Within a class, a field that has the same name as a field in the superclass hides the superclass's field.
public class Test {
public static void main(String[] args) {
Father father = new Son();
System.out.println(father.i); //why 1?
System.out.println(father.getI()); //2
System.out.println(father.j); //why 10?
System.out.println(father.getJ()); //why 10?
System.out.println();
Son son = new Son();
System.out.println(son.i); //2
System.out.println(son.getI()); //2
System.out.println(son.j); //20
System.out.println(son.getJ()); //why 10?
}
}
class Son extends Father {
int i = 2;
int j = 20;
@Override
public int getI() {
return i;
}
}
class Father {
int i = 1;
int j = 10;
public int getI() {
return i;
}
public int getJ() {
return j;
}
}
Can someone explain the results for me?
In java, fields are not polymorphic.
Father father = new Son();
System.out.println(father.i); //why 1? Ans : reference is of type father, so 1 (fields are not polymorphic)
System.out.println(father.getI()); //2 : overridden method called
System.out.println(father.j); //why 10? Ans : reference is of type father, so 2
System.out.println(father.getJ()); //why 10? there is not overridden getJ() method in Son class, so father.getJ() is called
System.out.println();
// same explaination as above for following
Son son = new Son();
System.out.println(son.i); //2
System.out.println(son.getI()); //2
System.out.println(son.j); //20
System.out.println(son.getJ()); //why 10?