameclass_name ( const class_name ) = delete; (4) (since C++11)

[edit]Explanation

  1. Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
  2. Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
  3. Forcing a copy assignment operator to be generated by the compiler.
  4. 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 of has a copy assignment operator whose parameters are B or const B& or constvolatile B&;
  • each non-static data member of of class type or array of class type has a copy assignment operator whose parameters are M or const M& or constvolatile 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 .(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 is defined as deleted if any of the following is true:

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

Otherwise, it is defined as defaulted.

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

  • has a non-static data member of non-class type (or array thereof) that is const;
  • has a non-static data member of a reference type;
  • 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);
  • 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 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);
  • has no virtual member functions;
  • has no virtual base classes;
  • the copy assignment operator selected for every direct base of is trivial;
  • the copy assignment operator selected for every non-static class type (or array of class type) member of is trivial;
  • has no non-static data members of volatile-qualified type.
(since C++14)

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 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

Run this code

Output:

#include <iostream>#include <memory>#include <string>#include <algorithm>   struct A {int n;std::string s1;// user-defined copy assignment, copy-and-swap form A& operator=(A other){std::cout<<"copy assignment of A\n";std::swap(n, other.n);std::swap(s1, other.s1);return*this;}};   struct B : A {std::string s2;// implicitly-defined copy assignment};   struct C {std::unique_ptr<int[]> data;std::size_t size;// non-copy-and-swap assignment C& operator=(const C& other){// check for self-assignmentif(&other == this)return*this;// reuse storage when possibleif(size != other.size){ data.reset(new int[other.size]); size = other.size;}std::copy(&other.data[0], &other.data[0]+ size, &data[0]);return*this;}// note: copy-and-swap would always cause a reallocation};   int main(){ A a1, a2;std::cout<<"a1 = a2 calls "; a1 = a2;// user-defined copy assignment   B b1, b2; b2.s1="foo"; b2.s2="bar";std::cout<<"b1 = b2 calls "; b1 = b2;// implicitly-defined copy assignmentstd::cout<<"b1.s1 = "<< b1.s1<<" b1.s2 = "<< b1.s2<<'\n';}
a1 = a2 calls copy assignment of A b1 = b2 calls copy assignment of A b1.s1 = foo b1.s2 = bar

[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
CWG 2171 C++14 operator=(X&)=default was non-trivial made trivial

The requirement looks wrong. The reason is that virtual methods and copy constructors don't work together. Virtual methods are useful if you use the object polymorphically, that is via reference/pointer to a base class. But copy constructor always constructs object of the static type it is written as, so if you give it polymorphic object, it will only create the base class, not the subclass you wanted. This is called slicing.

So you want to either:

  • Have a templated framework that knows the type at compile time and uses copy-constructor. But this does not need any virtual methods, nor for that matter a common base type, since the code will be compiled with the specific template parameter. Compiler will complain if the type substituted does not support the operations the template tries to use, but you can slightly improve diagnostics by using some explicit checks, e.g. from Boost.Concept Check.

  • Have a framework using polymorphic objects with base class, but that can't create copies with copy constructor, because it does not know the type to construct. There are three ways out:

    1. Have a pure virtual method, that will be implemented to do in each concrete subclass. You should probably wrap that in or immediately to avoid leaking the copies.
    2. Instead of copies, hand out references or smart pointers, probably in this case. It would probably be either const references/smart pointers or typed with interface that only contains methods that can't affect invariants in the framework.
    3. Make the method that inserts the objects a template, that would construct a cloner and store it along with the object. Cloner is a function (you can have a functor or a class with multiple helper methods like this). To ensure you really have correct type, the insert method should probably wrap the call to as well similar to e.g. .

By the way, nothing of this has anything to do with rule of three. Because rule of three says that if default copy constructor, default assignment operator or default destructor are not good enough for the class, than none of them is. But not because compiler would require it, but because the logic probably does. But for most classes that want to be copyable they are good enough. Because in most cases you hide the resource handling in some smart pointer and just let the default copy/assignment/destructor call to it.

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