Java method inheritance problem when overloading the method

Question

Here is my simple code to check Java inheritance & Method overloading. It gives compile error in my IDE. Could you please give an idea about this.? Error line commented in the code. If I comment that line program works fine and provide the given output.

class Bird {
    void sing() {
        System.out.println("I am Singing");
    }
}
class Peacock extends Bird {
    void sing() {
        System.out.println("I am Singing COO COO");
    }

    public void sing(String adverb) {
        System.out.println("I am Singing " + adverb);
    }
}
public class OverLoadingDemo {

    public static void main(String[] args) {
        Bird bird = new Peacock();
        bird.sing();//This return I am Singing COO COO
        bird.sing("Loudly");//ERROR The method sing() in the type Bird is not applicable for the arguments (String)
    }
}

If you want to call the sing(String) method, you need to call it on a type of variable which has that method. Bird does not. Even though at runtime your variable holds a reference to an object which does have that method, it's the type at compile-time that counts when resolving the method. — kaya3
– kaya3, Commented Dec 18, 2019 at 2:16

Andrew Merrill · Accepted Answer · 2019-12-18 02:16:26Z

5

The Java compiler sees that the type of the bird variable is Bird, and doesn't know that you are planning to store a reference to a Peacock in that variable. Since the Bird class does not have a method called sing that takes a String as an argument, it gives you an error.

If the Bird class had a sing method that took a String as an argument, then this code would compile. Even better, at runtime, it would use the version of the sing method (with a String argument) that was defined in the Peacock class since the variable bird would actually hold a Peacock object at that point.

answered Dec 18, 2019 at 2:16

Andrew Merrill

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WJS · Accepted Answer · 2019-12-18 02:31:33Z

2

When you assign an object to the type of it's ancestors or interfaces, only the methods of the ancestor hierarchy or interface are visible to the object.

In your case a Peacock is a Bird so it knows about Peacock things and Bird things. But a Bird object (even if assigned from a Peacock object) only knows about Bird things that are common to all Birds. Hence, it won't know (and can't tell) that it is a Peacock.

This is the is-a relationship. A Peacock is-a Bird in all cases. But a Bird is not a Peacock in all cases.

answered Dec 18, 2019 at 2:31

WJS

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