CODEWITHSUNDEEP
javaexception
Sam
Sam

Reputation: 841

Why there is Unhandled Exception in this code?

I came accross the following code:

class Animal{
    public void eat() throws Exception {}
}

class Dog extends Animal{
    public void eat(){} //no exception thrown
    public static void main(String[] args){
          Animal a = new Dog();
          Dog d = new Dog();
          d.eat();   //ok
          a.eat();  //does not compile!..(1)
    }
}

Here, (1) does not compile even though at runtime Dog's eat() method will be called. Why does this happen? What is the reason that Java supports this? Shouldn't this be filed as a bug?

Upvotes: 3

Views: 129

Answers (3)

Prashant
Prashant

Reputation: 2614

First thing that you are not following the rules of Method Overriding. if Base class method is throwing any Exception then child class method must throw equal or Low Level Exception. now this code will work fine because its following the method overriding rules. Also @Elliot is saying right thing, compiler doesn't aware of dog object(as Animal is referring) at compile time. it will resolve only at Run Time.

class Animal{
public void eat() throws Exception {}
}

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

Upvotes: 3

hqt
hqt

Reputation: 30276

Animal a = new Dog();

in OOP (Object Oriented Programming), this is called polymorphism. And in Java (and mostly in OOP support language such as C#), this behavior is compile checking. That means compiler at compile time just knows a is a animal, and cannot know a is a dog until runtime.

For example:

Animal a = new Dog();
a.bark(); // cannot. because compiler doesn't know a is a dog until runtime
((Dog)a).bark(); // yes. we say to compiler: a **must** be a dog

Another example is:

    Animal a = new Cat();
    // yes. can compile. we force compiler to know this animal is a dog. 
    //but will run-time exception because in fact this is a cat
    ((Dog)a).bark();

Hope this help :)

Upvotes: 3

Elliott Frisch
Elliott Frisch

Reputation: 201447

Because you're using an Animal reference to refer to a Dog. And the signature of Animal.eat includes the Exception. The compiler knows that a Dog is a kind of Animal, but once you use an Animal reference it doesn't know that it's a Dog until runtime.

To put it another way, all Dog(s) are Animal(s) but not all Animal(s) are Dog(s).

Edit

You could have added a cast

((Dog) a).eat();  //would compile

At runtime, that will fail if a isn't in fact a Dog.

Upvotes: 7

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