与正在定义的类具有相同类型的静态constexpr成员 [英] static constexpr member of same type as class being defined

问题描述

尝试1：

我想要一个类C的静态constexpr成员类型C.这是可能在C ++ 11吗？

  struct Foo {
 constexpr Foo（）{} 
 static constexpr Foo f = Foo 
}; 
 constexpr Foo Foo :: f; 
 

g ++ 4.7.0说：无效使用不完全类型指的是 Foo（）

尝试2：

struct Foo {
constexpr Foo（）{}
static constexpr Foo f;
};
constexpr Foo Foo :: f = Foo（）;


现在问题是缺少一个初始化器 constexpr



尝试3：

  struct Foo {
 constexpr Foo（）{} 
 static const Foo f; 
}; 
 constexpr Foo Foo :: f = Foo（）; 
  

现在g ++抱怨重新声明 Foo :: f 不同于 constexpr 。

解决方案

（§9.4.2/ 3）[...]文字类型的静态数据成员可以在
类定义中使用constexpr说明符声明;如果是，其声明应指定一个括号或初始值化器，其中作为赋值语句的每个初始化子句都是一个常量表达式。 [...]

从上面的内容（以及在静态数据中没有关于非文字类型的单独语句成员声明），我相信， constexpr 的静态数据成员必须是文字类型（如§3.9/ 10定义） ，和，它必须将定义包含在声明中。后面的条件可以通过使用下面的代码来满足：

  struct Foo {
 constexpr Foo b $ b static constexpr Foo f {}; 
}; 
  

这与您的Attempt 1类似，但没有类外部定义。

但是，因为在声明/定义静态成员时 Foo 不完整，编译器不能检查它是否

请注意，有一个字面类型（如§3.9/ 10中定义），因此它拒绝代码。此后C ++ - 11文档（N3308），其中讨论了 constexpr ，并提出修正建议。具体来说，建议的措辞部分建议修改§3.9/ 10，这意味着包含不完全类型作为一种文字类型。如果该修正案在未来版本的标准中被接受，您的问题将得到解决。

I would like a class C to have a static constexpr member of type C. Is this possible in C++11?

Attempt 1:

struct Foo {
    constexpr Foo() {}
    static constexpr Foo f = Foo();
};
constexpr Foo Foo::f;

g++ 4.7.0 says: 'invalid use of incomplete type' referring to the Foo() call.

Attempt 2:

struct Foo {
    constexpr Foo() {}
    static constexpr Foo f;
};
constexpr Foo Foo::f = Foo();

Now the problem is the lack of an initializer for the constexpr member f inside the class definition.

Attempt 3:

struct Foo {
    constexpr Foo() {}
    static const Foo f;
};
constexpr Foo Foo::f = Foo();

Now g++ complains about a redeclaration of Foo::f differing in constexpr.

解决方案

If I interpret the Standard correctly, it isn't possible.

(§9.4.2/3) [...] A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. [...]

From the above (along with the fact that there is no separate statement about non-literal types in static data member declarations), I believe it follows that a static data member that is constexpr must be a literal type (as defined in §3.9/10), and it must have its definition included in the declaration. The latter condition could be satisfied by using the following code:

struct Foo {
  constexpr Foo() {}
  static constexpr Foo f {};
};

which is similar to your Attempt 1, but without the class-external definition.

However, since Foo is incomplete at the time of declaration/definition of the static member, the compiler can't check whether it is a literal type (as defined in §3.9/10), so it rejects the code.

Note that there is this post-C++-11 document (N3308) which discusses various problems of the current definition of constexpr in the Standard, and makes suggestions for amendments. Specifically, the "Proposed Wording" section suggests an amendment of §3.9/10 that implies the inclusion of incomplete types as one kind of literal type. If that amendment was to be accepted into a future version of the Standard, your problem would be solved.

这篇关于与正在定义的类具有相同类型的静态constexpr成员的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！