为什么boost :: variant不提供operator！= [英] Why does boost::variant not provide operator !=

问题描述

给定两个完全相同的 boost :: variant 实例 a code> b ，则允许使用表达式 （a == b） / p>

但 （a！= b） 似乎未定义。为什么是这样？

解决方案

我认为它只是没有添加到库中。 Boost.Operators不会真正帮助，因为任一变体都将从boost :: operator :: equality_comparable派生。 David Pierre是正确的说你可以使用它，但是你的回答也是正确的，新的操作符！=将不能被ADL找到，所以你需要一个使用操作符。

我会在boost-users邮件列表中询问这个。

从@ AFoglia的注释编辑：

七个月后，我正在学习Boost.Variant，我对这个更好的解释列出了遗漏列表。

http://boost.org/Archives/boost/2006/06/105895.php

操作符== 调用 operator == 当前在变体中的实际类。同样，调用 operator！= 也应该调用类的 operator！= 。 （因为理论上可以定义一个类，所以 a！= b 不等于！（a == b）。）因此，这将增加另一个要求变量中的类有一个 operator！= 。 （有一个关于你是否可以在邮件列表线程中做这个假设的争论。）

Given two identical boost::variant instances a and b, the expression ( a == b ) is permitted.

However ( a != b ) seems to be undefined. Why is this?

解决方案

I think it's just not added to the library. The Boost.Operators won't really help, because either variant would have been derived from boost::operator::equality_comparable. David Pierre is right to say you can use that, but your response is correct too, that the new operator!= won't be found by ADL, so you'll need a using operator.

I'd ask this on the boost-users mailing list.

Edit from @AFoglia's comment:

Seven months later, and I'm studying Boost.Variant, and I stumble over this better explanation of the omission lists.

http://boost.org/Archives/boost/2006/06/105895.php

operator== calls operator== for the actual class currently in the variant. Likewise calling operator!= should also call operator!= of the class. (Because, theoretically, a class can be defined so a!=b is not the same as !(a==b).) So that would add another requirement that the classes in the variant have an operator!=. (There is a debate over whether you can make this assumption in the mailing list thread.)

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