decltype(void())中的void()到底是什么意思? [英] What does the void() in decltype(void()) mean exactly?

问题描述

这是这个问题的跟进，更确切地说是评论的此答案.

This is a follow-up of this question, more precisely of the comments of this answer.

decltype(void())中的void()到底代表什么?
它代表函数类型，表达式还是其他?

What does the void() in decltype(void()) represent exactly?
Does it represent a function type, an expression or whatever?

推荐答案

使用超链接的C ++语法， decltype(void())的解析是:

decltype( expression )
decltype( assignment-expression )
decltype( conditional-expression )

...许多涉及操作顺序的步骤都在这里...

... lots of steps involving order of operations go here ...

decltype( postfix-expression )
decltype( simple-type-specifier ( expression-listopt ) )
decltype( void() )

所以void()在这里是 expression 的一种，尤其是

So void() is a kind of expression here, in particular a postfix-expression.

具体请引用2011 ISO C ++标准的5.2.3节[expr.type.conf]第2段:

Specifically, quoting section 5.2.3 [expr.type.conf] paragraph 2 of the 2011 ISO C++ standard:

表达式T()，其中T是简单类型说明符或 typename-specifier 用于非数组完整对象类型或(可能具有cv限定)void类型，可创建 指定的类型，它是值初始化的(8.5；不进行初始化 完成void()情况.

The expression T(), where T is a simple-type-specifier or typename-specifier for a non-array complete object type or the (possibly cv-qualified) void type, creates a prvalue of the specified type, which is value-initialized (8.5; no initialization is done for the void() case).

因此，void()是类型void的表达式，就像int()是类型int的表达式(值0)一样.显然，void表达式没有任何值，但是这里是decltype的操作数，因此不对其进行求值. decltype仅引用其操作数的类型，而不是其值.

So void() is an expression of type void, just as int() is an expression of type int (with value 0). Clearly a void expression has no value, but here it's the operand of decltype, so it's not evaluated. decltype refers only to its operand's type, not its value.

decltype(void())只是引用类型void的冗长方式.

decltype(void()) is simply a verbose way of referring to the type void.

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