继承：为什么在继承和提供的变量之间的行为有差别？

问题描述

例如，在这段代码中，如果 line [a] 被注释掉，则输出为0.

For example, in this piece of code, if line [a] is commented out, the output is 0.

#include<iostream>
using namespace std;

class A {
    public:
        int x;
        A() { x = 10; }
};

class B : public A {
    public:
        int x;  // <--------- [a]
        B() { x = 0; }
};

int main() {
    A* ab = new B;
    cout << ab->x << endl;
}

$ g++ inh2.cpp
$ ./a.out
10
$

我有两个问题：

1. 在上述情况下， ab-> gt; x 如何解析为 10 对象是类B ，因此应该 0 。


2. 为什么注释 Line [a] 更改代码的行为？我的推理是， x 将继承，这应该导致相同的行为。

1. How does ab->x resolve to 10 in the above case? The object is of type class B, and thus should value to 0.
2. Why does commenting Line [a] change the behaviour of the code? My reasoning is that x would have anyways been inherited, which should result in same behaviour.

---

我对上述Q＃1的推理：

---

My reasoning for Q #1 above:

• ab 指向 B类的对象的内存位置。

• ab points to the memory location of an object of class B. It is a physical object in the sense that all the variables with their values are assigned memory.

变量 x 在此对象内存储值 0 。

Variable x within this object stores value 0.

当 ab- ; x 完成后，ab告诉我们对象的内存位置，我们去查看里面的内容，发现x是0.所以我们应该打印0。

When ab->x is done, ab tells us the memory location of the object, and we go look inside it to find that x is 0. So we should print 0.

我在哪里错了？

推荐答案

1. 是的，它是 B 类型，但是你将它指定为 A ，因此它使用在A上定义的 x （当我们处理一个指向A的指针时，我们不知道<$ c

1. Yes, it is of type B, but you are assigning it as a pointer to an A, and therefore it is using the x defined on A (as when we're dealing with a pointer to A, we don't know that B even exists, even though that's what you allocated).

当你注释掉这一行时，在构造过程中阶段， A 的构造函数先调用，然后 B 的构造函数设置 x （在其基类中）为0.此时只有一个 x ，Bs构造函数最后调用。

When you comment out the line, during the construction phase, As constructor is called first, then Bs constructor, which sets x (in its base class) to 0. There is only one x at this point, and Bs constructor is called last.

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