静态成员初始化期间访问私有静态函数

Question

我有一个带有静态成员的 class。这将使用相同 class 的私有静态函数进行初始化。

#include <iostream>
#include <string>

class A
{
public:
    static std::string const s;

private:
    static std::string make()
    {
        return "S";
    }
};

std::string const A::s = A::make();

int main()
{
    std::cout << A::s << std::endl;
    // std::cout << A::make() << std::endl; // <-- Does not work
    return 0;
}

我的问题是：由于哪条规则允许这样做？显然注释部分不起作用，因为不允许我从 class 外部访问私有函数。那么为什么私有静态成员在启动时的初始化是一个特例呢？（附带说明：这条规则的目的是什么？是否允许这种情况？）

我知道初始化静态成员的其他机制（如此处：Initializing private static members）。但在我的例子中，成员是常量，据我所知，唯一的设置方法是在定义的地方直接初始化。

Answer 1

因为静态数据成员的初始化被认为是 class 特征的一部分，即使静态数据成员是在命名空间范围内定义的（在 class 定义之外）。

根据标准，class.static.data#note-1：

[Note 1: The initializer in the definition of a static data member is in the scope of its class ([basic.scope.class]). — end note]

[Example 1:
class process {
  static process* run_chain;
  static process* running;
};

process* process::running = get_main();
process* process::run_chain = running;
The definition of the static data member run_chain of class process inhabits the global scope; the notation process::run_chain indicates that the member run_chain is a member of class process and in the scope of class process. In the static data member definition, the initializer expression refers to the static data member running of class process. — end example]

静态成员初始化期间访问私有静态函数

Access to private static function during static member initialization

c++

initialization

language-lawyer

static-methods

static-members