This question already has answers here:
Why bark method can not be called
(5 answers)
Closed 6 years ago.
The Superclass reference variable can hold the subclass object, but using that variable you can access only the members of the superclass, so to access the members of both classes it is recommended to always create reference variable to the subclass.
class Animal {
public void move() {
System.out.println("Animals can move");
}
}
class Dog extends Animal {
public void move() {
System.out.println("Dogs can walk and run");
}
public void bark() {
System.out.println("Dogs can bark");
}
}
public class TestDog {
public static void main(String args[]) {
Animal a = new Animal(); // Animal reference and object
Animal b = new Dog(); // Animal reference but Dog object
a.move(); // runs the method in Animal class
b.move(); // runs the method in Dog class
b.bark();
}
}
output:
TestDog.java:26: error: cannot find symbol
b.bark();
^
symbol: method bark()
location: variable b of type Animal
1 error
What I do not understand here is why is the object 'b' able to access the Dog.move() and not Dog.bark() because the statement mentioned above says it can access only the members of the superclass and not the subclass.Following this logic the output of b.move() should be "Animals can move" and not "Dogs can walk and run".But that is not case.Can anyone help me with this?Thanks in advance!
Congratulations - you just discovered polymorphism.
In Java the classes are bound dynamically. That is if you are invoking a method the implementation of the object is invoked (in your case the Dog) and not the method of the reference type (in your case the Animal).
This allows overwriting methods and replace or fulfill their implementation.
On the other hand, you can only access methods that are available in the type you are referencing, not the implementing type (in your case the Animal). To invoke the methods of the instance, you would have to use it as the reference type (in your case the Dog).
In your question Animal is a parent class which doesn't have bark() method so that method isn't overridden. If you were able to access bark() from parent class without declaring either abstract method or defining it, then that would be violation of the Polymorphism principle.
If you really want to access it that way, then you can either define a abstract public void bark(); in your parent or access that method by typecasting like this
((Dog) b).bark();
This will not compile since Animal does not have a method called bark.
Think of it this way, all dogs are animals, but not all animals are dogs. All dogs bark, but not all animals bark.
This code is wrong, as the line b.bark(); will give you a compiler error, because b is only defined as an Animal, which cannot bark().
If you change Animal b = new Dog(); to Dog d = new Dog(); it will work properly.
You've got inheritance mixed up. Dog can do what Animal can do, not vice versa.
class Animal {
public void move() {
System.out.println("Animals can move");
}
}
class Dog extends Animal {
#Override public void move() {
System.out.println("Dogs can walk and run");
}
public void bark() {
System.out.println("Dogs can bark");
}
public void moveSuper() {
super.move();
}
}
public class TestDog {
public static void main(final String args[]) {
final Animal a = new Animal(); // Animal reference and object
a.move(); // runs the method in Animal class
final Dog d = new Dog(); // Animal reference but Dog object
d.move(); // runs the method in Dog class
d.bark();
d.moveSuper();
}
}
Related
This question already has an answer here:
Overloading method invoke issue
(1 answer)
Closed 1 year ago.
class Animal{
void eat(Animal animal){
System.out.println("animal eats animal");
}
}
public class Dog extends Animal{
void eat(Dog dog){
System.out.println("dog eats dog");
}
public static void main(String[] args) {
Animal a = new Dog();
Dog b = new Dog();
a.eat(b);
b.eat(b);
}
}
In the above code, the output will be
animal eats animal
dog eats dog
Why this happened?
Probably you expect to see twice "dog eats dog". This does not happen because the two methods have a different signature. Therefore, Dog#eat(Dog) does not override Animal#eat(Animal) but provides a more specific eat method instead.
If you add #Override to void eat(Dog dog) there will be an error. Using this annotation is good practice because it denotes that the annotated method should override a method declaration in a supertype. If the method does not do that (as in your example) you get the following error to make you aware if it:
Method does not override method from its superclass
If you want to override the eat method in Dog, you need to provide the same signature:
#Override
void eat(Animal animal) { // instead of eat(Dog dog)
System.out.println("dog eats dog");
}
It is simply because Java does not support contravariant parameters. On the other hand, it supports covariant return types.
Due to the support for covariant return types, a subclass override can have a more specific return type in the hierarchy while overriding the base class method, like the code below is valid:
class Animal {
protected Animal getAnimal() {
System.out.println("Animal");
return this;
}
}
class Dog extends Animal {
#Override
protected Dog getAnimal() {
System.out.println("Dog");
return this;
}
}
In the above example, you can observe that Dog.getAnimal() returns a more specific Dog instead of the base Animal but it is still considered an override because Java supports covariant return types.
On the other hand, if you do that with parameters:
class Animal {
protected void petAnimal(Animal animal) {
System.out.println("Petting Animal");
}
}
class Dog extends Animal {
#Override
protected void petAnimal(Dog dog) {
System.out.println("Petting Dog");
}
}
This is not an override but an overload.
Hence both the petAnimal() methods (one with Animal as parameter and another with Dog as parameter) are treated as two different methods. Remember, parameters are part of the method signature whereas return types are not.
The second example does not even work as the #Override annotation finds out that the method is not an override. Whenever you want to ensure that you override, use the #Override annotation, it will let you know if you are not overriding the method. #Override can also be used when implementing interfaces.
It's based on the concept of Inheritance and Polymorphism
Overriding happens when the sub-class has the same signature methods as that of the superclass. In your code, in the below subclass *method, the parameter being passed is a Dog type object and in the superclass i.e Animal, the parameter passed is an Animal type object.
void eat(Dog dog){
System.out.println("dog eats dog");
}
So you can change the above method as below to see the overriding effect: -
void eat(Animal dog){
System.out.println("dog eats dog");
}
As suggested by #Mat, it's best to use #Override annotation because it will help the java compiler to find the issue at the compile time itself.
Below I'm trying to explain the concept of how inheritance and polymorphism are working once you change the signature of the eat method in Dog class: -
Inheritance is a way to base one class on another class, like a template built from an existing template. You could create a class called 'Dog' that acts as a template for all Dog objects. We could then create another class called 'Animal' that is a parent class of our 'Dog' class. All Dogs are animals, but not all animals are dogs. Our Animal class could define functionality for all Animals and then the Dog class could take all this functionality, without re-writing it, by extending/inheriting from the Animal class. The Dog class could then add more functionality, more variables, and methods, that are specific only to Dog objects.
The Dog class extends the Animal class, this is inheritance. The Dog class is overwriting the Animal class eat method.
When we say Animal a = new Dog();, we declare a variable a that is declared as an Animal type, but it initialized as a Dog object. This is polymorphism. Because the Dog extends from the Animal class, we can treat it as an Animal, and declare it as an Animal variable type. We cannot do the reverse, because the Animal class does not extend from the Dog class (not all Animals are dogs)
This is because the a variable is being treated as an Animal data type. This is why you will be able to access all the Animal class methods but the methods with the same signature as Dog will be overridden by the implementation from the Dog class. Remember, the left side of the equals '=' symbol is the declaration and the right side is the initialization.
Also, that's why when you declare Dog b = new Dog(); and call the eat() method it calls the implementation from the Class Dog and not Animal as it's explicitly mentioned as object type Dog
class Animal{
void eat(Animal animal){
System.out.println("animal eats animal");
}
}
public class Dog extends Animal{
void eat(Dog dog){
System.out.println("dog eats dog");
}
public static void main(String[] args) {
Animal a = new Dog(); //We use this when we don't know the exact runtime type of an object
//Parent can hold any child but only parent specific methods will be called.
Dog b = new Dog();
a.eat(b); //Parent method will be called i.e Animal.eat(...)
b.eat(b); //Dog Class method will be called i.e Dog.eat(...)
}
}
First of all, the eat() not being overridden in the child class Dog
Usually, method overloading doesn't necessarily need inheritance and can be achieved within the same class. However, in this code the eat() method is overloaded by child class Dog.
Overloaded methods are differentiated by the number and the type of the arguments passed into the method.
So, at compile time it always picks the most specific class implementation based on its type.
Example program:
class Animal {
public void eat() {
System.out.println(" Animal eats");
}
class Dog extends Animal {
public void eat(String s) {
System.out.println(" Dog eats" + s);
}
public class Demo {
public static void main(String args[]){
Animal a = new Dog();
Dog d = (Dog) a;
a.eat();
d.eat("Meat");
}
}
My question is why a.eat("Meat") is not reachable? Since a is Object during the time it's referring to Dog Object, it should allow to call eat("meat").
Can anyone clarify where am I going wrong?
You're trying to call method of subclass Dog through variable of base class Animal.
Java has strong typing, so if you declare variable of class Animal you can access only to methods and fields of Animal and its superclasses.
You can call eat("Meat") from Animal using casting ((Dog) a).eat("Meat") but you should avoid as much as possible such constructions.
Btw, your approach looks like a function overloading rather than polymorphism.
P.S. Maybe this article about strong typing will be helpful for you. And this one about difference between overloading and polymorphism.
Dog d = (Dog) a;
Here you are casting the Animal a to a Dog. However, this does not change the type of a. a is still an Animal hence you can only call eat() without a string.
why do I have to write like this
animal a = new cat();
cat b = (cat)a;
b.makesound();
a.makesound();
but not like this
animal a = new cat();
animal b = new cat();
a.makesound();
b.makesound();
they give the same outputs
This is the entire code.
public class DownCasting
{
public static void main(String args[])
{
animal a = new cat();
cat b = (cat)a;
b.makesound();
a.makesound();
}
}
class animal
{
public void makesound()
{
System.out.println('w');
}
}
class cat extends animal
{
public void makesound()
{
System.out.println("meow");
}
}
so downcasting just give an object second name? or it has other uses?
In the second variant, you create two cats.
In the first, you assign the additional name b to the already existing animal a.
so downcasting just give an object second name? or it has other uses?
You might want to have a list of animals, f.ex. a dog:
class dog extends animal
{
public void makesound()
{
System.out.println("woof");
}
}
could be combined in the same list with cats.
Casting is the process of forcefully making a variable behave as a variable of another type. If a class shares an IS-A or inheritance relationship with another class or interface, their variables can be cast to each other’s type.
public class DownCasting
{
public static void main(String args[])
{
animal a = new cat();
cat b = (cat)a;
b.makesound();
a.makesound();
}
}
class animal
{
public void makesound()
{
System.out.println('w');
}
}
class cat extends animal
{
public void makesound()
{
System.out.println("meow");
}
}
In the line animal a = new cat(); you are using a variable of type Animal to refer an object of type Cat, and a.makesound(); will print "meow" because with inheritance, the instance methods bind at runtime.
However, if you try to do cat b = a; (without casting) the Java Compiler will try to use a variable of type Cat to reference an object of type Animal, and that is not possible (you cannot use a variable of a sub-class to reference (execute) methods from a super-class), so it is necessary cast the variable to indicate to the java compiler that it can be sure that despite a is a variable of type Animal, is actually a Cat.
Let me give you an example to make this more clear:
I have your two classes with some modifications:
class animal
{
public void makesound()
{
System.out.println('w');
}
void sleep(){
System.out.println("ZZZ");
}
}
class cat extends animal
{
public void makesound()
{
System.out.println("meow");
}
void play(){
System.out.println("I´m playing");
}
}
both classes now have a method that the other class does not have.
Now let's use it:
public class DownCasting
{
public static void main(String args[])
{
animal a = new cat();
}
}
If I do a.play(); it will give me a compilation error because Java searchs for the method play() inside the Animal Class and cannot find it... Why java does that? as I already said: "With inheritance, the instance methods bind at runtime", so if I want to call that method I have to cast the a variable... ((Cat)a).play();
But what happen if you try to do it the other way around?
Cat c = new Animal(); //This line will never compile...
So, I can cast it like this in order to make it compile:
Cat c = (Cat)new Animal();
c.play();
BUT, in runtime it will throw a java.lang.ClassCastException why?, because you cannot use a variable of a sub-class to reference (execute) methods from a super-class.
To sum up, remember that the type of the object reference variable and the type of the object being referred to may be different.
But there are rules on how different these can be.
To this topic I will recommend you OCA Java SE 7 Programmer I Certification Guide: Prepare for the 1ZO-803 exam specifically the chapter 6.
why do I have to write like this
to answer this question you need to know first what can make a sound?
the cat or the animal?
you do this :
animal a = new cat();
a.makesound();
when the method makesound is declared in animal
if not then you need to cast
Using this as an example:
class Animal{
public void move(){
System.out.println("Animals can move");
}
}
class Dog extends Animal{
public void move(){
System.out.println("Dogs can walk and run");
}
}
public class TestDog{
public static void main(String args[]){
Animal a = new Animal(); // Animal reference and object
Animal b = new Dog(); // Animal reference but Dog object
a.move();// runs the method in Animal class
b.move();//Runs the method in Dog class
}
}
Is it right here to say upon calling b.move() the method "move() under the Dog class" has overridden the "move() under the Animal class" as the Dog object takes precedence when calling the same method when referenced by an Animal type?
Ive noticed many websites don't explain this, rather they just throw out examples without talking through the stuff line by line. Just want to clear up my terminology confusion.
A side note, would it be possible to have a Dog object but invoke move() which is under the Animal class? for example having something like:
Dog doggy = new Dog();
doggy.move()
>>>
Animals can move
>>>
Would this be possible? would ((Animal) doggy).move() accomplish this?
Absolutely, it is correct to say upon calling b.move() the method "move() under the Dog class" has overridden the "move() under the Animal class".
For the second question, you should implement the class Dog as:
public class Dog extends Animal {
public void move(){
super.move();
}
}
For the third question, the answer is "NO".
((Animal) doggy).move()
This is simply 'redundant' and give the output "move() under the Dog class".
You can do it like this
class Animal{
public void move(){
System.out.println("Animals can move");
}
}
class Dog extends Animal{
public void move(){
System.out.println("Dogs can walk and run");
}
public void moveParent() {
super.move();
}
}
public class Main{
public static void main(String args[]){
Animal a = new Animal(); // Animal reference and object
Animal b = new Dog(); // Animal reference but Dog object
a.move();// runs the method in Animal class
b.move();//Runs the method in Dog class
Dog doggy = new Dog();
doggy.moveParent();
}
}
Use super keyword to call upon the parents member functions or data members.
like: super.move(); in this case you parent function will be called.
From the oracle docs
If two or more independently defined default methods conflict, or a default method conflicts with an abstract method, then the Java compiler produces a compiler error. You must explicitly override the supertype methods.
So basically, if you're in the sub-class calling a method which is in the superclass, you won't be able to call the super-class' method unless you use super.function(). Read up more on it here
It's principle object oriented programming (aka OOP) - polymorphism. Dog, Cat, Elephant are all animal.
Animal d = new Dog();
Animal c = new Cat();
Animal t = new Tiger()
It must not care about, always right. :)
I have some question about upcast/downcast.
I created an abstract super class Animal, subclass Dog and subclass BigDog. and I also give abstract method in Animal, and override it in Dog and BigDog.
abstract public class Animal {
abstract public void greeting();
}
public class Dog extends Animal {
#Override
public void greeting() {
System.out.println("Woof!");
}
}
public class BigDog extends Dog {
#Override
public void greeting() {
System.out.println("Woow!");
}
}
now my test code:
public class TestAnimal {
public static void main(String[] args) {
Animal animal2 = new Dog();
Animal animal3 = new BigDog();
// Downcast
Dog dog2 = (Dog) animal2; //cast Animal class to Dog class, legit
BigDog bigDog2 = (BigDog) animal3; //cast Animal to BigDog, legit;
Dog dog3 = (Dog) animal3; //Animal Class contains BigDog cast into Dog?
dog2.greeting();
dog3.greeting(); //in which class the method is called?
}
}
I understand the relationship between superclass/subclass and how cast works. My question is, however, can you cast a superclass into a specific subclass, knowing there's a class in between? for example, if I have an Animal class object contains a BigDog object, can I cast the object to Dog? what if there are methods in BigDog that do not exist in Dog?
in short, you can certainly say a superclass object is a subclass object, but why can you invert?
On second thought,
I'm guessing this: I'm asking JVM cast an Animal class reference to Dog and link the new Dog reference to the BigDog object, rather than really casting the BigDog object.
So I can invoke all Dog and Animal methods on that Dog reference (to BigDog), but none of the BigDog methods, unless it was overridden in BigDog.
What Java checks when invoking a method is: if the reference (DOG) has the reference, and if the object(BigDog) has an override. if not, Dog method is called, otherwise, BigDog method is called.
Can anyone confirm my guess?
You can always cast to a specific subclass, unless the compiler is smart enough to know for certain that your cast is impossible.
The best way to cast to a subclass is to check if it can be done:
if ( doggy instanceof BigDog ) {
doSomethingWithBigdog( (BigDog) doggy );
} else if ( doggy instanceof SmallDog ) {
doSomethingWithSmalldog( (SmallDog) doggy );
} else {
// Neither a big dog nor a small dog
}
...
private void doSomethingWithBigdog( BigDog dog ) {
...
}
private void doSomethingWithSmalldog( SmallDog dog ) {
...
}
Keep in mind that casting is evil. Sometimes necessary, but often (not always) it can be designed away by implementing methods on the base class, or by not assigning a Dog to an Animal variable but to keep it a Dog.
If I have an Animal class object contains a BigDog object, can I cast the object to Dog? what if there are methods in BigDog that do not exist in Dog?.
Simply you will get compiler error.Since you can't call a method that is not declared in parent and declared in child class using parent reference
There is no method whose signature will match with these method calls :
dog2.greeting(dog3);
dog3.greeting(dog2);
so, Its pretty much a compilation failure.
You need to know about Dynamic Method Dispatch.
here are few links 1,2,3 go through them.
First correct the source code, so it will compile. The proper usage of the methods: dog2.greeting(); and dog3.greeting(); or add method public void greeting(Animal animal);.
dog3.greeting(); - invoking method greeting() for dog3. dog3 has the same reference as animal3. animal3 has reference of BigDog so method greeting() is invoked to the class BigDog and the output is Woow!
When you inherit Dog from class Animal, then class Dog have all methods from class Animal.