JavaCzyHerbata.pl

Tomku, nie ma tam błędu. Spójrz na poniższy przykład.

class Animal { }

class Cat extends Animal { }

class Dog extends Animal { }

public class AnimalTester {

	void say(Animal a) {
		System.out.println("Say like Animal");
	}

	void say(Cat c) {
		System.out.println("Say like Cat");
	}

	void say(Dog d) {
		System.out.println("Say like Dog");
	}

	void say(Object o) {
		System.out.println("Say like Object");
	}


	public static void main(String[] args) {
		AnimalTester animalTester = new AnimalTester();

		// different reference types
		Animal animalReferenceOfAnimal = new Animal();
		Cat catReferenceOfCat = new Cat();
		Dog dogReferenceOfDog = new Dog();

		animalTester.say(animalReferenceOfAnimal);
		animalTester.say(catReferenceOfCat);
		animalTester.say(dogReferenceOfDog);

		// same reference type
		Animal animalReferenceOfCat = new Cat();
		Animal animalReferenceOfDog = new Dog();

		animalTester.say(animalReferenceOfCat);
		animalTester.say(animalReferenceOfDog);

		// same (Object) reference type
		Object objectReferenceOfAnimal = new Animal();
		Object objectReferenceOfCat = new Cat();
		Object objectReferenceOfDog = new Dog();

		animalTester.say(objectReferenceOfAnimal);
		animalTester.say(objectReferenceOfCat);
		animalTester.say(objectReferenceOfDog);
	}
}

Po uruchomieniu programu powinno pojawić się coś takiego:

Say like Animal
Say like Cat
Say like Dog
Say like Animal
Say like Animal
Say like Object
Say like Object
Say like Object

Jak widzisz, typ referencji (po lewej stronie zmiennej) decyduje, która przeciążona metoda zostanie wywołana.