Skip to content

Copy Constructor Vs Assignment Operator Syntax

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

In the C++programming language, the assignment operator, , is the operator used for assignment. Like most other operators in C++, it can be overloaded.

The copy assignment operator, often just called the "assignment operator", is a special case of assignment operator where the source (right-hand side) and destination (left-hand side) are of the same class type. It is one of the special member functions, which means that a default version of it is generated automatically by the compiler if the programmer does not declare one. The default version performs a memberwise copy, where each member is copied by its own copy assignment operator (which may also be programmer-declared or compiler-generated).

The copy assignment operator differs from the copy constructor in that it must clean up the data members of the assignment's target (and correctly handle self-assignment) whereas the copy constructor assigns values to uninitialized data members.[1] For example:

My_Arrayfirst;// initialization by default constructorMy_Arraysecond(first);// initialization by copy constructorMy_Arraythird=first;// Also initialization by copy constructorsecond=third;// assignment by copy assignment operator

Return value of overloaded assignment operator[edit]

The language permits an overloaded assignment operator to have an arbitrary return type (including ). However, the operator is usually defined to return a reference to the assignee. This is consistent with the behavior of assignment operator for built-in types (returning the assigned value) and allows for using the operator invocation as an expression, for instance in control statements or in chained assignment. Also, the C++ Standard Library requires this behavior for some user-supplied types.[2]

Overloading copy assignment operator[edit]

When deep copies of objects have to be made, exception safety should be taken into consideration. One way to achieve this when resource deallocation never fails is:

  1. Acquire new resources
  2. Release old resources
  3. Assign the new resources' handles to the object
classMy_Array{int*array;intcount;public:My_Array&operator=(constMy_Array&other){if(this!=&other)// protect against invalid self-assignment{// 1: allocate new memory and copy the elementsint*new_array=newint[other.count];std::copy(other.array,other.array+other.count,new_array);// 2: deallocate old memorydelete[]array;// 3: assign the new memory to the objectarray=new_array;count=other.count;}// by convention, always return *thisreturn*this;}// ...};

However, if a no-fail (no-throw) swap function is available for all the member subobjects and the class provides a copy constructor and destructor (which it should do according to the rule of three), the most straightforward way to implement copy assignment is as follows:[3]

public:voidswap(My_Array&other)// the swap member function (should never fail!){// swap all the members (and base subobject, if applicable) with otherusingstd::swap;// because of ADL the compiler will use // custom swap for members if it exists// falling back to std::swapswap(array,other.array);swap(count,other.count);}My_Array&operator=(My_Arrayother)// note: argument passed by value!{// swap this with otherswap(other);// by convention, always return *thisreturn*this;// other is destroyed, releasing the memory}

Assignment between different classes[edit]

C++ supports assignment between different classes, both via implicit copy constructor and assignment operator, if the destination instance class is the ancestor of the source instance class:

classAncestor{public:inta;};classDescendant:publicAncestor{public:intb;};intmain(){Descendantd;Ancestora(d);Ancestorb(d);a=d;return0;}

Copying from ancestor to descendant objects, which could leave descendant's fields uninitialized, is not permitted.

See also[edit]

References[edit]

External links[edit]

  1. ^Stroustrup, Bjarne (2000). The C++ Programming Language (3 ed.). Addison-Wesley. p. 244. ISBN 978-0-201-70073-2. 
  2. ^Working Draft, Standard for Programming Language C++, Section 17.6.3.1, Table 23; http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2012/n3337.pdf
  3. ^Sutter, H.; Alexandrescu, A. (October 2004), C++ Coding Standards, Addison-Wesley, ISBN 0-321-11358-6