静态类成员的初始化.为什么是 constexpr?

Initialisation of static class member. Why constexpr?(静态类成员的初始化.为什么是 constexpr?)

问题描述

当我想要一个静态指针作为类的成员时，我需要 constexpr 来使用 nullptr 进行初始化.

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;
}

谁能解释我为什么需要这样做?我找不到静态变量必须在编译时存在的确切原因.

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

那是因为你在类定义中初始化它.这只允许用于常量整数和枚举类型(总是)和 constexpr 数据成员(自 C++11 起).通常，您会在定义它的位置(在类之外)对其进行初始化，如下所示:

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

class Application {
    private:
        static Application* app;
}

Application.cpp

Application* Application::app = nullptr;

请注意，即使在 constexpr 情况下，您也需要提供类外定义，但它不能包含初始化程序.不过，我相信第二种情况是您真正想要的.

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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