我可以同时为类编写复制和移动赋值运算符吗? [英] Can I write both copy and move assignment operators for a class?

问题描述

这些是我的原型，

MyClass& operator=(MyClass rhs);   // copy assignment
MyClass& operator=(MyClass &&rhs); // move assignment

但是当我打电话

MyClass a, b;
a = std::move(b);

，有一个错误.

556 IntelliSense: more than one operator "=" matches these operands:
        function "MyClass::operator=(MyClass rhs)"
        function "MyClass::operator=(MyClass &&rhs)"
        operand types are: MyClass = MyClass    

编译器返回:

Error   56  error C2593: 'operator =' is ambiguous  

推荐答案

重载解析度不明确，因为当您传递右值时，MyClass和MyClass &&都可以由它直接初始化.

Overload resolution is ambiguous because when you pass an rvalue, both MyClass and MyClass && can be directly initialised by it.

如果要提供复制和移动分配的其他实现，则常规方法是通过const引用获取复制分配运算符的参数:

If you want to provide a different implementation of copy and move assignment, the customary way is to take the copy assignment operator's parameter by const reference:

MyClass& operator=(const MyClass &rhs);   // copy assignment
MyClass& operator=(MyClass &&rhs); // move assignment

像这样，移动赋值运算符对于(非常量)右值参数严格来说是更好的匹配，因此它是由重载分辨率选择的.

Done like this, the move assignment operator is a strictly better match for a (non-const) rvalue argument and so it's chosen by overload resolution.

另一种称为复制和交换的方法是只提供赋值运算符，按值获取参数，然后使用swap来实现它:

An alternative approach, called copy-and-swap, is to provide just the assignment operator, take the parameter by value, and use swap to implement it:

MyClass& operator=(MyClass rhs)
{
  swap(*this, rhs);
  return *this;
};

这将复制/移动构造函数重新用于分配.它要求您已经实现了一个swap函数，该函数应该不被抛出.

This reuses the copy/move constructor for the assignment. It requires you to have implemented a swap function, which should be non-throwing.

这种方法的缺点是，有时手动执行副本分配比执行复制构建并进行移动要便宜.

The downside of this approach is that sometimes, manually implementing copy assignment can be cheaper than performing copy construction followed by a move.

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