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Move Constructors and Move Assignment Operators (C++)

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This topic describes how to write a move constructor and a move assignment operator for a C++ class. A move constructor enables the resources owned by an rvalue object to be moved into an lvalue without copying. For more information about move semantics, see Rvalue Reference Declarator: && .

This topic builds upon the following C++ class, MemoryBlock , which manages a memory buffer.

The following procedures describe how to write a move constructor and a move assignment operator for the example C++ class.

To create a move constructor for a C++ class

Define an empty constructor method that takes an rvalue reference to the class type as its parameter, as demonstrated in the following example:

In the move constructor, assign the class data members from the source object to the object that is being constructed:

Assign the data members of the source object to default values. This prevents the destructor from freeing resources (such as memory) multiple times:

To create a move assignment operator for a C++ class

Define an empty assignment operator that takes an rvalue reference to the class type as its parameter and returns a reference to the class type, as demonstrated in the following example:

In the move assignment operator, add a conditional statement that performs no operation if you try to assign the object to itself.

In the conditional statement, free any resources (such as memory) from the object that is being assigned to.

The following example frees the _data member from the object that is being assigned to:

Follow steps 2 and 3 in the first procedure to transfer the data members from the source object to the object that is being constructed:

Return a reference to the current object, as shown in the following example:

Example: Complete move constructor and assignment operator

The following example shows the complete move constructor and move assignment operator for the MemoryBlock class:

Example Use move semantics to improve performance

The following example shows how move semantics can improve the performance of your applications. The example adds two elements to a vector object and then inserts a new element between the two existing elements. The vector class uses move semantics to perform the insertion operation efficiently by moving the elements of the vector instead of copying them.

This example produces the following output:

Before Visual Studio 2010, this example produced the following output:

The version of this example that uses move semantics is more efficient than the version that does not use move semantics because it performs fewer copy, memory allocation, and memory deallocation operations.

Robust Programming

To prevent resource leaks, always free resources (such as memory, file handles, and sockets) in the move assignment operator.

To prevent the unrecoverable destruction of resources, properly handle self-assignment in the move assignment operator.

If you provide both a move constructor and a move assignment operator for your class, you can eliminate redundant code by writing the move constructor to call the move assignment operator. The following example shows a revised version of the move constructor that calls the move assignment operator:

The std::move function converts the lvalue other to an rvalue.

Rvalue Reference Declarator: && std::move

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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

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 && .

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-hand 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 may result in the argument being left empty. This is not, however, a guarantee. A move assignment is less, not more restrictively defined than ordinary assignment; where ordinary assignment must leave two copies of data at completion, move assignment is required to leave only one.

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;

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

The implicitly-declared (or defaulted on its first declaration) move assignment operator has an exception specification as described in dynamic exception specification (until C++17) exception specification (since C++17)

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.

Deleted implicitly-declared move assignment operator

The implicitly-declared or defaulted move assignment operator for class T is defined as deleted if 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);

Trivial move assignment operator

The move 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 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) member 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.

Implicitly-defined move assignment operator

If the implicitly-declared move 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 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 direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).

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 move assignment operator (same applies to copy assignment ).

See assignment operator overloading for additional detail on the expected behavior of a user-defined move-assignment operator.

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Move Semantics: The Basics

Move semantics was a new addition to C++ that arrived with C++ 11. Newer language standards have continued to expand and improve it. The feature is quite simple, but often misunderstood. I’m often reminded of that when I’m interviewing programmers.

Who will Find this Article Useful?

In this article I will cover all the basics of move semantics. This information will be useful for people that have never used move semantics but for people that are somehow familiar with the topic and use cases but don’t necessary use move semantics very often. The article will also be useful for people that want to dig deeper into the feature. And it will be useful for people that want to refresh their memory.

To benefit from this article to the fullest, you will need a good understanding of how copy constructors and assignment operators work, and you need to have a basic understanding of types and type casting.

1 – The case for Move Semantics

Before we look at the definition of move semantics, let us look at an example that will set up the premise of why move semantics exist.

1.1 – Setting the example

Let us assume that we are going to implement a struct that represents a memory block on the heap. The only thing this struct is responsible for is allocating the memory on construction and deallocating it on destruction. We can imagine that this struct is used by a complex memory allocation system. Let’s say that system is managing many instances of our struct, in other words, it is managing many memory buffers.

Here is the declaration of our struct:

We need to keep the example simple. And variables that could be needed for a robust solution, like used or free bytes and masks of all sorts, are out of the scope of this example.

1.2 – Using the Memory Buffer

Now let us also imagine that there is a free utility function that creates an array of buffers and returns the array. We can assume that this function is used by the complex memory allocation system in order to pre-allocate memory pools of different sizes.

Before C++ 11, no one would write this function in this way. Output parameters as pointer or reference will be used instead. This will make the function more complicated to implement, maintain and use. To explain why this was the case, let us have a look of how the “CreateMemoryPool” function is used. Let us assume that a bunch of pools of various sizes are created by the memory system in an “InitializePools” member function.

We will assume that we are not using any C++ 11 or newer additions to the language. We will also assume that for whatever reason the compiler will decide not to use Copy Elision to omit the copy of the return value of the “CreateMemoryPool(…)” function.

With all of these assumptions in play, calling the “CreateMemoryPool(…)” function will create a temporary return variable that will be used to assign to the member variables via the copy assignment operator of the std::vector. In other words, we are going to do a deep copy. This is very suboptimal way to write such code.

1.3 – The problem: deep copies can be very expensive

The problem with this scenario is that the local pool variable in the “CreateMemoryPool(…)” function, that stored the result of the function, will be copied into the m_PagePool and then destroyed. We are wasting a lot of cycles that way by executing extra instructions. We are also wasting cycles by allocating and deallocating memory (which is quite slow, generally speaking).

Before C++ 11, we would often reach for one of the following to minimize the allocations/deallocations caused by local and temporary variables:

• Pass a pointer or a reference to the m_PagePool as input to the “CreateMemoryPool(…)” and let the function directly push back into the m_PagePool. In this way “CreateMemoryPool(…)” can return void, and the function is going to work directly on the variable that will store the final result of the function. And in this way we avoid copies. The first drawback is the extra parameter that is passed to the function, adding complexity. For example, doing this creates ambiguity when it comes to responsibility for the input state of that vector. Is it the callers responsibility to ensure that the vector is empty when invoking the function or will the function itself clear the vector? The second is the fact that passing a non-constant pointer or reference of the internal variable m_PagePool makes our code less safe because anyone can write code that does something bad to m_PagePool and the caller of “CreateMemoryPool(…)” loses all guarantees.
• We can change the result of the function to return a vector of pointers to memory buffers, like so: vector<MemoryBuffer*>. This way we only copy the pointers when the “CreateMemoryPool(…)” function returns and there is no deep copying going on and no extra allocations/deallocations. The drawback to this is now the owner of the m_PagePool needs to worry about manually deallocating the pointers held by the vector, because the vector won’t do it in it own. Of course, we can use a vector of some of smart pointers to automate that part as well, but that also adds complexity.

1.4 – Move semantics solves the possible performance issue with deep copies

What we really want to do is keep the “CreateMemoryPool(…)” function implementation as it is, simple and safe, and move the result of the function directly into the m_PagePool variable without doing a deep copy, as if we are transferring water from one glass directly into another. Furthermore, we are doing even more copying by using the “push_back” function of the vector class and we want the push_back to also behave as if we are transferring water from one glass directly into another. C++ 11 and move semantics allows us to do exactly that. In the following sections, we are going to explore what move semantics are.

2 – Definition of Move Semantics

Move semantics are typically used to “steal” the resources held by another variable of the same type (e.g. pointers to dynamically-allocated objects, file descriptors, TCP sockets, I/O streams, running threads, etc.) and leave the source variable in some valid state, rather than making a deep copy of them.

The two most common use cases for move semantics are :

• Performance: converting an expensive deep copy operation into a cheap move, like moving the data from a source string into a target string without having to allocate/deallocate.
• Ownership semantics: Implementing user defined data types for which copying is not allowed but moving is, like the std::unique_ptr.

3 – Lvalues, Rvalues and Xvalues

To understand how move semantics work under the hood, we need to define a few other terms. Let’s start by looking at a pseudo-code that represents an assignment statement:

The left side of the statement is what historically is called Lvalue. And the right side of the statement is what historically is called Rvalue. Of course, because of the general nature of C++ things are often more complicated. Rvalues can appear on the left side, and Lvalues on the right side. Why that is the case is not relevant for this article.

3.1 – Easy way to differentiate between Lvalues and Rvalues

You might think that the fact that Lvalues and Rvalues can appear on either side is very confusing. Fortunately, there is a very simple rule we can use to remove all this confusion. A rule that will allow us to easily classify Lvalues and Rvalues in C++.

If a variable has an identifier that you have chosen yourself, then it is an Lvalue. And if a variable does not have a name that you have deliberately selected, then it is an Rvalue. To demonstrate this rule, let’s look at the following code snippets:

Here is another example:

3.2 – Double ampersands(&&) and the new operators

Prior to C++ 11, Rvalues and Lvalues were indistinguishable. The following example code would trigger the copy constructor and the copy assignment operator of the std::vector.

This is the Copy Constructor and the Copy Assignment Operator that would have been used in both cases:

To solve this problem, C++ 11 introduced Rvalue references, denoted by a double ampersands (&&). With this addition, we can now have two different copy constructors and assignment operators. One for Lvalues and one for Rvalues.

The following code snippet shows the declaration of the copy and the move constructors:

The following code snippet shows the declaration of the copy and the move assignment operators:

We can now revisit the example from above. And we can examine how in the cases where we have an Rvalue on the right hand side, the new move constructor and move assignment operators will be called instead of the copy constructor and copy assignment.

4 – Declaring Move Constructor and Move Assignment Operator

Now we know that C++ 11 introduced Rvalue references, denoted by a double ampersands (&&). And we know that we can declare a move constructor and a move assignment operator using the Rvalue type. Let’s declare them for our MemoryBuffer struct:

5 – Defining Move Constructor and Move Assignment Operator

We now know how to declare the move constructor and move assignment operator. It is now time to define them and actually implement the useful move semantics.

What we are doing in the move constructor and the move assignment is identical, except we also need to take care of self assignment in the move assignment operator. Let’s examine what we are doing in the move constructor:

• In the initializer list, we are copying the source m_Buffer pointer into our own internal m_Buffer pointer. This is a simple pointer copy and not a deep copy. At this point both the source and the internal m_Buffer pointers point to the same memory address.
• Then, also in the initializer list, we are copying the variable holding the size of the source buffer into the internal m_SizeInBytes variable.
• Then, in the move constructor body, we set the source buffer pointer to nullptr. Effectively leaving the source pointer in a valid state, but also at this point completing the process of stealing the resource. After this the internal m_Buffer pointer points to where the source m_Buffer pointer used to point.
• Finally, we reset the m_SizeInBytes of the source to 0, effectively leaving the source MemoryBuffer in a valid state.

For completeness, here is the the entire MemoryBuffer struct:

6 – Move Semantics in Action

With the new knowledge we acquired about Lvalues and Rvalues, as well as with the implementation of the move constructor and move assignment operator, we can now revisit the example we started with and see move semantics in action.

First, let us make a small change in the “CreateMemoryPool(…)” function.

To take advantage of the new move semantics and variadic templates, in C++ 11 a new version of the vector::push_back was added to the standard library. I recommend that you check the documentation, but in a nutshell, emplace_back will construct the new MemoryBuffer right in the memory location where it will be stored by the vector and in this way reducing the need to deep copy.

And finally, let us look at the “InitializePools” member function. Notice that the code is unchanged, but the compiler will call the move assignment operator and avoid deep copy and the extra allocations/deallocations, that are quite slow.

7 – Automatic vs. Manual Move

Looking at the “MemoryAllocationSystem::Initialize()” function, we can easily conclude that if move semantics were available to the compiler and if the types in question (std::vector and MemorBuffer in this case) supports move semantics, the compiler will detect the opportunity to move instead of copy. The reason for this is that the compiler will be able to see that the local variable created and returned by the “CreateMemoryPool(…)” is in fact an Rvalue.

The compiler is unable to detect every move opportunity. We can create a situation where we, as programmers, might have the intent to move the content of one array into another array (for whatever reasons) but the compiler will not be able to automatically detect our intent to move and will fallback to a deep copy. This of course happens when we have the intention to move an Lvalue into another Lvalue. Here is an example:

By default, the compiler will correctly assume that we ask for a deep copy. And if we really want to move, then we need to help the compiler a bit and turn the right hand Lvalues into Rvalue. To explicitly tell the compiler what our intent is, C++ 11 introduced a helper function: std::move . To utilize this new helper function, we can change our example to the following:

All std::move does is a cast. It casts the “ pool1 ” variable from an Lvalue reference (&) to an Rvalue reference (&&). Doing so tells the compiler which assignment operator it should call, because the variable type changes. And of course this is all based on standard function overloading rules. In its core, the assignment operator is a function, which is overloaded for two different input types (Lvalue reference and Rvalue reference).

8 – Xvalue

The following is a bit of an extra context and not absolutely necessary for this article, but I would rather mention it. The “ pool1 ” variable, in the example above, is technically an Xvalue after the std::move is executed. For all intent and purposes, you can treat it as an Lvalue, but the language implementers need a technical term for it. It means expiring value and it denotes an Lvalue object that can be reused after a move from it was executed. There are exactly three things that can be done to an Xvalue:

• Copy assign to the Xvalue, turning it into an Lvalue.
• Move assign to the Xvalue, turning it into an Lvalue.
• Call the destructor when the Xvalue goes out of scope.

9 – Special Operators’ Rules

We all know that sometimes the compiler will generate constructors and other special operators on its own. The good news is that there are rules for when the compiler will do that. And now with the new move constructor and move assignment operator, these rules have been updated. For completeness, it is worth mentioning the most important rules here.

Rule 1 : Default move operations are generated only if no copy operations or destructor is user defined. This is because C++ assumes that if you need a destructor or you manually implemented copy operations, or both, then you are dealing with some sort of resource that needs special treatment. And because it is a resource, you need to implement the move operations yourself because you, as the expert, know best how this resources should behave.

Rule 2 : Default copy operations are generated only if there is no user defined move operations. The reasons here are are the same as in Rule 1. Note that =default and =delete count as user defined.

Rule 3 : You don’t need to implement move semantics or even copy semantics manually if all of the internal member variables are of types that are movable or copyable. For example, if we were to change the “char* m_Buffer” in our MemoryBuffer class to “unique_ptr<char> m_Buffer”, then we do not need to implement the move semantics ourselves, because the unique_ptr<T> class already supports move semantics.

10 – Parameters Convention

Move semantics is an optimization that is only applicable to some use cases. In general prefer simple and conventional ways of passing information. The following table illustrates an accepted convention for passing parameters and what types you should use. When in doubt, always check it.

11 – Forwarding a Reference

For completeness, I need to cover one last topic. One more function was introduced together with std::move in C++ 11. And it was std::forward.

std::forward is pretty simple as it has only one use case. Its purpose is to preserve the value type, regardless if it is Lvalue or Rvalue, and pass it on. This is also called perfect forwarding. Typically a function accepting an Rvalue will attempt to move it and invalidate it, and when this is not the case, we can use std::forwad to pass on the Rvalue down the call-stack. The following example illustrates the use case.

For completeness, it is worth mentioning that In the code example above the input “T&& inArg” is called Forwarding reference . Forwarding references are a special kind of references that preserve the value category of a function argument. In this case calling “std::forward<T>(inArg)” will preserve the input argument and will forward it as Lvalue it if the input is an Lvalue, or it will forward it as Rvalue if the input is an Rvalue.

12 – Conclusion

In general, the concept of move semantics is quite simple and as programmers we should learn how and when to use it. The concept is built based on two fundamental C++ concepts: types and function/operator overloading. The following is a list of takeaway:

• Pre-C++ 11, value semantics sometimes lead to unnecessary and possibly expensive deep copy operations.
• C++ 11 introduces Rvalue references to distinguish from Lvalue references. And it also introduces std::move to cast from Lvalue to Rvalue.
• From C++ 11 on, temporary variables are treated as Rvalues.
• Moving POD or structs composed of PODs will not give you any benefits. To get benefits from move semantics you need some kind of resource (e.g. pointers to dynamically-allocated objects, file descriptors, TCP sockets, I/O streams, running threads, etc.).

13 – Source Code

The source code containing the MemoryBuffer example is here (under MoveSemantics ). You do not need a Github account to download it. When you open the repository, click on the Code button and choose Donwload ZIP .

14 – Credit

Special thanks to my colleagues for spending their personal time proofreading, challenge this article and helping me to make it better. It is a pleasure working with every one of you!

• Anton Andersson : Senior Gameplay Programmer at IO Interactive.
• Nils Iver Holtar : Lead AI Programmer at IO Interactive.

Please consider sharing this post if you find the information useful.

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21.12 — Overloading the assignment operator

The copy assignment operator (operator=) is used to copy values from one object to another already existing object .

Related content

As of C++11, C++ also supports “Move assignment”. We discuss move assignment in lesson 22.3 -- Move constructors and move assignment .

Copy assignment vs Copy constructor

The purpose of the copy constructor and the copy assignment operator are almost equivalent -- both copy one object to another. However, the copy constructor initializes new objects, whereas the assignment operator replaces the contents of existing objects.

The difference between the copy constructor and the copy assignment operator causes a lot of confusion for new programmers, but it’s really not all that difficult. Summarizing:

• If a new object has to be created before the copying can occur, the copy constructor is used (note: this includes passing or returning objects by value).
• If a new object does not have to be created before the copying can occur, the assignment operator is used.

Overloading the assignment operator

Overloading the copy assignment operator (operator=) is fairly straightforward, with one specific caveat that we’ll get to. The copy assignment operator must be overloaded as a member function.

This prints:

This should all be pretty straightforward by now. Our overloaded operator= returns *this, so that we can chain multiple assignments together:

Issues due to self-assignment

Here’s where things start to get a little more interesting. C++ allows self-assignment:

This will call f1.operator=(f1), and under the simplistic implementation above, all of the members will be assigned to themselves. In this particular example, the self-assignment causes each member to be assigned to itself, which has no overall impact, other than wasting time. In most cases, a self-assignment doesn’t need to do anything at all!

However, in cases where an assignment operator needs to dynamically assign memory, self-assignment can actually be dangerous:

First, run the program as it is. You’ll see that the program prints “Alex” as it should.

Now run the following program:

You’ll probably get garbage output. What happened?

Consider what happens in the overloaded operator= when the implicit object AND the passed in parameter (str) are both variable alex. In this case, m_data is the same as str.m_data. The first thing that happens is that the function checks to see if the implicit object already has a string. If so, it needs to delete it, so we don’t end up with a memory leak. In this case, m_data is allocated, so the function deletes m_data. But because str is the same as *this, the string that we wanted to copy has been deleted and m_data (and str.m_data) are dangling.

Later on, we allocate new memory to m_data (and str.m_data). So when we subsequently copy the data from str.m_data into m_data, we’re copying garbage, because str.m_data was never initialized.

Detecting and handling self-assignment

Fortunately, we can detect when self-assignment occurs. Here’s an updated implementation of our overloaded operator= for the MyString class:

By checking if the address of our implicit object is the same as the address of the object being passed in as a parameter, we can have our assignment operator just return immediately without doing any other work.

Because this is just a pointer comparison, it should be fast, and does not require operator== to be overloaded.

When not to handle self-assignment

Typically the self-assignment check is skipped for copy constructors. Because the object being copy constructed is newly created, the only case where the newly created object can be equal to the object being copied is when you try to initialize a newly defined object with itself:

In such cases, your compiler should warn you that c is an uninitialized variable.

Second, the self-assignment check may be omitted in classes that can naturally handle self-assignment. Consider this Fraction class assignment operator that has a self-assignment guard:

If the self-assignment guard did not exist, this function would still operate correctly during a self-assignment (because all of the operations done by the function can handle self-assignment properly).

Because self-assignment is a rare event, some prominent C++ gurus recommend omitting the self-assignment guard even in classes that would benefit from it. We do not recommend this, as we believe it’s a better practice to code defensively and then selectively optimize later.

The copy and swap idiom

A better way to handle self-assignment issues is via what’s called the copy and swap idiom. There’s a great writeup of how this idiom works on Stack Overflow .

The implicit copy assignment operator

Unlike other operators, the compiler will provide an implicit public copy assignment operator for your class if you do not provide a user-defined one. This assignment operator does memberwise assignment (which is essentially the same as the memberwise initialization that default copy constructors do).

Just like other constructors and operators, you can prevent assignments from being made by making your copy assignment operator private or using the delete keyword:

Note that if your class has const members, the compiler will instead define the implicit operator= as deleted. This is because const members can’t be assigned, so the compiler will assume your class should not be assignable.

If you want a class with const members to be assignable (for all members that aren’t const), you will need to explicitly overload operator= and manually assign each non-const member.

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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 && .

[ 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-hand 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 may result in the argument being left empty. This is not, however, a guarantee. A move assignment is less, not more restrictively defined than ordinary assignment; where ordinary assignment must leave two copies of data at completion, move assignment is required to leave only one.

[ 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;

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

The implicitly-declared (or defaulted on its first declaration) move assignment operator has an exception specification as described in dynamic exception specification (until C++17) exception specification (since C++17)

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 if 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);

[ edit ] Trivial move assignment operator

The move 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 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) member 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 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 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 direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).

[ 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 move assignment operator (same applies to copy assignment ).

See assignment operator overloading for additional detail on the expected behavior of a user-defined move-assignment operator.

[ edit ] Example

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Move Constructors in C++ with Examples

• Dynamic Constructor in C++ with Examples
• How to Define the Default Constructor in C++?
• Constructor Delegation in C++
• Constructor Overloading in C++
• Constructors in C++
• Copy Constructor in C++
• std::is_move_constructible in C++ with Examples
• std::is_copy_constructible in C++ with Examples
• std::is_nothrow_move_constructible in C++ with Example
• std::is_constructible template in C++ with Examples
• std::is_trivially_move_constructible in C++ with Examples
• std::is_trivially_constructible in C++ with Examples
• std::is_nothrow_copy_constructible in C++ with Examples
• std::is_default_constructible in C++ with Examples
• Default Methods in C++ with Examples
• std::allocator() in C++ with Examples
• Constructors in Objective-C
• How to create a List with Constructor in C++ STL
• What is conversion constructor in C++?

Prerequisites: l-value and r-value references in C++ , Copy Constructor in C++ .

What is a Move Constructor?  

The copy constructors in C++ work with the l-value references and copy semantics(copy semantics means copying the actual data of the object to another object rather than making another object to point the already existing object in the heap). While move constructors work on the r-value references and move semantics(move semantics involves pointing to the already existing object in the memory).

On declaring the new object and assigning it with the r-value, firstly a temporary object is created, and then that temporary object is used to assign the values to the object. Due to this the copy constructor is called several times and increases the overhead and decreases the computational power of the code. To avoid this overhead and make the code more efficient we use move constructors.

Why Move Constructors are used?

Move constructor moves the resources in the heap, i.e., unlike copy constructors which copy the data of the existing object and assigning it to the new object move constructor just makes the pointer of the declared object to point to the data of temporary object and nulls out the pointer of the temporary objects. Thus, move constructor prevents unnecessarily copying data in the memory.  

Work of move constructor looks a bit like default member-wise copy constructor but in this case, it nulls out the pointer of the temporary object preventing more than one object to point to same memory location.

Below is the program without declaring the move constructor:

Explanation:   The above program shows the unnecessarily calling copy constructor and inefficiently using the memory by copying the same data several times as it new object upon each call to copy constructor.

Syntax of the Move Constructor:

This unnecessary use of the memory can be avoided by using move constructor. Below is the program declaring the move constructor:

Explanation:   The unnecessary call to the copy constructor is avoided by making the call to the move constructor. Thus making the code more memory efficient and decreasing the overhead of calling the move constructor.

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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
• (until C++14) 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&&) .

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)
• (until C++14) T has a non-static data member or a direct or virtual base without a move assignment operator that is not trivially copyable.
• (until C++14) T has a direct or indirect virtual base class

(since C++14) A deleted implicitly-declared move assignment operator is ignored by overload resolution

[ edit ] Trivial move assignment operator

The move 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
• 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) member 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 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 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 direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).

[ 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

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 && .

[ 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-hand 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 may result in the argument being left empty. This is not, however, a guarantee. A move assignment is less, not more restrictively defined than ordinary assignment; where ordinary assignment must leave two copies of data at completion, move assignment is required to leave only one.

[ 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;

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

[ edit ] Trivial move assignment operator

The move 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 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) member 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 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 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 direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).

[ 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 move assignment operator (same applies to copy assignment ).

[ edit ] Copy-and-swap idiom

Move assignment operator can be expressed in terms of move constructor, destructor, and the swap() member function, if one is provided, at the cost of one additional call to the move constructor of T , which is often acceptable:

For non-throwing swap(), this form provides strong exception guarantee . For lvalue arguments, this form automatically invokes the copy constructor. It is sometimes referred to as "unified assignment operator" (as in, both copy and move). See assignment operator overloading for further detail on the use and trade-offs.

[ edit ] Example

[ edit ] defect reports.

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.