Inheritance in c++ base constructor called

Asked Dec 12 '16 at 02:31

Active Dec 12 '16 at 02:31

Viewed 16 times

Why is my base class constructor called this this instance?

#include <iostream>
using namespace std;

class A {
public:
    A() {
        cout << "A" << endl;
    }
};

class B : public A {
public:
    B() {
        cout << "B" << endl;
    }
};

int main() {
    B b;
    getchar();
    return 0;
}

The output is

Why is A called before B is? And why is A even called in the first place?

asked Dec 12 '16 at 02:31

Yusuf Jama

2

Because that's how constructors and classes work, in C++. – Sam Varshavchik Dec 12 '16 at 02:33
@SamVarshavchik so does it always call the base constructor and work it's way up each derived step? – Yusuf Jama Dec 12 '16 at 02:34
See the linked question for the explanation. – Sam Varshavchik Dec 12 '16 at 02:34

Inheritance in c++ base constructor called

0 Answers0