c++ - 除标识符外，非限定标识符如何根据左值和右值进行分类？

Question

似乎标准并没有明确讲到一些unqualified-id的表达类别。另一方面，作为 unqualified-id 之一的 identifier 分类如下:

The result is an lvalue if the entity is a function, variable, or data member and a prvalue otherwise.

那么，其他unqualified-id呢——operator-function-id, conversion-function-id, literal -operator-id, ~class-name, ~decltype-specifier, template-id?

Answer 1

这已被确认为 Core Working Group issue 536几乎是 10 年前提交的。所以可能它还没有被认为重要到需要修复。

More importantly, some kinds of id-expressions are not described by 5.1.1 [expr.prim.general]. The structure of this section is that the result, type, and lvalue-ness are specified for each of the cases it covers [...]

This treatment leaves unspecified all the non-identifier unqualified-ids (operator-function-id, conversion-function-id, and template-id) [...]

-- CWG 536

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从各种编译器的行为来看，我们可以假设其意图是所有这些表达式都遵循标识符规则:

The type of the expression is the type of the identifier. The result is the entity denoted by the identifier. The result is an lvalue if the entity is a function, variable, or data member and a prvalue otherwise.

-- post-N4296 [expr.prim.general]p13

顺便说一句:除了template-id 之外的所有这些表达式都只能引用函数。 Template-id 也可以引用变量。 template-id和普通的identifier都可以引用类和枚举，这让我有点困惑(可以有int这样的表达式吗？它们是纯右值吗？)