定义 static constexpr auto class 变量

Question

由于未定义的引用，以下代码无法 link：

// file.h
struct S {
    static constexpr auto x = 0;
};

// file.cpp
int main() {
    auto const & ref = S::x;
}

根据 Why doesn't the C++11 'auto' keyword work for static members? 的建议，这似乎适用于 clang 3.5：

// file.h
struct S {
    static constexpr auto x = 0;
};

// file.cpp
constexpr decltype(S::x) S::x;

int main() {
    auto const & ref = S::x;
}

它实际上是有效的 C++ 吗？这似乎违反了函数遵循的 "auto everywhere or nowhere" 规则（您可以转发声明一个 returns auto 的函数，然后将其定义为 return auto，但您不能将 auto 与非 -汽车）。

Answer 1

auto类型说明符有两个相关但独立的目的

[dcl.spec.auto] / 1

The auto and decltype(auto) type-specifiers are used to designate a placeholder type that will be replaced later by deduction from an initializer. The auto type-specifier is also used to introduce a function type having a trailing-return-type or to signify that a lambda is a generic lambda.

对于您的静态成员，类型由初始化程序确定，因此 x 在其声明末尾已经具有类型 int。

[dcl.spec.auto] / 4

The type of a variable declared using auto or decltype(auto) is deduced from its initializer.

您提到的规则仅适用于函数和函数模板，与声明变量时auto的使用无关。

[dcl.spec.auto] / 13

Redeclarations or specializations of a function or function template with a declared return type that uses a placeholder type shall also use that placeholder, not a deduced type.

定义 static constexpr auto class 变量

Defining static constexpr auto class variable

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