Question

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

inh2.cpp

#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;
}

results from gcc

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

I have two questions:

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.

---

My reasoning for Q #1 above:

• 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.

• Variable x within this object stores value 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.

Where am I wrong here?

Answer 1

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).

2. 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.