c++ - 静态成员初始化期间访问私有(private)静态函数

Question

我有一个带有静态成员的类。这将使用同一类的私有(private)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，因为不允许我从类外访问私有(private)函数。那么为什么私有(private)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，该成员是常量，据我所知，设置它的唯一方法是在定义位置直接初始化。

Answer 1

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

来自标准，class.static.data#note-1 :

[Note 1: The initializer in the definition of a static data member isin 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 processinhabits the global scope; the notation process​::​run_­chainindicates that the member run_­chain is a member of class process andin the scope of class process. In the static data member definition,the initializer expression refers to the static data member running ofclass process. — end example]