有没有人发现需要声明一个复制赋值操作符const的返回参数？ [英] Has anyone found the need to declare the return parameter of a copy assignment operator const?

问题描述

副本赋值运算符具有通常的签名：

  my_class& operator =（my_class const& rhs）; 
  my_class const& operator =（my_class const& rhs）; 
  
您只能定义一个或另一个，但不能同时定义。
解决方案
使复制赋值的返回类型为非const引用的主要原因是它是标准中的Assignable的要求。 p> 
 
 
如果你使返回类型为 const 引用，那么你的类将不能满足在任何标准库容器。
 
The copy assignment operator has the usual signature:
    my_class & operator = (my_class const & rhs);
Does the following signature have any practical use?
    my_class const & operator = (my_class const & rhs);
You can only define one or the other, but not both.
解决方案
The principle reason to make the return type of copy-assignment a non-const reference is that it is a requirement for "Assignable" in the standard.

If you make the return type a const reference then your class won't meet the requirements for use in any of the standard library containers.

                        
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