静态类成员的初始化。为什么是constexpr？ [英] Initialisation of static class member. Why constexpr?

问题描述

当我想有一个静态指针作为类的成员我需要 constexpr 用于初始化 nullptr 。

when I want to have a static pointer as a member of a class I need constexprfor the initialisation with nullptr.

class Application {
    private:
        constexpr static Application* app = nullptr;
}

有人可以解释我为什么需要这样做吗？我找不到在编译时必须存在静态变量的确切原因。

Can someone explain me why I need to do that? I cannot find the exact reason why it`s necessary that the static variable has to exist at compile time.

推荐答案

'在类定义内部初始化它。这只允许用于常量积分和枚举类型（always）和 constexpr 数据成员（自C ++ 11）。通常，您可以在定义它的地方（类外）初始化它，例如：

That's because you're initialising it inside the class definition. That's only allowed for constant integral and enumeration types (always) and for constexpr data members (since C++11). Normally, you'd initialise it where you define it (outside the class), like this:

Application.h b
$ b

Application.h

class Application {
    private:
        static Application* app;
}

Application.cpp

Application* Application::app = nullptr;

请注意，您需要提供类外定义，即使在 constexpr 大小写，但它不能包含初始值。不过，我相信第二种情况是你真正想要的。

Note that you need to provide the out-of-class definition even in the constexpr case, but it must not contain an initialiser then. Still, I believe the second case is what you actually want.

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