Question

I’ve known that if B is derived from A, then when I create a new object of B, for example b, the constructor of A will call first. When I destroy the object b, The destructor of B will call first. Then I have a question here, if there’re more than one constructor in the Base class, which constructor will call ? and why?

I’ve write one test program below, I guess it will call the default constructor in the Base class, But I’m not sure if it is just a coincidence?

#include <iostream>
using namespace std;

class A{
    public:
    A(int i){cout<<"A con with param"<<endl;}
    A(){cout<<"A con"<<endl;}
    ~A(){}
};

class B : public A
{
    public:
    B(int i){cout<<"B con with param"<<endl;}
    B(){cout<<"B con"<<endl;}
    ~B(){}
};

int main()
{
    B b(5);
    return 0;
}

I wonder if any boss can tell me the reason or any advise to figure out this problem?

Answer 1

If you write:

B(int i) { cout<<"B con with param"<<endl; }

then constructor A() (without arguments) will be called.

If you write:

B(int i): A(i) { cout<<"B con with param"<<endl; }

then constructor A(int) will be called.