[英]Why does boost::variant not provide operator !=
Given two identical boost::variant
instances a
and b
, the expression ( a == b )
is permitted. 给定两个相同的boost::variant
实例a
和b
,允许表达式( a == b )
。
However ( a != b )
seems to be undefined. 但是( a != b )
似乎未定义。 Why is this? 为什么是这样?
I think it's just not added to the library. 我认为它只是没有添加到库中。 The Boost.Operators won't really help, because either variant would have been derived from boost::operator::equality_comparable. Boost.Operators实际上并没有帮助,因为任何一个变体都是从boost :: operator :: equality_comparable派生的。 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. David Pierre说你可以使用它,但你的反应也是正确的,ADL找不到新的运算符!=所以你需要一个using运算符。
I'd ask this on the boost-users mailing list. 我在boost-users邮件列表上问这个问题。
Edit from @AFoglia's comment: 编辑来自@ AFoglia的评论:
Seven months later, and I'm studying Boost.Variant, and I stumble over this better explanation of the omission lists. 七个月后,我正在研究Boost.Variant,我偶然发现了这个遗漏名单的更好解释。
http://boost.org/Archives/boost/2006/06/105895.php http://boost.org/Archives/boost/2006/06/105895.php
operator==
calls operator==
for the actual class currently in the variant. operator==
为当前变体中的实际类调用operator==
。 Likewise calling operator!=
should also call operator!=
of the class. 同样,调用operator!=
也应该调用类的operator!=
。 (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!=
. (因为理论上,可以定义一个类,所以a!=b
与!(a==b)
。)这样就会增加另一个要求,即变量中的类有一个operator!=
。 (There is a debate over whether you can make this assumption in the mailing list thread.) (关于你是否可以在邮件列表主题中做出这个假设存在争议。)
This is a link to the answer from the author himself when this question was formulated on boost mailing list 当这个问题在boost邮件列表中制定时,这是作者自己答案的链接
Summarizing it, in the author opinion, implementing comparison operators (!= and <) would add more requirements on the types used to create the variant type. 总结一下,在作者看来,实现比较运算符(!=和<)会对用于创建变体类型的类型添加更多要求。
I don't agree with his point of view though, since != can be implemented in the same way as ==, without necessarily hiding the possible implementations of these operators for each of the types making up the variant 我不同意他的观点,因为!=可以以与==相同的方式实现,而不必为构成变体的每种类型隐藏这些运算符的可能实现
Because it doesn't need to. 因为它不需要。
Boost has an operators library which defines operator!= in term of operator== Boost有一个运算符库 ,它根据operator ==定义operator!=
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