Superclass variable referencing subclass objects

[S Kak]

Below is a code I wrote which did not run for obvious reasons. As I understand the process flow:

When I declared yourname below, Java created a reference variable. This reference variable is of type Person. Then I created an object of class Player and referenced yourname to this newly created object. At this point Java created space in memory for the object which has three fields (name, age and score). However, when I tried adding information to score, Java threw error "cannot find symbol". From a code point of view, I get it. yourname is of type Person which does not have score as a field so at compile time Java checks for score in class Person and does not find it and hence throws an error. So my question here is:

As yourname refers to an object which has all three fields, why does Java not allow adding information to score? Should the error not be thrown at Person yourname = new Player() line hence not allowing the programmer to create an object having fields more than the reference type in the first place?

Also when I print out System.out.println(yourname) alone, the output is Player@HashCode, which implies that yourname belongs to the class Player which has three fields.

class Person {
  String name;
  int age;
}

  class Player extends Person{
    int score;
  }

class Test{
  public static void main (String args[]){
  Person yourname = new Player();//should the error not been thrown here
  yourname.score=55; //cannot find symbol error
  System.out.println(yourname.score); //cannot find symbol error
  System.out.println(yourname);
  }
}

[Henry Wong]

S Kak wrote:
As yourname refers to an object which has all three fields, why does Java not allow adding information to score? Should the error not be thrown at Person yourname = new Player() line hence not allowing the programmer to create an object having fields more than the reference type in the first place?

The compiler/JVM does allow you to access the score field, you just can't do it with a Person reference. Just cast the reference to a Player reference, or just cast and use it directly.

Henry

[Campbell Ritchie]

Welcome to the Ranch

You should make the fields in both your classes private, and access them via methods. But, in the current code, you declare the variable as type Person. Now some Person objects have a score field and some don't. If you declare it as Person, you can do this:-

class Test{
  public static void main (String args[]){
  Person yourname = new Player();//should the error not been thrown here
  yourname.score=55; //cannot find symbol error
  System.out.println(yourname.score); //cannot find symbol error
  System.out.println(yourname);
  }
}

... or this:--

class Test{
  public static void main (String args[]){
  Person yourname = new Player();//should the error not been thrown here
  yourname = new Person();
  yourname.score=55; //cannot find symbol error
  System.out.println(yourname.score); //cannot find symbol error
  System.out.println(yourname);
  }
}

Now, when you get to line 5, you have an object which does not have a score field. What are you going to do? Are you going to allow that assignment because some Person objects do have a score field? And what will happen at runtime, an Exception or something?
You are notnow in the realms of “behaviour undefined”, and that is dangerous. How can you be sure your program will behave correctly? How can you even say what correct behaviour is?
The correct answer is that the program cannot run. And the best way to ensure that a program does not run is not to allow it to compile in the first place. A Player is a Person, so you would expect the assignment to a Player (line 3) to run correctly.

So, you would expect the compiler error when you try to access the score field.

[edit]Correct not to now.

[Knute Snortum]

Welcome to the Ranch!

Person yourname = new Player();//should the error not been thrown here

If the comment means that maybe the compile should throw an error at that point, then I would disagree.  I write this all the time:

List<String> myList = new ArrayList<>();

[Carey Brown]

By using constructors that also initialize the fields, and a generic toString() method, you can achieve your goals without needing a cast. If this doesn't work for you you may need an "instanceof" operator and a cast.

public class PersonTest
{
	public static void main( String args[] )
	{
		Person2 yourname = new Player( "Jane", 31, 55 );
		System.out.println( yourname );
	}
}

class Person2
{
	String name;
	int age;

	public Person2( String name, int age )
	{
		this.name = name;
		this.age = age;
	}

	public String toString()
	{
		return "name:" + name + " age:" + age;
	}
}

class Player extends Person2
{
	int score;

	public Player( String name, int age, int score )
	{
		super( name, age );
		this.score = score;
	}

	public String toString()
	{
		return super.toString() + " score:" + score;
	}
}

[Ganish Patil]

Carey Brown wrote:

System.out.println( yourname );

I kept pondering on that how is this printing correct output  instead of fullyqualifiedclassname@hashcode then got detail information in source code, It was really awesome. I never knew we could print in this manner also.