为什么C++可以使用未初始化的成员变量(引用或指针 *NOT 值复制*）来初始化其父类的成员变量

我有两个关于在C++中初始化成员变量的问题。

1. 为什么当Fruits的构造函数定义为explicit Fruits(std::string* f_n) :f_name_(f_n)时(或者当这些代码中的指针被替换为引用时)，代码explicit Apple(const char* name) :Fruits(&_name)工作？

派生类的成员变量

2. 是在其子基类的成员变量之后还是之前分配内存？

请参阅下面的代码。

// Apple.h c++
#pragma once
#include<iostream>
#include<cstring>
#include<string>
class Plant {
public:
Plant() {
std::cout << "Plant construction function called." << std::endl;
}
inline void show() {
std::cout << "name：" << s_name_ << std::endl;
}
void Init() {
std::cout << "Init assign value called." << std::endl;
s_name_ = this->getName();
}
virtual ~Plant() {
std::cout << "Plant deconstruction function called." << std::endl;
}
private:
std::string s_name_;
protected:
virtual std::string getName() = 0;
};
class Fruits: public Plant{
public:
explicit Fruits(std::string* f_n) :f_name_(f_n) {
std::cout << "Fruits construction function called."<< std::endl;
}
virtual ~Fruits() {
std::cout << "Fruits deconstruction function called." << std::endl;
}
protected:
std::string getName() override {
std::cout << "getName() function called." << std::endl; 
return *f_name_;
}
private:
std::string* f_name_;
};
class Apple:public Fruits {
private:
std::string _name;
public:
explicit Apple(const char* name) :Fruits(&_name) {
_name = name;
std::cout << "Apple constuction function called."<< std::endl;
Plant::Init();
}
~Apple() {
std::cout << "Apple deconstruction function called." << std::endl;
}
};
// main.cpp
#include<iostream>
#include "Apple.h"
using namespace std;
int main() {
{
cout << "test1:" << endl;
Apple t1("Apple1");
t1.show();
}
cin.get();
return 0;
}