Initialisation of static class member. Why constexpr?

问题

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;
}

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.

回答1:

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;

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.

回答2:

If you don't want it to be constexpr (and it's not an integer) then you need to initialise it outside of the class body:

class Application
{
private:
    static Application* app;
};

Application* Application::app = nullptr;

回答3:

Typically, you need to initialise a static member variable outside the class declaration, unless it is const. I think this explains it better than I could.

回答4:

Static variables do not need to "exist at compile time". But if you want to initialize a static variable inside the class, its value needs to be known at compile time.

However, I do not know the reason for this restriction.

来源：https://stackoverflow.com/questions/19319291/initialisation-of-static-class-member-why-constexpr

标签

c++

c++11

initialization

static-members

constexpr