What happen when you call an override method using super and this keyword?

Question

public class Superclass {



    void method0(){
        System.out.println("superclass");
    }
}


public class Subclass extends Superclass{

    void method0(){
        System.out.println("subclass");
    }

    void method1(){
        super.method0();
    }

    void method2(){
        this.method0();
    }
}


public class RunClass {

    public static void main(String[] args){
        new Subclass().method1();
        new Subclass().method2();
    }
}

the code above print out

superclass
superclass

while I expect it to print out

superclass
subclass

Isn't this.method0() refer to the method0 in subclass and print out subclass instead of superclass ?

Answer 1

super represents the instance of parent class. this represents the instance of current class. It will print
superclass
subclass

Answer 2

I ran your code and it gaves me

superclass
subclass

this what should printed every thing seems ok

Answer 3

new Subclass().method1();

executes the method1() of Subclass instance, which in turn calls super.method0(); ie parent class instance's method0() ie Superclass instance method0() .

new Subclass().method2();

executes the method2() of Subclass instance, which in turn calls this.method0(); ie this instance's method0() ie Subclass instance method0() .

super is used to access instance members of the parent class while this is used to access members of the current class.

Answer 4

First of all, it prints out the what you are expecting.

Second,

Isn't this.method0() refer to the method0 in subclass and print out subclass instead of superclass?

this => refer to current object, using it you can use it (kinda of pointer to itself, in general terms )

super => refer to super class object in an hierarchy, usually used to access the hidden members in subclass