这是为什么常量常量常量声明中无效 [英] Why is this const const const declaration valid
问题描述
有一个朋友问我解释为什么
常量常量常量常量const int的常量I = 0;
是有效的语法。我拒绝对这个问题的看法。虽然我很好奇，如果它只是一个语法的事情吗？
编辑。标签是C ++，我的朋友被引用的gcc，所以我也许应该补充的C标记。
从n1256：
6.7.3 / 4
如果同一合格网络连接器出现在同一不止一次
特定网络连接ER-合格音响器列表，无论是直接或经由一个或多个的typedef，
的行为是相同的，就好像它只出现一次。在C ++中，这应该给一个错误。
这是C ++标准的03：
7.1.5 / 1 [dcl.type]
作为一般规则，至多一个类型说明符中允许的
声明的完整DECL说明符 - 起。唯一的例外
此规则有以下几种：
- const或挥发性可与任何其它类型说明符进行组合。不过，多余的CV-预选赛被禁止时，通过使用类型定义（7.1.3）或模板类型参数（14.3），在这种情况下，多余的cv修饰符会被忽略，除了介绍
从N3797（C ++ 11）：
7.1.6 / 2 [dcl.type]
作为一般规则，至多一个类型特定网络连接器被允许在
声明的或以类型特定网络ER-SEQ完整DECL-Fi无线特定ER-SEQ
或尾随型特定网络ER-SEQ。唯一的例外是
以下内容：- 常量可以用除了本身的任何类型说明符相结合。
A friend asked me to explain why
const const const const const int const i = 0;is valid syntax. I declined to have any thoughts on the subject. Though I'm curious if it is just a grammar thing?
Edit. Tag is C++, my friend was referencing gcc, so I should probably add the C tag.
解决方案From n1256:
6.7.3/4
If the same qualifier appears more than once in the same specifier-qualifier-list,either directly or via one or more typedefs, the behavior is the same as if it appeared only once.
In C++, this should give an error.
From C++03 standard:
7.1.5/1 [dcl.type]
As a general rule, at most one type-specifier is allowed in the complete decl-specifier-seq of a declaration. The only exceptions to this rule are the following:
— const or volatile can be combined with any other type-specifier. However, redundant cv- qualifiers are prohibited except when introduced through the use of typedefs (7.1.3) or template type arguments (14.3), in which case the redundant cv-qualifiers are ignored.
From N3797 (C++11):
7.1.6/2 [dcl.type]
As a general rule, at most one type-specifier is allowed in the complete decl-specifier-seq of a declaration or in a type-specifier-seq or trailing-type-specifier-seq. The only exceptions to this rule are the following:
— const can be combined with any type specifier except itself.
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