`equality_comparable_with`是否需要要求`common_reference`? [英] Does `equality_­comparable_with` need to require `common_reference`?

问题描述

概念 equality_comparable_with< T，U> 旨在声明可以将类型为 T 和 U 的对象彼此比较，如果相等，则具有预期的含义.很好.

The concept equality_­comparable_with<T, U> is intended to declare that objects of type T and U can be compared equal to each other, and if they are, then this has the expected meaning. That's fine.

但是，该概念还需要存在 common_reference_t< T& U& . common_reference 及其附带功能的主要推动力似乎是启用代理迭代器，使其具有表示此类迭代器的 reference 和 value_type 之间的关系的地方.

However, this concept also requires common_reference_t<T&, U&> to exist. The primary impetus for common_reference and its attendant functionality seems to be to enable proxy iterators, to have a place to represent the relationship between reference and value_type for such iterators.

那太好了，但是...与测试一个 T 和一个 U 是否可以彼此相等有什么关系?为什么标准要求 T 和 U 具有共同的引用关系，只是为了让您比较它们是否相等?

That's great, but... what does that have to do with testing if a T and a U can be compared equal to one another? Why does the standard require that T and U have a common reference relationship just to allow you to compare them equal?

这会产生奇怪的情况，在这种情况下，很难拥有两种在逻辑上不具有合理可比性的公共参照关系的类型.例如， vector< int> 和 pmr :: vector< int> 在逻辑上应该是可比的.但这不是因为两个其他不相关的类型之间没有合理的公共引用.

This creates oddball situations where it's very difficult to have two types which don't reasonably have a common-reference relationship that are logically comparable. For example, vector<int> and pmr::vector<int> logically ought to be comparable. But they can't be because there's no reasonable common-reference between the two otherwise unrelated types.

推荐答案

这可以追溯到 Palo Alto报告，§3.3和D.2.

This goes all the way back to the Palo Alto report, §3.3 and D.2.

为使交叉类型概念在数学上合理，您需要定义交叉类型比较的含义.对于 equality_comparable_with ， t == u 通常表示 t 和 u 相等，但这甚至意味着什么两个不同类型的值相等?该设计表示，交叉类型相等性是通过将它们映射到普通(引用)类型来定义的(必须进行此转换才能保留值).

For the cross-type concepts to be mathematically sound, you need to define what the cross-type comparison means. For equality_comparable_with, t == u generally means that t and u are equal, but what does it even mean for two values of different types to be equal? The design says that cross-type equality is defined by mapping them to the common (reference) type (this conversion is required to preserve the value).

如果不希望使用 equality_comparable_with 的强大公理，则该标准使用仅博览会的概念 weakly-equality-comparable-with ，该概念不需要通用参考.然而，这是语义可憎".(用Casey Carter的话来说)，因此仅出于说明目的:它允许具有 t == u 和 t2 == u 但具有 t！=t2 (这实际上是哨兵所必需的).

Where the strong axioms of equality_comparable_with is not desirable, the standard uses the exposition-only concept weakly-equality-comparable-with, and that concept doesn't require a common reference. It is, however, a "semantic abomination" (in the words of Casey Carter) and is exposition-only for that reason: it allows having t == u and t2 == u but t != t2 (this is actually necessary for sentinels).

这篇关于`equality_comparable_with`是否需要要求`common_reference`?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！