超载'=='让你'！='？ [英] Does overloading '==' get you '!='?

问题描述

如果我为结构手动重载 == 运算符，那么我可以免费获得！= （可能定义为布尔相反），或者我必须手动重载它（即使只是 return！（this == rhs）？

If I manually overload the == operator for a structure, do I get the != operator for free (presumably defined to be the boolean opposite), or do I have to overload it manually (even if to just return !(this == rhs)?

编辑 - 问题不是我是否可以重载两个运算符，但是如果我已经重载了等于运算符，我是否必须重载不等式。无论如何，给出了好的答案。 / p>

Edit-The question is not whether or not I CAN overload both operators, but whether I must overload inequality if I've already overloaded the equality operator. Regardless, good answers have been given.

推荐答案

重载运算符== 不会给您 operator！= 。你必须手动执行它，并且规范的方式是以 operator == $ c>！（left == right）。

Overloading operator == does not give you operator !=. You have to do it manually and the canonical way is to implement it in terms of operator == as in !(left == right).

运算符的语义不是由标准决定的， operator == 表示等于重载 operator！= 不是这是一个好的做法，实际上应该劝阻。

The semantics of the operators are not dictated by the standard. You could very well overload operator == to mean equality yet overload operator != to something different like addition or even equality again (not that this is a good practice, in fact it should be discouraged. When in doubt, do as the ints do...).[Refer (1) Below]

在旁注中， Boost.Operators 可以帮助您为操作符提供规范的实现。还有 std :: rel_ops ，具有 operator！= 的规范实现。

On a side note, Boost.Operators can help you provide canonical implementations for operators. There is also std::rel_ops with a canonical implementation for operator !=.

（1）要了解详情，请阅读 操作员的三个基本规则在C ++中重载。

(1) To know more about it read Three basic rules of operator overloading in C++.

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