assignment operator move

22.3 — Move constructors and move assignment

In lesson 22.1 -- Introduction to smart pointers and move semantics , we took a look at std::auto_ptr, discussed the desire for move semantics, and took a look at some of the downsides that occur when functions designed for copy semantics (copy constructors and copy assignment operators) are redefined to implement move semantics.

In this lesson, we’ll take a deeper look at how C++11 resolves these problems via move constructors and move assignment.

Recapping copy constructors and copy assignment

First, let’s take a moment to recap copy semantics.

Copy constructors are used to initialize a class by making a copy of an object of the same class. Copy assignment is used to copy one class object to another existing class object. By default, C++ will provide a copy constructor and copy assignment operator if one is not explicitly provided. These compiler-provided functions do shallow copies, which may cause problems for classes that allocate dynamic memory. So classes that deal with dynamic memory should override these functions to do deep copies.

Returning back to our Auto_ptr smart pointer class example from the first lesson in this chapter, let’s look at a version that implements a copy constructor and copy assignment operator that do deep copies, and a sample program that exercises them:

In this program, we’re using a function named generateResource() to create a smart pointer encapsulated resource, which is then passed back to function main(). Function main() then assigns that to an existing Auto_ptr3 object.

When this program is run, it prints:

(Note: You may only get 4 outputs if your compiler elides the return value from function generateResource())

That’s a lot of resource creation and destruction going on for such a simple program! What’s going on here?

Let’s take a closer look. There are 6 key steps that happen in this program (one for each printed message):

• Inside generateResource(), local variable res is created and initialized with a dynamically allocated Resource, which causes the first “Resource acquired”.
• Res is returned back to main() by value. We return by value here because res is a local variable -- it can’t be returned by address or reference because res will be destroyed when generateResource() ends. So res is copy constructed into a temporary object. Since our copy constructor does a deep copy, a new Resource is allocated here, which causes the second “Resource acquired”.
• Res goes out of scope, destroying the originally created Resource, which causes the first “Resource destroyed”.
• The temporary object is assigned to mainres by copy assignment. Since our copy assignment also does a deep copy, a new Resource is allocated, causing yet another “Resource acquired”.
• The assignment expression ends, and the temporary object goes out of expression scope and is destroyed, causing a “Resource destroyed”.
• At the end of main(), mainres goes out of scope, and our final “Resource destroyed” is displayed.

So, in short, because we call the copy constructor once to copy construct res to a temporary, and copy assignment once to copy the temporary into mainres, we end up allocating and destroying 3 separate objects in total.

Inefficient, but at least it doesn’t crash!

However, with move semantics, we can do better.

Move constructors and move assignment

C++11 defines two new functions in service of move semantics: a move constructor, and a move assignment operator. Whereas the goal of the copy constructor and copy assignment is to make a copy of one object to another, the goal of the move constructor and move assignment is to move ownership of the resources from one object to another (which is typically much less expensive than making a copy).

Defining a move constructor and move assignment work analogously to their copy counterparts. However, whereas the copy flavors of these functions take a const l-value reference parameter (which will bind to just about anything), the move flavors of these functions use non-const rvalue reference parameters (which only bind to rvalues).

Here’s the same Auto_ptr3 class as above, with a move constructor and move assignment operator added. We’ve left in the deep-copying copy constructor and copy assignment operator for comparison purposes.

The move constructor and move assignment operator are simple. Instead of deep copying the source object (a) into the implicit object, we simply move (steal) the source object’s resources. This involves shallow copying the source pointer into the implicit object, then setting the source pointer to null.

When run, this program prints:

That’s much better!

The flow of the program is exactly the same as before. However, instead of calling the copy constructor and copy assignment operators, this program calls the move constructor and move assignment operators. Looking a little more deeply:

• Res is returned back to main() by value. Res is move constructed into a temporary object, transferring the dynamically created object stored in res to the temporary object. We’ll talk about why this happens below.
• Res goes out of scope. Because res no longer manages a pointer (it was moved to the temporary), nothing interesting happens here.
• The temporary object is move assigned to mainres. This transfers the dynamically created object stored in the temporary to mainres.
• The assignment expression ends, and the temporary object goes out of expression scope and is destroyed. However, because the temporary no longer manages a pointer (it was moved to mainres), nothing interesting happens here either.

So instead of copying our Resource twice (once for the copy constructor and once for the copy assignment), we transfer it twice. This is more efficient, as Resource is only constructed and destroyed once instead of three times.

Related content

Move constructors and move assignment should be marked as noexcept . This tells the compiler that these functions will not throw exceptions.

We introduce noexcept in lesson 27.9 -- Exception specifications and noexcept and discuss why move constructors and move assignment are marked as noexcept in lesson 27.10 -- std::move_if_noexcept .

When are the move constructor and move assignment called?

The move constructor and move assignment are called when those functions have been defined, and the argument for construction or assignment is an rvalue. Most typically, this rvalue will be a literal or temporary value.

The copy constructor and copy assignment are used otherwise (when the argument is an lvalue, or when the argument is an rvalue and the move constructor or move assignment functions aren’t defined).

Implicit move constructor and move assignment operator

The compiler will create an implicit move constructor and move assignment operator if all of the following are true:

• There are no user-declared copy constructors or copy assignment operators.
• There are no user-declared move constructors or move assignment operators.
• There is no user-declared destructor.

The implicit move constructor and implicit move assignment operator both do a memberwise move. That is, each member of the moved-from object is moved to the moved-to object.

The key insight behind move semantics

You now have enough context to understand the key insight behind move semantics.

If we construct an object or do an assignment where the argument is an l-value, the only thing we can reasonably do is copy the l-value. We can’t assume it’s safe to alter the l-value, because it may be used again later in the program. If we have an expression “a = b” (where b is an lvalue), we wouldn’t reasonably expect b to be changed in any way.

However, if we construct an object or do an assignment where the argument is an r-value, then we know that r-value is just a temporary object of some kind. Instead of copying it (which can be expensive), we can simply transfer its resources (which is cheap) to the object we’re constructing or assigning. This is safe to do because the temporary will be destroyed at the end of the expression anyway, so we know it will never be used again!

C++11, through r-value references, gives us the ability to provide different behaviors when the argument is an r-value vs an l-value, enabling us to make smarter and more efficient decisions about how our objects should behave.

Key insight

Move semantics is an optimization opportunity.

Move functions should always leave both objects in a valid state

In the above examples, both the move constructor and move assignment functions set a.m_ptr to nullptr. This may seem extraneous -- after all, if a is a temporary r-value, why bother doing “cleanup” if parameter a is going to be destroyed anyway?

The answer is simple: When a goes out of scope, the destructor for a will be called, and a.m_ptr will be deleted. If at that point, a.m_ptr is still pointing to the same object as m_ptr , then m_ptr will be left as a dangling pointer. When the object containing m_ptr eventually gets used (or destroyed), we’ll get undefined behavior.

When implementing move semantics, it is important to ensure the moved-from object is left in a valid state, so that it will destruct properly (without creating undefined behavior).

Automatic l-values returned by value may be moved instead of copied

In the generateResource() function of the Auto_ptr4 example above, when variable res is returned by value, it is moved instead of copied, even though res is an l-value. The C++ specification has a special rule that says automatic objects returned from a function by value can be moved even if they are l-values. This makes sense, since res was going to be destroyed at the end of the function anyway! We might as well steal its resources instead of making an expensive and unnecessary copy.

Although the compiler can move l-value return values, in some cases it may be able to do even better by simply eliding the copy altogether (which avoids the need to make a copy or do a move at all). In such a case, neither the copy constructor nor move constructor would be called.

Disabling copying

In the Auto_ptr4 class above, we left in the copy constructor and assignment operator for comparison purposes. But in move-enabled classes, it is sometimes desirable to delete the copy constructor and copy assignment functions to ensure copies aren’t made. In the case of our Auto_ptr class, we don’t want to copy our templated object T -- both because it’s expensive, and whatever class T is may not even support copying!

Here’s a version of Auto_ptr that supports move semantics but not copy semantics:

If you were to try to pass an Auto_ptr5 l-value to a function by value, the compiler would complain that the copy constructor required to initialize the function parameter has been deleted. This is good, because we should probably be passing Auto_ptr5 by const l-value reference anyway!

Auto_ptr5 is (finally) a good smart pointer class. And, in fact the standard library contains a class very much like this one (that you should use instead), named std::unique_ptr. We’ll talk more about std::unique_ptr later in this chapter.

Another example

Let’s take a look at another class that uses dynamic memory: a simple dynamic templated array. This class contains a deep-copying copy constructor and copy assignment operator.

Now let’s use this class in a program. To show you how this class performs when we allocate a million integers on the heap, we’re going to leverage the Timer class we developed in lesson 18.4 -- Timing your code . We’ll use the Timer class to time how fast our code runs, and show you the performance difference between copying and moving.

On one of the author’s machines, in release mode, this program executed in 0.00825559 seconds.

Now let’s run the same program again, replacing the copy constructor and copy assignment with a move constructor and move assignment.

On the same machine, this program executed in 0.0056 seconds.

Comparing the runtime of the two programs, (0.00825559 - 0.0056) / 0.00825559 * 100 = 32.1% faster!

Do not implement move semantics using std::swap

Since the goal of move semantics is to move a resource from a source object to a destination object, you might think about implementing the move constructor and move assignment operator using std::swap() . However, this is a bad idea, as std::swap() calls both the move constructor and move assignment on move-capable objects, which would result in an infinite recursion. You can see this happen in the following example:

This prints:

And so on… until the stack overflows.

You can implement the move constructor and move assignment using your own swap function, as long as your swap member function does not call the move constructor or move assignment. Here’s an example of how that can be done:

This works as expected, and prints:

Deleting the move constructor and move assignment

You can delete the move constructor and move assignment using the = delete syntax in the exact same way you can delete the copy constructor and copy assignment.

If you delete the copy constructor, the compiler will not generate an implicit move constructor (making your objects neither copyable nor movable). Therefore, when deleting the copy constructor, it is useful to be explicit about what behavior you want from your move constructors. Either explicitly delete them (making it clear this is the desired behavior), or default them (making the class move-only).

The rule of five says that if the copy constructor, copy assignment, move constructor, move assignment, or destructor are defined or deleted, then each of those functions should be defined or deleted.

While deleting only the move constructor and move assignment may seem like a good idea if you want a copyable but not movable object, this has the unfortunate consequence of making the class not returnable by value in cases where mandatory copy elision does not apply. This happens because a deleted move constructor is still declared, and thus is eligible for overload resolution. And return by value will favor a deleted move constructor over a non-deleted copy constructor. This is illustrated by the following program:

cppreference.com

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 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

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How to Implement Move Assignment Operator in C++?

In C+, the move assignment operator is used to transfer the ownership of the resources from one object to another when the object is assigned some rvalue references or using std::move. It allows efficient memory manipulation by avoiding unnecessary copies of the objects. In this article, we will learn how to implement a move assignment operator in C++.

Implement Move Assignment Operator in C++

The move assignment operator also overloads the assignment operator in its implementation like the copy assignment operator. The difference is in the type of argument. The move assignment operator takes the rvalue reference as its argument. To define the move assignment operator use the below syntax:

Syntax to Define Move Assignment Operator

Here, c lassName is the name of our class and && indicates that other is an rvalue reference that allows the function to bind to temporary objects.

C++ Program to Implement Move Assignment Operator

The below example demonstrates the implementation of the move assignment operator.

Time Complexity: O(1) Space Complexity: O(1)

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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 && .

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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Assignment operators

• 8 contributors

expression assignment-operator expression

assignment-operator : one of   =   *=   /=   %=   +=   -=   <<=   >>=   &=   ^=   |=

Assignment operators store a value in the object specified by the left operand. There are two kinds of assignment operations:

simple assignment , in which the value of the second operand is stored in the object specified by the first operand.

compound assignment , in which an arithmetic, shift, or bitwise operation is performed before storing the result.

All assignment operators in the following table except the = operator are compound assignment operators.

Assignment operators table

Operator keywords.

Three of the compound assignment operators have keyword equivalents. They are:

C++ specifies these operator keywords as alternative spellings for the compound assignment operators. In C, the alternative spellings are provided as macros in the <iso646.h> header. In C++, the alternative spellings are keywords; use of <iso646.h> or the C++ equivalent <ciso646> is deprecated. In Microsoft C++, the /permissive- or /Za compiler option is required to enable the alternative spelling.

Simple assignment

The simple assignment operator ( = ) causes the value of the second operand to be stored in the object specified by the first operand. If both objects are of arithmetic types, the right operand is converted to the type of the left, before storing the value.

Objects of const and volatile types can be assigned to l-values of types that are only volatile , or that aren't const or volatile .

Assignment to objects of class type ( struct , union , and class types) is performed by a function named operator= . The default behavior of this operator function is to perform a member-wise copy assignment of the object's non-static data members and direct base classes; however, this behavior can be modified using overloaded operators. For more information, see Operator overloading . Class types can also have copy assignment and move assignment operators. For more information, see Copy constructors and copy assignment operators and Move constructors and move assignment operators .

An object of any unambiguously derived class from a given base class can be assigned to an object of the base class. The reverse isn't true because there's an implicit conversion from derived class to base class, but not from base class to derived class. For example:

Assignments to reference types behave as if the assignment were being made to the object to which the reference points.

For class-type objects, assignment is different from initialization. To illustrate how different assignment and initialization can be, consider the code

The preceding code shows an initializer; it calls the constructor for UserType2 that takes an argument of type UserType1 . Given the code

the assignment statement

can have one of the following effects:

Call the function operator= for UserType2 , provided operator= is provided with a UserType1 argument.

Call the explicit conversion function UserType1::operator UserType2 , if such a function exists.

Call a constructor UserType2::UserType2 , provided such a constructor exists, that takes a UserType1 argument and copies the result.

Compound assignment

The compound assignment operators are shown in the Assignment operators table . These operators have the form e1 op = e2 , where e1 is a non- const modifiable l-value and e2 is:

an arithmetic type

a pointer, if op is + or -

a type for which there exists a matching operator *op*= overload for the type of e1

The built-in e1 op = e2 form behaves as e1 = e1 op e2 , but e1 is evaluated only once.

Compound assignment to an enumerated type generates an error message. If the left operand is of a pointer type, the right operand must be of a pointer type, or it must be a constant expression that evaluates to 0. When the left operand is of an integral type, the right operand must not be of a pointer type.

Result of built-in assignment operators

The built-in assignment operators return the value of the object specified by the left operand after the assignment (and the arithmetic/logical operation in the case of compound assignment operators). The resultant type is the type of the left operand. The result of an assignment expression is always an l-value. These operators have right-to-left associativity. The left operand must be a modifiable l-value.

In ANSI C, the result of an assignment expression isn't an l-value. That means the legal C++ expression (a += b) += c isn't allowed in C.

Expressions with binary operators C++ built-in operators, precedence, and associativity C assignment operators

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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

Assignment operators

Assignment operators modify the value of the object.

Explanation

copy assignment operator replaces the contents of the object a with a copy of the contents of b ( b is not modified). For class types, this is a special member function, described in copy assignment operator .

move assignment operator replaces the contents of the object a with the contents of b , avoiding copying if possible ( b may be modified). For class types, this is a special member function, described in move assignment operator . (since C++11)

For non-class types, copy and move assignment are indistinguishable and are referred to as direct assignment .

compound assignment operators replace the contents of the object a with the result of a binary operation between the previous value of a and the value of b .

Builtin direct assignment

The direct assignment expressions have the form

For the built-in operator, lhs may have any non-const scalar type and rhs must be implicitly convertible to the type of lhs .

The direct assignment operator expects a modifiable lvalue as its left operand and an rvalue expression or a braced-init-list (since C++11) as its right operand, and returns an lvalue identifying the left operand after modification.

For non-class types, the right operand is first implicitly converted to the cv-unqualified type of the left operand, and then its value is copied into the object identified by left operand.

When the left operand has reference type, the assignment operator modifies the referred-to object.

If the left and the right operands identify overlapping objects, the behavior is undefined (unless the overlap is exact and the type is the same)

In overload resolution against user-defined operators , for every type T , the following function signatures participate in overload resolution:

For every enumeration or pointer to member type T , optionally volatile-qualified, the following function signature participates in overload resolution:

For every pair A1 and A2, where A1 is an arithmetic type (optionally volatile-qualified) and A2 is a promoted arithmetic type, the following function signature participates in overload resolution:

Builtin compound assignment

The compound assignment expressions have the form

The behavior of every builtin compound-assignment expression E1 op = E2 (where E1 is a modifiable lvalue expression and E2 is an rvalue expression or a braced-init-list (since C++11) ) is exactly the same as the behavior of the expression E1 = E1 op E2 , except that the expression E1 is evaluated only once and that it behaves as a single operation with respect to indeterminately-sequenced function calls (e.g. in f ( a + = b, g ( ) ) , the += is either not started at all or is completed as seen from inside g ( ) ).

In overload resolution against user-defined operators , for every pair A1 and A2, where A1 is an arithmetic type (optionally volatile-qualified) and A2 is a promoted arithmetic type, the following function signatures participate in overload resolution:

For every pair I1 and I2, where I1 is an integral type (optionally volatile-qualified) and I2 is a promoted integral type, the following function signatures participate in overload resolution:

For every optionally cv-qualified object type T , the following function signatures participate in overload resolution:

Operator precedence

Operator overloading