Initializing static constexpr variables and classes inside a struct

Question

Here is my working code example:

#include <iostream>

template<typename B>
class b {
public:
    int y;

    constexpr b(int x) : y(x) {

    }

    constexpr void sayhi() {
        std::cout << "hi" << std::endl;
    }
};



template<int x>
struct A {
    static constexpr b<int> bee = x;
    static constexpr int y = x;         // this one is fine and usable already, I don't have to do something like what I did on member bee

    inline static void sayhi() {
        std::cout << y << std::endl;
    }
};

template<int x>
constexpr b<int> A<x>::bee;        // why do I have to do something like this, it will cause errors if I don't.

int main(int argc, char** argv) {
    A<30>::bee.sayhi();             // works fine
    A<30>::sayhi();                 // works fine

    return 0;
}

What my code does is simple, I have template struct A that has two static variables, namely a static constexpr int y and a static constexpr b<int> bee = x;. My template struct A will get the value of the argument which will be copied by x from the template parameter. My question is: how come when it comes to classes, I have to initialize the class by doing something like this:

template<int x>
constexpr b<int> A<x>::bee; 

If I don't use the code above, I get the undefined reference error. Wherein the int is already fine and accessible just from doing something like:

static constexpr int y = x;    

I am concerned why I don't have to forward declare it anymore.

Answer 1

A static constexpr member has a value upon its initialization inside the class { } scope, but it does not have a location in memory (an address) until it is defined outside the class { }. The reason is that you may decide to include some or all of its specializations in a link library (e.g. .o or .so), or whether to give effectively-inline linkage to specializations by default.

The out-of-class definition is required if the address of the object is ever used, which implies that it must exist as a global variable. On the other hand, if you want the constexpr member only to exist at compile time, prohibiting global storage allocation, then omitting the definition is a good choice.

By the way, it's not allowed to put the constexpr specifier on a function that can never be evaluated as a constant expression, such as sayhi which prints to std::cout. This is a "no diagnostic required (NDR)" rule, meaning that the compiler might not complain now but the next compiler version might.