Implicitly Defined Copy Assignment Operator

In the tradition of The Zen of Python, I’ve been thinking about pushing for explicit declarations of otherwise implicitly-defined member functions in C++, both in code that I write and in code that I review:

// Instances of this class should not be copied. MyClass(const MyClass&) = delete; MyClass& operator=(const MyClass&) = delete; // We are OK with the default semantics. OtherClass(const OtherClass&) = default; OtherClass& operator=(const OtherClass&) = default; OtherClass(OtherClass&&) = default; OtherClass& operator=(OtherClass&&) = default;

[Background: C++ specifies several member functions that the compiler will implicitly define for you in any class: the default constructor, the copy/move constructor(s), and the copy/move assignment operator(s). I say “implicitly define”, as though that always happens, but there are a number of constraints on when the compiler will do this. For the purposes of the discussion below, I’ll ignore the default constructor bit and focus on the copy/move constructor and assignment operator. (I will also happily ignore all the different variants thereof that can occur, e.g. when the compiler defines for you.) I think the arguments apply equally well to the default constructor case, but most classes I deal with tend to either declare their own default constructor or have several user-defined constructors anyway, which prohibit the compiler from implicitly declaring the default constructor.]

I think the argument for is more obvious and less controversial, so I’ll start there.  ‘ing functions you don’t want used is part of the API contract of the class.  Functions that shouldn’t be used shouldn’t be exposed to the user, and ensures that the compiler won’t implicitly define part of your API surface (and users thereby unknowingly violate API guarantees).  The copy constructor/assignment operator are the obvious candidates for , but using for the move constructor/assignment operator makes sense in some cases (e.g. RAII classes). Using gives you pleasant compiler error messages, and it’s clearer than:

private: MyClass(const MyClass&); MyClass& operator=(const MyClass&);

If you’re lucky, there might be a comment to the effect of .  I know which code I’d prefer to read. (Using also ensures you don’t accidentally use the not-defined members inside the class itself, then spend a while waiting for the linker errors to tell you about your screw-up.)

appears to be a little harder to argue for.  “Experienced” programmers always know which functions are provided by the compiler, right?

Understanding whether the compiler implicitly defines something requires looking at the entire class definition (including superclasses) and running a non-trivial decision algorithm. I sure don’t want each reader of the code to do that for two or four different member functions (times superclasses, too), all of which are reasonably important in understanding how a class is intended to be used.

Explicitly declaring what you intend can also avoid performance pitfalls. In reading through the C++ specification to understand when things were implicitly declared, I discovered that the same functions can also be implicitly deleted, including this great note: “When the move constructor is not implicitly declared or explicitly supplied, expressions that otherwise would have invoked the move constructor may instead invoke a copy constructor.” So, if the move constructor was implicitly declared at some point, but then was implicitly deleted through some change, expressions that were previously efficient due to moving would become somewhat less so due to copying. Isn’t C++ great?

Being explicit also avoids the possibility of meaning to define something, but getting tripped up by the finer points of the language:

template<typename T> class MyClass { public: // This does not define a copy constructor for MyClass<T>. template<typename U> MyClass(const MyClass<U>& aOther) : ... { ... } ... };

Comments could serve to notify the reader that we’re OK with the default definition, but if I could choose between encoding something in a place solely intended for humans, or a place both humans and the compiler will understand, I know which one I’d pick.

Tags: c++, code review, code style, mozilla

This entry was posted on Thursday, August 20th, 2015 at 12:14 pm and is filed under Uncategorized. You can follow any comments to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.

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.


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)


  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.


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.


Run this code


#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(&[0], &[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

One thought on “Implicitly Defined Copy Assignment Operator

Leave a Reply

Your email address will not be published. Required fields are marked *