如何声明和定义具有推断类型的静态成员？

问题描述

我需要用一个复杂的（很多模板参数）类型定义一个静态成员（不是constexpr）。因此，希望有这样的东西：

I need to define a static member (not constexpr) with a complex (many template parameters) type. Therefore it would be desired to have something like this:

struct X {
    static auto x = makeObjectWithComplexType();
};

但这不是C ++。因此，我尝试解决此问题，并认为下面的代码片段可以工作，但它无效：

But it is not C++. So I tried to workaround it, and thought the snippet below would work, but it does not:

#include <string>

struct X {
    static auto abc() {
        return std::string();
    }

    static decltype(abc()) x;
};

decltype(abc()) X::x;

int main() {}

失败并显示错误：错误：在扣除自动 *之前使用静态自动X :: abc（） *

有什么方法可以使上面的代码段去工作。还是有其他方法来定义具有推断类型的静态成员？

Is there any way to make the snippet above to work. Or is there any other way to define a static member with a deduced type?

推荐答案

如果您使用的是C ++ 17，则您可以这样做：

If you have C++17, then you can do this:

struct X {
    static inline auto x = makeObjectWithComplexType();
};

如果不这样做，不幸的是必须重复 makeObjectWithComplexType（）：

If you don't, you unfortunately have to repeat makeObjectWithComplexType():

struct X {
    static decltype(makeObjectWithComplexType()) x; // declaration
};

auto X::x = makeObjectWithComplexType(); // definition

请注意，clang可以成功编译该版本，但gcc和msvc不能。我不确定哪个编译器正确，所以我在。

Note, that clang successfully compiles this version, but gcc and msvc don't. I'm not sure which compiler is right, so I've asked it in a question.

如果您有兴趣，为什么您的解决方法不起作用，请查看以下问题：

If you're interested, why your workaround doesn't work, check out this question: Why can't the type of my class-static auto function be deduced within the class scope?

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