什么& =在C& operator =（const C&）& =默认;做？ [英] What does & = in C& operator=(const C&) & = default; do?

问题描述

SO的几个问题使用特定语法来声明默认赋值运算符。

规则三使用C ++ 11成为五规则？

  class C {
 C（const C&）= default; 
 C（C&&）= default; 
 C& operator =（const C&）& =默认; 
 C& operator =（C&& amp;）& =默认; 
 virtual〜C（）{} 
}; 
  

=用于赋值运算符。经过快速测试，默认赋值运算符声明似乎编译并给出了带或不带附加&符号的预期行为。

我看不到& c> c>

code < a href =http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2009/n2819.html =nofollow> ref限定词。

在这种特殊情况下，它使得你想分配给C的实例必须是一个非常量的值。

A few questions on SO use a particular syntax for declaring default assignment operators.

Rule-of-Three becomes Rule-of-Five with C++11?

class C {
  C(const C&) = default;
  C(C&&) = default;
  C& operator=(const C&) & = default;
  C& operator=(C&&) & = default;
  virtual ~C() { }
};

I'm confused by the & = used for the assignment operators. After a quick test, default assignment operator declarations seem to compile and give the expected behavior with or without the additional ampersand.

I don't see the & = syntax on cppreference.

解决方案

The & there is a ref qualifier.

In that particular case it makes it so that the instance of C you want to assign to must be a non-const lvalue.

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