Why is the derived class constructor is skipped?

Asked Apr 16 '20 at 19:50

Active Apr 16 '20 at 19:54

Viewed 74 times

I have a code like this

class Base {
public:
    Base() {};
};

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

int main() {
    Derived a(); // Only called base constructor
    Derived b;   // called derived constructor
}

What is the difference between these two calls? My understanding is that base constructor is always called before derived constructor? Why is the derived constructor skipped?

Thanks,

asked Apr 16 '20 at 19:50

Mira

4

"Only called base constructor" No it didn't. This is a "vexing parse". – Ben Voigt Apr 16 '20 at 19:51
1

Notice that `Derived a()` [does not call any constructor](https://godbolt.org/z/pSdZTv). – François Andrieux Apr 16 '20 at 19:52
1

@Gupta If you don't specify which base type constructor to use, it defaults to using the default constructor. Adding `Base()` has no effect, it's already implied. – François Andrieux Apr 16 '20 at 19:53

Why is the derived class constructor is skipped?

0 Answers0