移动分配和引用成员 [英] Move-assignment and reference member

问题描述

具有引用成员变量的类的复制分配是一个no-no，因为您不能重新分配引用。但是移动任务呢？我尝试简单地 move 它，但是，当然，当我只是想移动引用本身时，破坏源对象：

  class C 
 {
 public：
 C（X& x）：x_（x）{} 
 C（C& ;& other）：x_（std :: move（other.x_））{} 
 C& operator =（C&& other）
 {
 x_ = std :: move（other.x_）; 
} 
 private：
 X& X_; 
}; 
 
 X y; 
 C c1（y）; 
 
 X z; 
 C c2（z）; 
 
 c2 = c1; //破坏y以及z 
  

我应该只是不执行移动任务和坚持移动只有建设？这使得交换（C& amp;）很难实现。

解决方案

p>（作为OP建议的评论的回答）

一般来说，如果想用C ++中的引用做一些不平凡的东西， reference_wrapper< T> ，这本质上是 T& 的花式价值语义替代，完成它 - 它已经提供（重）分配和其他操作。我相信这将使移动构造函数和赋值接近平凡，如果不是微不足道（注意琐碎如 is_trivially _ * semantics）。

参考包装器作为TR1的一部分添加到C ++ 03中， >

文档： http://en.cppreference.com/ w / cpp / utility / functional / reference_wrapper

Copy-assignment for a class with a reference member variable is a no-no because you can't reassign the reference. But what about move-assignment? I tried simply moveing it but, of course, that destroys the source object when I just want to move the reference itself:

class C
{
public:
    C(X& x) : x_(x) {}
    C(C&& other) : x_(std::move(other.x_)) {}
    C& operator=(C&& other)
    {
        x_ = std::move(other.x_);
    }
private:
    X& x_;
};

X y;
C c1(y);

X z;
C c2(z);

c2 = c1; // destroys y as well as z

Should I just not be implementing move-assignment and sticking with move-construction only? That makes swap(C&, C&) hard to implement.

解决方案

(Posted as an answer from comment as suggested by the OP)

In general if one wants to do non-trivial stuff with references in C++, one would be using reference_wrapper<T>, which is essentially a fancy value-semantic stand-in for a T&, then one would be done with it - it already provides (re)assignment and other operations. I'm sure that would make move constructor and assignment near-trivial, if not trivial (note not trivial as in per the is_trivially_* semantics).

"Reference wrapper" is added to C++03 as part of TR1, and is part of C++11.

Documentation: http://en.cppreference.com/w/cpp/utility/functional/reference_wrapper

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