c++ - 如何强制初始化静态成员？

Question

考虑这个示例代码:

template<class D>
char register_(){
    return D::get_dummy(); // static function
}

template<class D>
struct Foo{
    static char const dummy;
};

template<class D>
char const Foo<D>::dummy = register_<D>();

struct Bar
    : Foo<Bar>
{
    static char const get_dummy() { return 42; }
};

( Also on Ideone .)

我希望 dummy 会在 Foo 的具体实例化后立即被初始化，而我在 Bar 中就有。 This question (以及最后的标准报价)解释得很清楚，为什么没有发生。

[...] in particular, the initialization (and any associated side-effects) of a static data member does not occur unless the static data member is itself used in a way that requires the definition of the static data member to exist.

有没有办法强制dummy被初始化(有效地调用register_)没有任何实例Bar 或 Foo (没有实例，所以没有构造函数技巧)并且没有 Foo 的用户需要以某种方式显式声明成员？不需要派生类做任何事情的额外 cookie。

---

编辑:Found a way对派生类的影响最小:

struct Bar
    : Foo<Bar>
{   //                              vvvvvvvvvvvv
    static char const get_dummy() { (void)dummy; return 42; }
};

不过，我仍然希望派生类不必这样做。 :|

Answer 1

考虑:

template<typename T, T> struct value { };

template<typename T>
struct HasStatics {
  static int a; // we force this to be initialized
  typedef value<int&, a> value_user;
};

template<typename T>
int HasStatics<T>::a = /* whatever side-effect you want */ 0;

也可以不引入任何成员:

template<typename T, T> struct var { enum { value }; };
typedef char user;

template<typename T>
struct HasStatics {
  static int a; // we force this to be initialized
  static int b; // and this

  // hope you like the syntax!
  user :var<int&, a>::value,
       :var<int&, b>::value;
};

template<typename T>
int HasStatics<T>::a = /* whatever side-effect you want */ 0;

template<typename T>
int HasStatics<T>::b = /* whatever side-effect you want */ 0;