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# Move assignment operator

From cppreference.com
< cpp‎ | language

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

class_name `&` class_name `:: operator=` ( class_name `&&` ) | (1) | (since C++11)
class_name `&` class_name `:: operator=` ( class_name `&&` ) = default; | (2) | (since C++11)
class_name `&` class_name `:: operator=` ( class_name `&&` ) = delete; | (3) | (since C++11)

### [edit] Explanation

1. Typical declaration of a move assignment operator
2. Forcing a move assignment operator to be generated by the compiler
3. 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) 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 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

Run this code

    #include <string>
    #include <iostream>
    #include <utility>

    struct A {
        std::string s;
        A() : s("test") {}
        A(const A& o) : s(o.s) { std::cout << "move failed!\n";}
        A(A&& o) : s(std::move(o.s)) {}
        A& operator=(const A&) { std::cout << "copy assigned\n"; return *this; }
        A& operator=(A&& other) {
             s = std::move(other.s);
             std::cout << "move assigned\n";
             return *this;
        }
    };

    A f(A a) { return a; }

    struct B : A {
         std::string s2;
         int n;
         // implicit move assignment operator B& B::operator=(B&&)
         // calls A's move assignment operator
         // calls s2's move assignment operator
         // and makes a bitwise copy of n
    };

    struct C : B {
        ~C() {}; // destructor prevents implicit move assignment
    };

    struct D : B {
        D() {}
        ~D() {}; // destructor would prevent implicit move assignment
        D& operator=(D&&) = default; // force a move assignment anyway
    };

    int main()
    {
        A a1, a2;
        std::cout << "Trying to move-assign A from rvalue temporary\n";
        a1 = f(A()); // move-assignment from rvalue temporary
        std::cout << "Trying to move-assign A from xvalue\n";
        a2 = std::move(a1); // move-assignment from xvalue

        std::cout << "Trying to move-assign B\n";
        B b1, b2;
        std::cout << "Before move, b1.s = \"" << b1.s << "\"\n";
        b2 = std::move(b1); // calls implicit move assignment
        std::cout << "After move, b1.s = \"" << b1.s << "\"\n";

        std::cout << "Trying to move-assign C\n";
        C c1, c2;
        c2 = std::move(c1); // calls the copy assignment operator

        std::cout << "Trying to move-assign D\n";
        D d1, d2;
        d2 = std::move(d1);
    }

Output:

    Trying to move-assign A from rvalue temporary
    move assigned
    Trying to move-assign A from xvalue
    move assigned
    Trying to move-assign B
    Before move, b1.s = "test"
    move assigned
    After move, b1.s = ""
    Trying to move-assign C
    copy assigned
    Trying to move-assign D
    move assigned

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