为什么在C ++ 20中不推荐使用std :: rel_ops :: operators？

问题描述：

根据 cppreference.com ， std :: rel_ops :: operator！=，>，< =，> = 在C ++ 20中将不推荐使用。

According to cppreference.com, std::rel_ops::operator!=,>,<=,>= will be deprecated in C++20.

背后的原理是什么？

答

在C ++ 20中，您会得到三向比较（运算符< => ），它会自动生成默认比较（如果提供）：

In C++20, you get three-way comparison (operator <=>), which automatically "generates" default comparisons if provided:

struct A {
   // You only need to implement a single operator.
   std::strong_ordering operator<=>(const A&) const;
};

// Compiler generates 4 relational operators (you need to default the
// three-way comparison operator to get == and !=).
A to1, to2;
if (to1 > to2) { /* ... */ } // ok
if (to1 <= to2) { /* ... */ } // ok, single call to <=>

与 std :: rel_ops相比，三元比较有多个优点，这可能是为什么不推荐使用 std :: rel_ops 运算符的原因。在我的头上：

There are multiple advantages of the three-way comparison over std::rel_ops, which is probably why std::rel_ops operators are deprecated. On top of my head:

它更加通用，因为取决于的返回类型运算符< => （ std :: strong_ordering ， std :: weak_ordering ，。 ..），仅生成相关的运算符。请参见 < compare> 标头以获取更多信息。

It is more versatile, since, depending on the return type of operator<=> (std::strong_ordering, std::weak_ordering, ...), only relevant operators are generated. See the <compare> header for more information.

使用命名空间std :: rel_ops进行不会带来大量模板操作符重载。

You do not bring a bunch of templated operator overloads by doing using namespace std::rel_ops.

您可以要求编译器通过默认方式为您生成三元运算符（自动运算符< =>（A const&）=默认值）—基本上，这将对基类和非静态数据成员进行字典比较，并且如果返回类型为 auto ，则会推断出正确的排序类型。

You can ask the compiler to generate the three-way operator for you by defaulting it (auto operator<=>(A const&) = default) — This will basically generate a lexicographic comparison of base classes and non-static data members, plus it will deduce the right type of ordering if the return type is auto.

为什么在C ++ 20中不推荐使用std :: rel_ops :: operators？

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