CODEWITHSUNDEEP
java
xheyhenry
xheyhenry

Reputation: 1069

Declared types and Actual types

I understand that when creating a new object such as this:

GeomObject tri = new Triangle();

is more general and allows for more resuability, but what happens when tri is created like this:

Triangle tri = new Triangle();

Since Triangle is a subclass of GeomObject, isn't tri still a GeomObject? How does the declared type affect compilation? Thanks

*add: An additional question: say I have 

Integer n1 = new Integer(3);
Object n2 = new Integer(4); 
System.out.println(n1.compareTo(n2));

I tried this out on Eclipse and I got errors even if I reversed n1 with n2. I thought that n2.compareTo(n1) would work because it would call the Object compareTo method and since Integer is an instance of object, it would be passable, but this is not the case. Could you explain this?

Upvotes: 4

Views: 6993

Answers (1)

Luiggi Mendoza
Luiggi Mendoza

Reputation: 85779

Since Triangle is a subclass of GeomObject, isn't tri still a GeomObject?

Yes it is. Use the instanceof operator to test this:

System.out.println( (tri instanceof Triangl) ); //prints true
System.out.println( (tri instanceof GeomObject) ); //prints true
System.out.println( (tri instanceof Object) ); //prints true because every class extends from Object

How does the declared type affect compilation?

It won't affect in any matter, just will make your code inflexible in case you need to use another implementation of GeomObject that is not a Triangle.

More info:

I thought that n2.compareTo(n1) would work because it would call Object#compareTo method

This is incorrect since Object class doesn't have a compareTo method.

On the other hand, n1.compareTo(n2) won't work since you're passing an Object to the compareTo method when Integer#compareTo receives another Integer class type.

Note that when declaring this:

Object n2 = new Integer(4);
  • The variable type will be Object, no matter if you initialize it as Integer or another class e.g. String.
  • Only the overridden methods will behave as defined in the object reference run time type, this means if your n2 variable holds an Integer, only the methods overridden in class Integer from class Object will behave as defined in the Integer class, all the other methods, fields, even the variable itself will behave as Object. In case of Integer class, these methods are equals, hashCode and toString.
  • The link provided above: What does it mean to "program to an interface"? explains the benefits of using an interface (or abstract class or a generic class) to generalize the work through generic interfaces/classes instead of direct implementation. Note that there are cases where this approach won't apply e.g. your current example using Object when you should use Integer. Note that Object class is way too generic (at least for this case), so I won't recommend using Object directly at least that you understand what you're really doing.

Upvotes: 11

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