how to understand super in java

Question

I encountered a problem that confused me, it is the keyword 'super', my test code is like this:

package test;

public class Parent {
   private String name;

   public Parent(){
        this.name = "parent";      
   }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public void showName(){
        System.out.println(this.name);
    }

}


public class Child extends Parent{

    public Child(){
        this.setName("Child");
    }

    public void showName(){
        System.out.println(super.getClass().toString());
        System.out.println(super.toString());

        super.showName();
        System.out.println(super.getName());
    }
}

public class Test {

    public static void main(String[] args) {
        Child d = new Child();
        d.showName();    
    }
}

so the result is like this:

class test.Child
test.Child@2207d8bb
Child
Child

my understanding about 'super' is that it is a reference to the parent instance of current instance, so my expecting output is like 'Parent', from the result , I am wrong, its like the current instance calls the parent method, and 'super' is not parent instance, is my understanding right ? and is there a way that I can get parent instance only initializing the Child class ?

score 8 · Accepted Answer · answered Aug 29 '13 at 07:38

my understanding about 'super' is that it is a reference to the parent instance of current instance

No - there's no such thing as a "parent instance". When you create an instance, there's only one object created - an instance of Child, which inherits all the fields and methods from Parent as well.

super is used in two ways:

To refer to the superclass implementation of a method, typically when you've overridden the method in the subclass. This is what you're doing in Child.showName - it's calling Parent.showName, but on the same instance (because there is only one instance)
To call a superclass constructor from a subclass constructor

sara · Answer 2 · 2013-08-29T07:45:43.517

2

From the javadocs, getClass returns the runtime class of an object.

The runtime class of your object is "Child".

As you did not override getClass() (you can't because it's final), so super.getClass() acts exactly like getClass(). getClass method of the Object class is called.

If you want to print the Parent, call getClass().getSuperclass()

edited Aug 29 '13 at 07:45

answered Aug 29 '13 at 07:40

sara

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score 0 · Answer 3 · answered Sep 02 '13 at 07:24

You have used four print statements invoked using Child object:

System.out.println(super.getClass().toString()); Since the getClass() is final method thus it cant be overridden, so when you invoke super.getClass(), you actually invoke the Object class's getClass() which returns the name of the class of the actual invoking object.
System.out.println(super.toString()); Here again the toString() method of Object class is invoked which returns "getClass().getName() + "@" + Integer.toHexString(hashCode());" However, if you override this method in your parent class you will definitely get that version executed
super.showName(); this method returns "Child".... here's why: When you create an object of Child class the contructor runs as follows-

public Child(){ super(); // this is implicitly called by the compiler even if you did not write this this.setName("Child"); }

So following action take place - 1. Parent class's contructor is called which sets the 'name' to 'parent'. 2. Child class's contructor now overwrites this valus with 'Child'. So the value of instance variable 'name' is 'Child'.
System.out.println(super.getName()); Here as said in point 3 the there is only one object and one instance variable 'name' whose value is 'Child'.

score -1 · Answer 4 · answered May 20 '14 at 07:56

class Animal {
  void eat() {
    System.out.println("animal : eat");
  }
}

class Dog extends Animal {
  void eat() {
    System.out.println("dog : eat");
  }
  void anotherEat() {
    super.eat();
  }
}

public class Test {
  public static void main(String[] args) {
    Animal a = new Animal();
    a.eat();
    Dog d = new Dog();
    d.eat();
    d.anotherEat();
  }

}

The output is going to be

animal : eat dog : eat animal : eat

The third line is printing "animal:eat" because we are calling super.eat(). If we called this.eat(), it would have printed as "dog:eat".

Doesn't address the actual question. – user207421 May 20 '14 at 08:11 — user207421, May 20 '14 at 08:11

how to understand super in java

4 Answers4

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