`decltype`含有多个参数的含义 [英] Meaning of `decltype` with more than one argument

问题描述

我找到了一个 decltype 相关问题及其答案其中
OP和回答者似乎假定
decltype（t，u）是取决于 t 和 u 。

I have found a decltype-related question and its answers in which both the OP and the answerers seem to assume that decltype(t,u) is something that depends on the types of both t and u.

链接问题中的 decltype（t，u）
是与
兼容的最小类型 t 和 u ，在某种意义上最小和兼容。
如果它的含义类似于§5（9）中的常见类型或者 std :: common_type c>从§20.9.7.6（3），那么意思是
decltype（true？t：u）。

I suppose that the intended meaning of the decltype(t,u) in the linked question is that of a minimal type compatible with both t and u, in some sense of minimal and compatible. If it is meant like a common type as in §5(9) or in the definition of std::common_type from §20.9.7.6(3), then the intended meaning would be decltype(true ? t : u).

但是，似乎C ++ 11只定义了一个参数 decltype 。
也就是说，解析器指定接受 decltype（ 表达式 ）
其中 可以是以逗号分隔的 assignment-expression 列表。
从§7.1.6.2（4）重新读取 decltype 的规范，
我必须解释 decltype ，u）为 decltype（（u））。

However, it seems that C++11 defines only a one-argument decltype. That said, the parser is specified to accept decltype( expression ) where expression can be a comma-separated list of assignment-expression. Re-reading the specification of decltype from §7.1.6.2(4), I have to interpret decltype(t,u) as decltype((u)).

是 decltype（t，u）实际上？我缺少一些东西？

Question: So what is decltype(t,u) actually? Am I missing something?

编辑：作为答案收敛于这是逗号运算符，我仍然想看到
你的判断， decltype（（u））（声明类型的引用）或 decltype（u） 。

As answers converge on "this is the comma operator", I'd still like to see your judgement on whether it's decltype((u)) (a reference to the declared type) or decltype(u) (the declared type). I suspect the former.

推荐答案

这里的逗号是运算符，不是参数分隔符

Comma here is the operator, not argument separator

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