为什么除非使用静态模板字段初始化？

问题描述

编辑：有一个副本，但我会离开这，因为我个人很难找到它。此外，这里的答案帮助我：

假设有以下类：

template<class X>
struct A
{
   static bool x;
   static bool foo()
   {
      cout << "here";
      return true;
   }
};

template<class X>
bool A<X>::x = A<X>::foo();

我会假设当我专门 A ，静态字段 x 将被初始化。但是，以下内容：

I would have assumed that when I specialize A, the static field x would get initialized. However, the following:

A<int> a;
//no output

不会导致调用 foo 。如果我尝试访问成员，行为是预期的：

doesn't result in a call to foo. If I try to access the member, the behavior is as expected:

A<int> a;
bool b = a.x;
//output: here

编辑：我如何确保<$

How can I make sure A::x is initialized without accessing it?

推荐答案

这个想法是参考（14.7.1.2）：

This think is the reference(14.7.1.2) :

除非类模板或成员模板的成员已被显式实例化或明确专门化，否则成员的特化当在需要成员定义存在的上下文中引用特化时，隐含地实例化;特别是，静态数据成员的初始化（和任何相关的副作用）不会发生，除非静态数据成员本身以需要静态数据成员的定义的方式使用。

Unless a member of a class template or a member template has been explicitly instantiated or explicitly specialized, the specialization of the member is implicitly instantiated when the specialization is referenced in a context that requires the member definition to exist; 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.

template<class X, bool y>
struct A
{
    static cosnt bool x = y;
    static bool foo()
    {
       cout << "here";
       return true;
    }
 };

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