Move assignment operator

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[edit]

一招赋值操作符类Toperator=，T&&，const T&&，或volatile T&&类型的一个参数的名称const volatile T&&是一个非模板的非静态成员函数。与公众移动赋值运算符是A型MoveAssignable.

移动赋值操作符典型的“窃取”的参数（例如动态分配的对象的指针，文件描述符，TCP套接字I / O流，正在运行的线程等）的资源，而不是让它们的副本，以及离开的说法在一些有效的，但其他不确定的状态。例如，将分配从std::string或从std::vector叶的右手边参数为空.

[编辑] 隐式声明的移动赋值运算符

如果没有提供用户定义的移动赋值运算符的类类型（struct，class，或union），以下是正确的：

有没有用户声明的拷贝构造函数
Original:
there are no user-declared copy constructors
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有没有用户声明的move构造函数
Original:
there are no user-declared move constructors
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有没有用户声明的拷贝赋值运算符
Original:
there are no user-declared copy assignment operators
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有没有用户声明析构函数
Original:
there are no user-declared destructors
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隐式声明的移动赋值操作符就不会被定义为删除
Original:
the implicitly-declared move assignment operator would not be defined as deleted
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那么编译器会声明的举动分配，运营商inline public类成员的签名

一个类可以有多个移动赋值运算符，如： T& T::operator=(const T&&)和T& T::operator=(T&&)。如果一些用户自定义的移动赋值操作符都存在，用户仍然可以强制生成的隐式声明的移动赋值操作符与关键字default.

由于一些赋值运算符（移动或复制）的任何类，基类的赋值运算符总是宣称永远是隐藏的。如果一个声明是用来把从基类中的赋值运算符，它的参数类型可以是相同的参数类型的隐式派生类的赋值运算符，声明还隐藏着由隐声明.

[编辑] 删除隐式声明的移动赋值运算符

隐式声明的或默认的移动赋值操作符类T被定义为“删除”以下是正确的：

T有一个非静态数据成员是const
Original:
T has a non-static data member that is const
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T有一个非静态数据成员的引用类型.
Original:
T has a non-static data member of a reference type.
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T有一个非静态数据成员不能移动分配（已删除，无法访问，或暧昧的举动赋值运算符）
Original:
T has a non-static data member that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
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T有直接或虚基类，不能移动分配（已删除，无法访问，或暧昧的举动赋值运算符）
Original:
T has direct or virtual base class that cannot be move-assigned (has deleted, inaccessible, or ambiguous move assignment operator)
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T有一个非静态数据成员或直接或虚基类，没有一招赋值运算符，是不平凡复制的.
Original:
T has a non-static data member or a direct or virtual base without a move assignment operator that is not trivially copyable.
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T有直接或间接的虚基类
Original:
T has a direct or indirect virtual base class
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[编辑] 琐碎的举动赋值运算符

隐式声明的移动赋值操作符类T以下是微不足道的，如果是真实的

T没有虚成员函数
Original:
T has no virtual member functions
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T没有虚基类
Original:
T has no virtual base classes
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选择每一个直接基础T的移动赋值运算符是微不足道的
Original:
The move assignment operator selected for every direct base of T is trivial
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选择每一个非静态类类型（或类类型的数组）星期四会员T的移动赋值运算符是微不足道的
Original:
The move assignment operator selected for every non-static class type (or array of class type) memeber of T is trivial
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一个平凡的举动赋值运算符执行相同的简单复制assignmentoperator的行动，就是使对象表示，如果由std::memmove的副本。所有兼容的数据类型与C语言（POD类型）是平凡的移动可分配的

[编辑] 隐式定义的移动赋值运算符

如果隐式声明的移动赋值运算符是不会被删除或微不足道的，它的定义，函数体生成和编译的编译器。 union类型，隐式定义的移动赋值操作符复制对象表示（std::memmove）的。对于非工会类类型（class和struct），在移动分配运营商进行完整的成员，明智的举动分配的对象的基础和非静态成员，在他们的初始化顺序，使用内置的在分配的标量和移动赋值操作符类的类型.

[编辑] 注释

如果复制和移动赋值运算符重载解析选择移动分配的说法是“右值”（或者“prvalue”如一个无名的临时或“xvalue”的结果std::move ），并选择拷贝赋值的说法是“左值”（命名对象或函数/操作符返回左值参考）。如果仅仅是拷贝赋值，所有的参数类别选择（只要它采取它的参数的值或为const，因为右值绑定到const引用），这使得拷贝赋值移动分配的回退，当移动是不可用

复制和交换赋值运算符

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

执行分配一个额外通话费用的右值参数相当于移动的移动构造函数T，这往往是可以接受的.

[编辑] 为例

#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 E\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 E
move assigned

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class_name `&` class_name `:: operator=` ( class_name `&&` )	(1)	（C++11 起）

class_name `&` class_name `:: operator=` ( class_name `&&` ) = default;	(2)	（C++11 起）

class_name `&` class_name `:: operator=` ( class_name `&&` ) = delete;	(3)	（C++11 起）

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