assignment operator is implicitly deleted

• Windows Programming
• UNIX/Linux Programming
• General C++ Programming
• Assignment operator implicitly deleted

  Assignment operator implicitly deleted

About Jonathan Boccara

Hello, my name is Jonathan Boccara, I'm your host on Fluent C++. I have been a developer for 10 years. My focus is on how to write expressive code . I wrote the book The Legacy Code Programmer's Toolbox . I'm happy to take your feedback, don't hesitate to drop a comment on a post, follow me or get in touch directly !

Jonathan Boccara's blog

Recent Posts

• Usage First, Implementation After: A Principle of Software Development
• Design Patterns VS Design Principles: Factory method
• How to Store an lvalue or an rvalue in the Same Object
• Copy-Paste Developments
• Design Patterns VS Design Principles: Abstract Factory
• How to Generate All the Combinations from Several Collections

How to Make a Copyable Object Assignable in C++

Some types in C++ have a copy constructor that doesn’t have the same semantics as their assignment operator ( operator= ).

Take references, for example. References can be copied:

But it doesn’t do the same thing as assigning to them:

With the copy, r2 points to the same thing as r1 , but with the assignment r2 still points to the same object it was pointing to before.

Or take the example of copying a lambda:

The above code compiles fine.

Now if we add the following line:

It doesn’t compile. As the compiler (clang) says:

Lambdas don’t even have an operator= to begin with (except in C++20 where they do if they don’t capture anything).

Right. But is any of this a problem?

Why we need operator=

After all, the behaviour of the references makes some sense, and why on earth would we like to assign on a lambda we’ve just created?

However, there is a case when the absence of operator= becomes a problem: when the object that doesn’t have an operator= is a member of a class. It makes it difficult for that class to have an operator= itself. For one thing, the compiler is not going to write it for you.

Even for references, the compiler won’t generate an operator= for a class if one of its members is a reference. It assumes that you’d better write it yourself to choose what to do with the reference member.

This problem came up in a project I’ve been working on, the pipes library . This library has classes that have lambdas as data members, and passes objects of those classes as output iterators of STL algorithms. And in Visual Studio, the STL in debug mode calls the operator= on output iterators in the _Recheck function. So the class that contains a lambda needs an operator= .

Haven’t you ever faced too the situation where the compiler couldn’t write the operator= you needed because of a problematic data member?

The standard has us covered for references

In C++11, and equivalently in Boost long before that, std::reference_wrapper<T>  has the same behaviour as a reference (you initialize it with a reference, and it even has a operator T& ) with one exception: it has an operator= that rebinds the reference.

This means that after calling operator= between two std::reference_wrapper s, they point to the same object:

The fact that std::reference_wrapper<T> has an operator= allows the compiler to generate an operator= for the classes that contains it. And the fact that it rebinds gives the operator= of the containing class a natural behaviour.

Why is this behaviour natural? Because it is consistent with the copy of the reference: in both case, the two reference(_wrapper)s point to the same object after the operation.

The general case

Even if the case of references is solved with std::reference_wrapper , the case of the lambda remains unsolved, along with all the types that have a copy constructor and no operator= .

Let’s design a component, inspired from std::reference_wrapper , that would add to any type an operator= which is consistent with its copy constructor.

If you have an idea on how to name this component, just leave a comment below at the bottom of the post. For the time being, let’s call it assignable .

assignable needs an operator= that relies on the copy constructor of its underlying type. Fortunately, we know how to implement that with a std::optional , like we saw in How to Implement operator= When a Data Member Is a Lambda :

But now that we’ve written the copy assignment operator, the compiler is going to refrain from generating the move constructor and the move assignment operator. It’s a shame, so let’s add them back:

Now that we’ve written all this, we might as well write the copy constructor too. The compiler would have generated it for us, but I think it looks odd to write everything except this one:

Finally, in order to hide from its users the fact that assignable contains an optional , let’s add constructors that accepts a T :

Giving access to the underlying value

Like optional , assignable wraps a type to add an extra feature, but its goal is not to mimic the interface of the underlying object. So we should give access to the underlying object of assignable . We will define a get()  member function, because  operator*  and operator->  could suggest that there is an indirection (like for pointers and iterators).

The underlying object of the  assignable happens to to be the underlying object of the optional inside of the assignable :

We don’t check for the nullity of the optional, because the interface of assignable is such that all the paths leading to those dereferencing operators guarantee that the optional has been initialized.

Which gives us food for thought: optional is not the optimal solution here. It contains a piece of information that we never use: whether the optional is null or not.

A better solution would be to create a component that does placement news like optional, but without the possibility to be null.

Lets keep this as food for thought for the moment. Maybe we’ll come back to it in a later article. Please leave a comment if you have thoughts on that.

Making the assignable callable

std::reference_wrapper has a little known feature that we explored in How to Pass a Polymorphic Object to an STL Algorithm : it has an operator() that calls its underlying reference when it’s callable.

This is all the more relevant for assignable since our motivating case was a lambda.

If we don’t implement operator() , we’d have to write code like this:

Whereas with an operator() , calling code becomes more natural, resembling the one of a lambda:

Let’s do it then!

We rely on C++14 decltype(auto) . Note that we could also implement this in C++11 the following way:

The case of assignable references

Now we have implemented an assignable<T> that works when T is a lambda.

But what if T is a reference?

It can happen in the case of a function reference. In that case, we need exactly the same features as the ones we needed with the lambda.

However, assignable<T> doesn’t even compile when T is a reference. Why? Because it uses an std::optional<T> and optional references didn’t make it in the C++ standard .

Luckily, implementing assignable for references is not difficult. In fact, it’s a problem already solved by… std::reference_wrapper !

So we need to create a specialization of assignable<T> when T is a reference. It would be great if we could just write this:

But this is not possible in C++.

Instead we have to implement a type that wraps std::reference_wrapper and relies on its behaviour:

This way, we can use assignable on reference types.

Putting it all together

In summary, here is all the code of assignable all put together:

And classes can use it as a data member this way:

For such as class, the compiler would be able to generate a operator= as long as Function has a copy constructor, which many classes–including lambdas–do.

Thanks to Eric Niebler for the inspiration, as assignable was inspired from techniques I’ve seen in range-v3 , which is my go-to model for library implementation.

If you have any piece of feedback on assignable , I’d love to hear it in a comment below!

You will also like

• How to Pass a Polymorphic Object to an STL Algorithm
• How to Implement operator= When a Data Member Is a Lambda
• An Alternative Design to Iterators and Ranges, Using std::optional
• Why Optional References Didn’t Make It In C++17
• Pointers, References and Optional References in C++
• Smart Output Iterators: A Symmetrical Approach to Range Adaptors

The C++ implicit assignment operator is a non-ref-qualified member, even if the base class’s assignment has a ref-qualifier

Raymond Chen

September 16th, 2021 6 0

Consider the following C++ class:

This defines a class for which you can assign to an lvalue reference, but not to an rvalue reference.

Assigning to an rvalue is not generally useful, since the object has no name, and consequently it is difficult to do anything with the assigned-to object afterward.¹

Great, we got rid of assignment to a temporary, which we’ve seen has been a source of confusion .

Now consider this:

We created a derived class and inherited the assignment operator from it. Do you expect the inherited assignment operator to block rvalues?

You probably guessed that the answer is no , seeing as I gave it away in the title.

The reason is that I lied when I said that the assignment operator was inherited. It was not inherited. It was implicitly declared .

The rules for implicit declaration of the copy assignment operator are spelled out in [class.copy.assign] , paragraphs 2 and 4. The short version is that a class is eligible for an implicitly-declared copy assignment operator if its base classes and non-static members all have a copy assignment operator. (Analogous rules apply for the implicitly-declared move assignment operator.)

The catch is that the implicitly-declared copy assignment and move assignment operators are declared as an unqualified assignment operator, regardless of the reference-qualifications of the base classes and members. In our example, we get

The lack of a ref-qualification means that this assignment operator applies equally to lvalues and rvalues.

Our attempt to block rvalue assignment fails to propagate to derived classes!

In order to repair this, each derived class must redeclare its assignment operator as lvalue-only.

Oh, we’ve only started our journey down the rabbit-hole.

At least for now, explicitly declaring a copy assignment operator does not cause the implicitly-declared copy/move constructors to disappear, but the behavior is noted as deprecated in the C++ language specification, with the note that a future version of the language may indeed delete them.

To make sure you don’t run into trouble in the future, you’ll want to declare them explicitly.

Great, we’ve restored the copy and move constructors.

But explicitly declaring any constructors causes us to lose the implicitly-declared default constructor.

We’ll have to bring that back too.

The same exercise applies if we also want to block the move assignment operator to rvalues, but it’s more urgent because an explicit declaration of a move assignment operator does delete both the copy and move constructors even in C++20.

Phew, that was annoying.

¹ I mean, I guess you could do this:

but it seems pointless to go to the effort of asking Get­Base to create you a Base object, only to overwrite it with your own. You may as well just create your own temporary.

Or, if you didn’t even mean to create a temporary, just use the original value:

Discussion is closed. Login to edit/delete existing comments.

> Because explicitly declaring an assignment operator causes the implicitly-declared copy/move constructors to disappear.

Are you sure about that? Seems to [work for me]( https://godbolt.org/z/veb97xxrT ). Also, if that were true, then `Derived(Derived const&) = default;` wouldn’t have worked either, as that requires `Base` to have a copy constructor, and you claim that it doesn’t have one.

A nit: in many examples in the article, `Derived` isn’t actually derived from `Base`.

You’re right. I confused it with the move assignment operator. I’ve revised the article. Thanks.

There are still many instances where `Derived` is not actually derived from `Base`. And there’s still the issue that, in the world where `Derived` wouldn’t have an implicit copy constructor, `Base` wouldn’t have one either, and then `Derived(Derived const&) = default;` would default it as deleted.

Sometimes C++ seems like … all a big mistake

> Great, we got rid of assignment to a temporary, which we’ve seen has been a source of confusion.

Being able to do assignment into a temporary can be marginally useful. I can think on one use-case for it : let’s assume we have a matrix of bools whose underlying layout is like vector of bools. Then calling matrix(i, j) will yield a temporary (proxy) object whose assignment will result in mutating the corresponding bit in the matrix buffer.

Well, we all know we should not imitate vector of bools but proxy objects are a thing, especially with the new addition of ranges in the standard. Being able to refine behavior for these cases is especially useful when implementing custom adapters/facades. And not being able to inherit such behavior will most probably be really annoying…

This browser is no longer supported.

Upgrade to Microsoft Edge to take advantage of the latest features, security updates, and technical support.

Explicitly Defaulted and Deleted Functions

• 8 contributors

In C++11, defaulted and deleted functions give you explicit control over whether the special member functions are automatically generated. Deleted functions also give you simple language to prevent problematic type promotions from occurring in arguments to functions of all types—special member functions, and normal member functions and nonmember functions—which would otherwise cause an unwanted function call.

Benefits of explicitly defaulted and deleted functions

In C++, the compiler automatically generates the default constructor, copy constructor, copy-assignment operator, and destructor for a type if it doesn't declare its own. These functions are known as the special member functions , and they're what make simple user-defined types in C++ behave like structures do in C. That is, you can create, copy, and destroy them without extra coding effort. C++11 brings move semantics to the language and adds the move constructor and move-assignment operator to the list of special member functions that the compiler can automatically generate.

This is convenient for simple types, but complex types often define one or more of the special member functions themselves, and this can prevent other special member functions from being automatically generated. In practice:

If any constructor is explicitly declared, then no default constructor is automatically generated.

If a virtual destructor is explicitly declared, then no default destructor is automatically generated.

If a move constructor or move-assignment operator is explicitly declared, then:

No copy constructor is automatically generated.

No copy-assignment operator is automatically generated.

If a copy constructor, copy-assignment operator, move constructor, move-assignment operator, or destructor is explicitly declared, then:

No move constructor is automatically generated.

No move-assignment operator is automatically generated.

Additionally, the C++11 standard specifies the following additional rules:

• If a copy constructor or destructor is explicitly declared, then automatic generation of the copy-assignment operator is deprecated.
• If a copy-assignment operator or destructor is explicitly declared, then automatic generation of the copy constructor is deprecated.

In both cases, Visual Studio continues to automatically generate the necessary functions implicitly, and doesn't emit a warning by default. Since Visual Studio 2022 version 17.7, C5267 can be enabled to emit a warning.

The consequences of these rules can also leak into object hierarchies. For example, if for any reason a base class fails to have a default constructor that's callable from a deriving class—that is, a public or protected constructor that takes no parameters—then a class that derives from it can't automatically generate its own default constructor.

These rules can complicate the implementation of what should be straight-forward, user-defined types and common C++ idioms—for example, making a user-defined type noncopyable by declaring the copy constructor and copy-assignment operator privately and not defining them.

Before C++11, this code snippet was the idiomatic form of noncopyable types. However, it has several problems:

The copy constructor has to be declared privately to hide it, but because it's declared at all, automatic generation of the default constructor is prevented. You have to explicitly define the default constructor if you want one, even if it does nothing.

Even if the explicitly defined default constructor does nothing, the compiler considers it to be nontrivial. It's less efficient than an automatically generated default constructor and prevents noncopyable from being a true POD type.

Even though the copy constructor and copy-assignment operator are hidden from outside code, the member functions and friends of noncopyable can still see and call them. If they're declared but not defined, calling them causes a linker error.

Although this is a commonly accepted idiom, the intent isn't clear unless you understand all of the rules for automatic generation of the special member functions.

In C++11, the noncopyable idiom can be implemented in a way that is more straightforward.

Notice how the problems with the pre-C++11 idiom are resolved:

Generation of the default constructor is still prevented by declaring the copy constructor, but you can bring it back by explicitly defaulting it.

Explicitly defaulted special member functions are still considered trivial, so there's no performance penalty, and noncopyable isn't prevented from being a true POD type.

The copy constructor and copy-assignment operator are public but deleted. It's a compile-time error to define or call a deleted function.

The intent is clear to anyone who understands =default and =delete . You don't have to understand the rules for automatic generation of special member functions.

Similar idioms exist for making user-defined types that are nonmovable, that can only be dynamically allocated, or that can't be dynamically allocated. Each of these idioms have pre-C++11 implementations that suffer similar problems, and that are similarly resolved in C++11 by implementing them in terms of defaulted and deleted special member functions.

Explicitly defaulted functions

You can default any of the special member functions—to explicitly state that the special member function uses the default implementation, to define the special member function with a nonpublic access qualifier, or to reinstate a special member function whose automatic generation was prevented by other circumstances.

You default a special member function by declaring it as in this example:

Notice that you can default a special member function outside the body of a class as long as it's inlinable.

Because of the performance benefits of trivial special member functions, we recommend that you prefer automatically generated special member functions over empty function bodies when you want the default behavior. You can do this either by explicitly defaulting the special member function, or by not declaring it (and also not declaring other special member functions that would prevent it from being automatically generated.)

Deleted functions

You can delete special member functions and normal member functions and nonmember functions to prevent them from being defined or called. Deleting of special member functions provides a cleaner way of preventing the compiler from generating special member functions that you don't want. The function must be deleted as it's declared; it can't be deleted afterwards in the way that a function can be declared and then later defaulted.

Deleting of normal member function or nonmember functions prevents problematic type promotions from causing an unintended function to be called. This works because deleted functions still participate in overload resolution and provide a better match than the function that could be called after the types are promoted. The function call resolves to the more-specific—but deleted—function and causes a compiler error.

Notice in the preceding sample that calling call_with_true_double_only by using a float argument would cause a compiler error, but calling call_with_true_double_only by using an int argument wouldn't; in the int case, the argument will be promoted from int to double and successfully call the double version of the function, even though that might not be what you intend. To ensure that any call to this function by using a non-double argument causes a compiler error, you can declare a template version of the deleted function.

Was this page helpful?

Coming soon: Throughout 2024 we will be phasing out GitHub Issues as the feedback mechanism for content and replacing it with a new feedback system. For more information see: https://aka.ms/ContentUserFeedback .

Submit and view feedback for

Additional resources

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)

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 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 (until C++14) ;
• 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 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.

• Pages with unreviewed CWG DR marker
• 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 31 March 2017, at 13:00.
• This page has been accessed 272,825 times.
• Privacy policy
• About cppreference.com
• Disclaimers

Move assignment operator

A move 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 && , const T && , volatile T && , or const volatile T && . A type with a public move assignment operator is MoveAssignable .

[ edit ] Syntax

[ edit ] explanation.

• Typical declaration of a move assignment operator
• Forcing a move assignment operator to be generated by the compiler
• Avoiding implicit move assignment

The move assignment operator is called whenever it is selected by overload resolution , e.g. when an object appears on the left side of an assignment expression, where the right-hand side is an rvalue of the same or implicitly convertible type.

Move assignment operators typically "steal" the resources held by the argument (e.g. pointers to dynamically-allocated objects, file descriptors, TCP sockets, I/O streams, running threads, etc), rather than make copies of them, and leave the argument in some valid but otherwise indeterminate state. For example, move-assigning from a std:: string or from a std:: vector leaves the right-hand side argument empty.

[ edit ] Implicitly-declared move assignment operator

If no user-defined move assignment operators are provided for a class type ( struct , class , or union ), and all of the following is true:

• there are no user-declared copy constructors
• there are no user-declared move constructors
• there are no user-declared copy assignment operators
• there are no user-declared destructors
• the implicitly-declared move assignment operator would not be defined as deleted

then the compiler will declare a move assignment operator as an inline public member of its class with the signature T& T::operator= T(T&&)

A class can have multiple move assignment operators, e.g. both T & T :: operator = ( const T && ) and T & T :: operator = ( T && ) . If some user-defined move assignment operators are present, the user may still force the generation of the implicitly declared move assignment operator with the keyword default .

Because some assignment operator (move or copy) 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 move assignment operator

The implicitly-declared or defaulted move assignment operator for class T is defined as deleted in any of the following is true:

• T has a non-static data member that is const
• T has a non-static data member of a reference type.
• T has a non-static data member that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
• T has direct or virtual base class that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
• T has a non-static data member or a direct or virtual base without a move assignment operator that is not trivially copyable.
• T has a direct or indirect virtual base class

[ edit ] Trivial move assignment operator

The implicitly-declared move assignment operator for class T is trivial if all of the following is true:

• T has no virtual member functions
• T has no virtual base classes
• The move assignment operator selected for every direct base of T is trivial
• The move assignment operator selected for every non-static class type (or array of class type) memeber of T is trivial

A trivial move assignment operator performs the same action as the trivial copy assignment operator, that is, makes a copy of the object representation as if by std:: memmove . All data types compatible with the C language (POD types) are trivially move-assignable.

[ edit ] Implicitly-defined move assignment operator

If the implicitly-declared move assignment operator is not deleted or trivial, it is defined (that is, a function body is generated and compiled) by the compiler. For union types, the implicitly-defined move assignment operator copies the object representation (as by std:: memmove ). For non-union class types ( class and struct ), the move assignment operator performs full member-wise move assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and move assignment operator for class types.

[ 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.

The copy-and-swap assignment operator

T & T :: operator = ( T arg ) {     swap ( arg ) ;     return * this ; }

performs an equivalent of move assignment for rvalue arguments at the cost of one additional call to the move constructor of T, which is often acceptable.

[ edit ] Example