Why instance of base class chooses to execute base method of parent constructor?

Question

Here is the code:

class Food {
    Food() { printFlavor(); }
    void printFlavor() { System.out.println("bland"); }
}
class Pepper extends Food {
    void printFlavor() { System.out.println("spicy"); }
}
public class Lunch {
    public static void main(String[] args) {
        Food lunch = new Pepper();
    }
}

When I ensure to run its own method of its own class with this keyword, it still runs base method.

class Food {
    Food() { this.printFlavor(); }
    void printFlavor() { System.out.println("bland"); }
}
class Pepper extends Food {
    void printFlavor() { System.out.println("spicy"); }
}
public class Lunch {
    public static void main(String[] args) {
        Food lunch = new Pepper();
    }
}

My question is not how to resolve this issue. My question is why parent constructor chooses to call base method instead of parent method?

Does this answer your question? Calling base class overridden function from base class method — Amongalen
– Amongalen, Commented Jul 17, 2020 at 6:57

Shubham Kumar · Accepted Answer · 2020-07-17 07:48:05Z

0

The reason is whenever you create an object of child class ,the constructor of parent class gets called first(by default) followed by the child class constructor.

To run the parent class method , you need to create parent class object Food obj = new Food. The object type that you are creating is not of Food Class , it is of Pepper Class.

edited Jul 17, 2020 at 7:48

answered Jul 17, 2020 at 6:56

Shubham Kumar

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2 Comments

Nay Sie Over a year ago

So, for method, base methods are called first and then parent methods????

Shubham Kumar Over a year ago

No just for constructor , the base class constructor is called first then the child class cons is called

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