C++ 类静态私有属性的初始化问题...

Question

class A{
public:
    int id;
    static A getA() {
        return self;
    }
private:
    static A self;
    A(){
        id = 3;
        std::cout << "A" << " ";
    }

};

代码如上, 类似这样的类, 我要如何初始化self呢?

同时我发现, 我如果这样写的话, 它会默认调用可以初始化A()初始化self, 屏幕上能显示 A第一行3, 但是如果我在头文件中倒数第二行 A::self 改成 A::self(); 就报错, 说我重定义了self, 他说之前self是一个属性, 这里我重定义它为一个函数...

#include <iostream>
#include "A.h"

using namespace std;

int main(void){
    cout << "第一行";
    cout << A::self.x;
}


#ifndef LAB_A_H
#define LAB_A_H

#include <iostream>

class A{
public:
    int x = 3;

    static A self;
private:
    A(){
        std::cout << "A";
    }
};

A A::self;

#endif //LAB_A_H

问题在于 :

1. 对于这种成员的初始化, 如果使用默认的构造函数就不能加括号是吗?(我尝试过用一个带参数的构造函数, 是可以编译运行的)

2. 如果我去掉最后一行, 也就是不对这个静态属性self进行初始化, 只要我main.cpp中不用到这个类, 我就算include这个头文件(显然是错误的), 也可以正常运行, 这又是为什么呢?

Answer 1

If you can change this class, then the standard approach is

static A& getA() {
     static A a;
     return a;
}

a will only be initialized once, when getA is called for the first time. This is thread-safe since C++11.

If you can't change it, then you can only initialize this self outside the class. And your constructor is private, so it cannot be initialized.

Regarding your default constructor without ():
Yes, parentheses will cause ambiguity, and the compiler will understand that this is a function signature

  
 A a(); // 声明一个返回值为A，不含参数的函数

Answer 2

Just write this in a.cpp:

// a.cpp
#include "a.h"

A A::self;

Tested it:

// main.cpp
#include "a.h"

int main() {
    A a = A::getA();
    std::cout << "| static A.id:" << a.id << std::endl;
    return 0;
}

// output:
A | static A.id:3