Let's take this simple Java code:
public class Animal {
public void eat() {
System.out.println("Generic Animal Eating Generically");
}
}
public class Horse extends Animal {
public void eat() {
System.out.println("Horse eating hay ");
}
public void eat(String s) {
System.out.println("Horse eating " + s);
}
}
I'm trying to figure out which version of the three eat() methods will run. Now, when I type
Animal a = new Animal();
a.eat();
The output is "Generic Animal Eating Generically", which is completely understandable.
The same thing happens when I type:
Horse h = new Horse();
h.eat();
The output is "Horse eating hay", which is, again, completely logical.
Here's where it gets confusing for me though. When I type:
Animal ah = new Horse();
ah.eat();
I get:
Horse eating hay
I expected the compiler to invoke the eat() method from the Animal class reference, not the Horse object reference.
So my question is, how can I know for sure which method the compiler is going to invoke when I have a generic reference variable
types referring to an object type (like this one: Animal horse = new Horse();
I expected the compiler to invoke the eat() method from the Animal class reference, not the Horse object reference.
Let's correct this statement first. The variable ah is a reference of type Animal and the statement new Horse() creates an instance of type Horse and assigns it to an Animal reference.
Now that the terminologies are clear, this behavior is expected and is termed as runtype-polymorphism or dynamic method dispatch. At compile time, eat() is resolved based on the reference type which is of type Animal, but at runtime, the method that will be called is based on the instance type which is Horse.
how can I know for sure which method the compiler is going to invoke when I have a generic reference variable types referring to an object type
You could follow these simple steps :
Check the method being called. ah.eat() is calling the method eat.
See if a method with the exact same signature (with the exception being return type covariance) is present in both the parent and child class. (method overriden or not?)
Check the reference type. In Animal ah = new Horse(), the reference type is Animal that is the parent class
Check the instance type. In Animal ah = new Horse(), the instance type is Horse which is the child class.
If all the above conditions are satisfied, you are looking at runtype polymorphism and the method from the child class will be called. In any other scenario, the method to be called will be resolved based on the reference type.
It would also pay to understand that a child class inherits methods from its parents. Lets say that you delete the public void eat() method from Horse class, you are no longer Overrding the eat() method; however, the public void eat(String s) method in Horse is still said to Overload the inherited eat method from Animal. Next, lets add a public void eat(String s) method in Animal. With this addition, you are now Overloading the eat method in Animal and Overrding it in Horse class. No matter how you change the code, the 4 steps mentioned above will always help you decide which method will be called.
This is called dynamic binding in Java. The explicite object type is used not the reference type.
It is not possible to call the overriden super method and the overriding method using a single method, see: How to call the overridden method of a superclass. You could add a method to your horse, which delegates the call to the animal like:
public class Horse extends Animal {
public void animalEat() {
super.eat();
}
public void eat() {
System.out.println("Horse eating hay ");
}
}
This is happens because of method overriding. In method overriding, the reference type does not matter, it is the object type that matters. Animal ah is simply a reference to the object and the actual object is of type Horse. So, Horse's method will be called instead of reference type Animal's method.
Ohkay,
new keyword will create instance of given class...
new Horse();
Now Horse class already is a child of Animal. So following will be instantiate.
public void eat() {
System.out.println("Horse eating hay ");
}
Now you are trying to store that object in Animal's object.
It means Object Of Horse is stored in Animal's Object.
Animal ah = new Horse();
So in Animal's object member of Horse is already stored.
That is the reason that compiler is printing child class method values.
Related
Here I am trying to converting superclass object to subclass. I am getting the runtime error as "class can not be cast".
Eg :
class Animal {}
class Cat : Animal() {}
class abc {
fun abcd(): Animal {
return Animal()
}
fun getData() {
val cat: Cat = abcd() as Cat //Giving me runtime error.
}
}
You can't cast a base class 'instance' to a descendant class, because a base class does not necessarily implement the behaviors of its descendants neither knows anything about them.
In your specific example the method abcd() returns an instance of the base class Animal, and therefore such can't be cast to Cat, since Animal may not have any of the behaviors defined in Cat.
An example, imagine you had also a Dog class, and both Cat and Dog implement different methods such as dog.fetch() and cat.jump(). Such behaviors don't exist in the base class Animal, and therefore it can't be explicitly cast to a specific animal.
The opposite is valid, so casting Cat to Animal, because Cat inherits the behaviors of its base class Animal.
Instead, what you can do is to instantiate a Cat in abcd(), and still return Animal:
fun abcd(): Animal {
return Cat()
}
This is valid, and the casting will work. But, you must pay attention to avoid potential ClassCastException's at runtime if mixing up derived classes, for example if instantiating a Dog while the return type is Animal and try to use it as Cat.
Small remark: I'm assuming the reason Animal isn't open in your example is just a copy/paste mistake, as it clearly needs such keyword to allow inheritance.
Maybe what you are trying to do is something like creating a type and based on what sub-type then do something, like this:
sealed class Animal
data class Cat(val...) : Animal()
data class Dog(val...) : Animal()
class YourMapper {
fun animal(condition: Type): Animal {
return when(condition) {
... -> Dog(...)
... -> Cat(...)
}
}
fun getData(condition: Type): Animal {
return animal(condition)
}
And then the usage is
val data = YourMapper().getData(condition)
when(data) {
is Dog -> {/*do something with your dog*/}
is Cat -> {/*do something with your cat*/}
}
I know this is an old question, but it's the first Google hit for the search "Kotlin how to cast superclass to subclass", so for prosperity:
How to cast to subcalss in Kotlin
use the "as" keyword as in the original question:
if(animal is Cat) {
Cat cat = animal as Cat
}
The original question
Short answer, you can't cast a superclass object to a subclass object. Casting only changes the type of the reference to the object, the object itself remains unchanged.
The Animal class in the question should almost certainly be marked as abstract. That way there's no possibility to accidentally instantiate non-specific animals, which is what happens in the question and causes the exception.
An Animal reference variable can absolutely be cast to a Cat, provided the object it references is a Cat. But in the question a non-specific Animal is instantiated, and then later attempted to cast to a Cat, but as it is not referencing a Cat object, this understandably throws an exception.
So unlike, say, casting an Int to a Double where the cast seemingly changes the type of the object, casting object references doesn't actually "change" the object, only how finely-grained your reference to it is.
Casting cannot turn a Dog into a Cat, it can only change you from examining an animal as a generic Animal, viewing properties commont to ALL animals, to examining it as a Cat or a Dog and additionally having access to the properties only Cats or Dogs have.
I have the code:
class Father{
String name="father";
void f(){System.out.print("father class");}
}
class Son extends Father{
String name = "son";
void f(){System.out.print("son class");}
void f2(){}
}
public class Test {
public static void main(String[] args) {
Father s = new Son();
System.out.println(s.name);// outputs father
s.f();// outputs "son class"
s.f2();// s does not have f2
}
}
My question is, what is the difference between doing Father s = new Father() or, Father s = new Son() or, Son s = new Son()?
As well, why does s.f2 in the example cause an error? Must Father implement f2()?
I think it is easier to explain with an animal example:
class Animal {
void printName() {
System.out.println("Animal");
}
}
class Dog extends Animal{
#Override
void printName() {
System.out.println("Dog");
}
}
class Cat extends Animal{
#Override
void printName() {
System.out.println("Cat");
}
void meow() {
System.out.println("meow");
}
}
When you extend classes, the child class can override parent's methods and can have its own methods. In my Animal example the generic Animal object can only give its name, but the Cat object can give its name and also meow. Obviously, the meow method is specific to Cat as we know that Dogs can't do meow and Animals in general.
When you do
Animal animal = new Cat();
You actually create an instance of the Cat but use it as a general Animal. Thus, your animal instance only has methods which are available in the Animal class but the execution of the methods overridden by Cat class will be delegated to the Cat class.
if you want to execute Cat's specific methods then you need to cast your Animal to the Cat
(Cat) animal.meow();
In your example to call f2() method you need to cast your father object to the son first
(Son)s.f2();
What you're dealing with is reference type (variable type) and object type (what's actually being referred to). The Java compiler needs some kind of guarantee that the object being referred to can run the method you're calling. To do this, it looks to the reference type. When executed, the method run is that of the object type.
Simply put:
Father f = new Father(); //Treated as a Father, behaves like a Father
Son s = new Son(); //Treated as a Son, behaves like a Son
Father q = new Son(); //Treated as a Father, behaves like a Son (sounds like my own father)
If you cast q to a Son by saying (Son)q, it will be treated as a Son by the compiler, unless the object isn't actually a Son, in which case you'll get a ClassCastException.
Let's take a simpler concept, since your hierarchy implies that all Sons are Fathers, but not all Fathers are Sons (which ain't quite true).
Let's take the abstract class Number and any of its children - for brevity, we can use Integer, Float and BigInteger.
Suppose we declare this:
Number num = Float.NaN;
We now have a Float instance which is referenced by a Float. We can do anything we want to that instance, but only in the context of a Number.
There's a useful method for Float called isNan which would allow us to see if our float actually is a number. In the context of Number...that method doesn't exist.
There are advantages to doing it like this - if you don't need the specificity of the child reference, you can refer to everything by its parent class (or interface). This also uncouples you from the child's API should you want to be uncoupled from it (see developing to an interface).
OK I see where is the confusion here.
In java you can override methods but not class variables
keep that rule in mind
So when you did
Father s = new Son();
the object "s" is of type father
and as we said the variables inside it wont be overwritten just the methods
So the the final result is an object that has the member variables from the Father class (the "name" variable) and the method from son class (since father had only 1 method and son overridden it).
and for why the f2 does not work
it is because the object "s" is of type Father not son (it is father object that has 1 method of it overwritten by son class other than that it will keep be father object) and Father does not have f2 method thats why you get the compile error
s.f2() is the syntax error because you told JVM that s is Father, not Son.
In the code, it can't find f2 method in Father class
class Father{
String name="father";
void f(){System.out.print("father class");}
}
But it doesn't mean the code is wrong, just the JVM doesn't like it.
if you change s.f2() to
(Son)s.f2();
It will works
What is the difference between up-casting and down-casting with respect to class variable?
For example in the following program class Animal contains only one method but Dog class contains two methods, then how we cast the Dog variable to the Animal Variable.
If casting is done then how can we call the Dog's another method with Animal's variable.
class Animal
{
public void callme()
{
System.out.println("In callme of Animal");
}
}
class Dog extends Animal
{
public void callme()
{
System.out.println("In callme of Dog");
}
public void callme2()
{
System.out.println("In callme2 of Dog");
}
}
public class UseAnimlas
{
public static void main (String [] args)
{
Dog d = new Dog();
Animal a = (Animal)d;
d.callme();
a.callme();
((Dog) a).callme2();
}
}
Upcasting is casting to a supertype, while downcasting is casting to a subtype. Upcasting is always allowed, but downcasting involves a type check and can throw a ClassCastException.
In your case, a cast from a Dog to an Animal is an upcast, because a Dog is-a Animal. In general, you can upcast whenever there is an is-a relationship between two classes.
Downcasting would be something like this:
Animal animal = new Dog();
Dog castedDog = (Dog) animal;
Basically what you're doing is telling the compiler that you know what the runtime type of the object really is. The compiler will allow the conversion, but will still insert a runtime sanity check to make sure that the conversion makes sense. In this case, the cast is possible because at runtime animal is actually a Dog even though the static type of animal is Animal.
However, if you were to do this:
Animal animal = new Animal();
Dog notADog = (Dog) animal;
You'd get a ClassCastException. The reason why is because animal's runtime type is Animal, and so when you tell the runtime to perform the cast it sees that animal isn't really a Dog and so throws a ClassCastException.
To call a superclass's method you can do super.method() or by performing the upcast.
To call a subclass's method you have to do a downcast. As shown above, you normally risk a ClassCastException by doing this; however, you can use the instanceof operator to check the runtime type of the object before performing the cast, which allows you to prevent ClassCastExceptions:
Animal animal = getAnimal(); // Maybe a Dog? Maybe a Cat? Maybe an Animal?
if (animal instanceof Dog) {
// Guaranteed to succeed, barring classloader shenanigans
Dog castedDog = (Dog) animal;
}
Downcasts can be expressed more succinctly starting from Java 16, which introduced pattern matching for instanceof:
Animal animal = getAnimal(); // Maybe a Dog? Maybe a Cat? Maybe an Animal?
if (animal instanceof Dog castedDog) {
// now castedDog is available here as in the example above
}
Down-casting and up-casting was as follows:
Upcasting: When we want to cast a Sub class to Super class, we use Upcasting(or widening). It happens automatically, no need to do anything explicitly.
Downcasting : When we want to cast a Super class to Sub class, we use
Downcasting(or narrowing), and Downcasting is not directly possible in Java, explicitly we have to do.
Dog d = new Dog();
Animal a = (Animal) d; //Explicitly you have done upcasting. Actually no need, we can directly type cast like Animal a = d; compiler now treat Dog as Animal but still it is Dog even after upcasting
d.callme();
a.callme(); // It calls Dog's method even though we use Animal reference.
((Dog) a).callme2(); // Downcasting: Compiler does know Animal it is, In order to use Dog methods, we have to do typecast explicitly.
// Internally if it is not a Dog object it throws ClassCastException
Autoboxing-vs-Casting
Upcasting and downcasting are important part of Java, which allow us to build complicated programs using simple syntax, and gives us great advantages, like Polymorphism or grouping different objects. Java permits an object of a subclass type to be treated as an object of any superclass type. This is called upcasting. Upcasting is done automatically, while downcasting must be manually done by the programmer, and i'm going to give my best to explain why is that so.
Upcasting and downcasting are NOT like casting primitives from one to other, and i believe that's what causes a lot of confusion, when programmer starts to learn casting objects.
Polymorphism: All methods in java are virtual by default. That means that any method can be overridden when used in inheritance, unless that method is declared as final or static.
You can see the example below how getType(); works according to the object(Dog,Pet,Police Dog) type.
Assume you have three dogs
Dog - This is the super Class.
Pet Dog - Pet Dog extends Dog.
Police Dog - Police Dog extends Pet Dog.
public class Dog{
public String getType () {
System.out.println("NormalDog");
return "NormalDog";
}
}
/**
* Pet Dog has an extra method dogName()
*/
public class PetDog extends Dog{
public String getType () {
System.out.println("PetDog");
return "PetDog";
}
public String dogName () {
System.out.println("I don't have Name !!");
return "NO Name";
}
}
/**
* Police Dog has an extra method secretId()
*/
public class PoliceDog extends PetDog{
public String secretId() {
System.out.println("ID");
return "ID";
}
public String getType () {
System.out.println("I am a Police Dog");
return "Police Dog";
}
}
Polymorphism : All methods in java are virtual by default. That means that any method can be overridden when used in inheritance, unless that method is declared as final or static.(Explanation Belongs to Virtual Tables Concept)
Virtual Table / Dispatch Table : An object's dispatch table will contain the addresses of the object's dynamically bound methods. Method calls are performed by fetching the method's address from the object's dispatch table. The dispatch table is the same for all objects belonging to the same class, and is therefore typically shared between them.
public static void main (String[] args) {
/**
* Creating the different objects with super class Reference
*/
Dog obj1 = new Dog();
` /**
* Object of Pet Dog is created with Dog Reference since
* Upcasting is done automatically for us we don't have to worry about it
*
*/
Dog obj2 = new PetDog();
` /**
* Object of Police Dog is created with Dog Reference since
* Upcasting is done automatically for us we don't have to worry
* about it here even though we are extending PoliceDog with PetDog
* since PetDog is extending Dog Java automatically upcast for us
*/
Dog obj3 = new PoliceDog();
}
obj1.getType();
Prints Normal Dog
obj2.getType();
Prints Pet Dog
obj3.getType();
Prints Police Dog
Downcasting need to be done by the programmer manually
When you try to invoke the secretID(); method on obj3 which is PoliceDog object but referenced to Dog which is a super class in the hierarchy it throws error since obj3 don't have access to secretId() method.In order to invoke that method you need to Downcast that obj3 manually to PoliceDog
( (PoliceDog)obj3).secretID();
which prints ID
In the similar way to invoke the dogName();method in PetDog class you need to downcast obj2 to PetDog since obj2 is referenced to Dog and don't have access to dogName(); method
( (PetDog)obj2).dogName();
Why is that so, that upcasting is automatical, but downcasting must be manual? Well, you see, upcasting can never fail.
But if you have a group of different Dogs and want to downcast them all to a to their types, then there's a chance, that some of these Dogs are actually of different types i.e., PetDog, PoliceDog, and process fails, by throwing ClassCastException.
This is the reason you need to downcast your objects manually if you have referenced your objects to the super class type.
Note: Here by referencing means you are not changing the memory address of your ojects when you downcast it it still remains same you are just grouping them to particular type in this case Dog
I know this question asked quite long time ago but for the new users of this question.
Please read this article where contains complete description on upcasting, downcasting and use of instanceof operator
There's no need to upcast manually, it happens on its own:
Mammal m = (Mammal)new Cat(); equals to Mammal m = new Cat();
But downcasting must always be done manually:
Cat c1 = new Cat();
Animal a = c1; //automatic upcasting to Animal
Cat c2 = (Cat) a; //manual downcasting back to a Cat
Why is that so, that upcasting is automatical, but downcasting must be manual? Well, you see, upcasting can never fail. But if you have a group of different Animals and want to downcast them all to a Cat, then there's a chance, that some of these Animals are actually Dogs, and process fails, by throwing ClassCastException.
This is where is should introduce an useful feature called "instanceof", which tests if an object is instance of some Class.
Cat c1 = new Cat();
Animal a = c1; //upcasting to Animal
if(a instanceof Cat){ // testing if the Animal is a Cat
System.out.println("It's a Cat! Now i can safely downcast it to a Cat, without a fear of failure.");
Cat c2 = (Cat)a;
}
For more information please read this article
Better try this method for upcasting, it's easy to understand:
/* upcasting problem */
class Animal
{
public void callme()
{
System.out.println("In callme of Animal");
}
}
class Dog extends Animal
{
public void callme()
{
System.out.println("In callme of Dog");
}
public void callme2()
{
System.out.println("In callme2 of Dog");
}
}
public class Useanimlas
{
public static void main (String [] args)
{
Animal animal = new Animal ();
Dog dog = new Dog();
Animal ref;
ref = animal;
ref.callme();
ref = dog;
ref.callme();
}
}
Maybe this table helps.
Calling the callme() method of class Parent or class Child.
As a principle:
UPCASTING --> Hiding
DOWNCASTING --> Revealing
1.- Upcasting.
Doing an upcasting, you define a tag of some type, that points to an object of a subtype (Type and subtype may be called class and subclass, if you feel more comfortable...).
Animal animalCat = new Cat();
What means that such tag, animalCat, will have the functionality (the methods) of type Animal only, because we've declared it as type Animal, not as type Cat.
We are allowed to do that in a "natural/implicit/automatic" way, at compile-time or at a run-time, mainly because Cat inherits some of its functionality from Animal; for example, move(). (At least, cat is an animal, isn't it?)
2.- Downcasting.
But, what would happen if we need to get the functionality of Cat, from our type Animal tag?.
As we have created the animalCat tag pointing to a Cat object, we need a way to call the Cat object methods, from our animalCat tag in a some smart pretty way.
Such procedure is what we call Downcasting, and we can do it only at the run-time.
Time for some code:
public class Animal {
public String move() {
return "Going to somewhere";
}
}
public class Cat extends Animal{
public String makeNoise() {
return "Meow!";
}
}
public class Test {
public static void main(String[] args) {
//1.- Upcasting
// __Type_____tag________object
Animal animalCat = new Cat();
//Some animal movement
System.out.println(animalCat.move());
//prints "Going to somewhere"
//2.- Downcasting
//Now you wanna make some Animal noise.
//First of all: type Animal hasn't any makeNoise() functionality.
//But Cat can do it!. I wanna be an Animal Cat now!!
//___________________Downcast__tag_____ Cat's method
String animalNoise = ( (Cat) animalCat ).makeNoise();
System.out.println(animalNoise);
//Prints "Meow!", as cats usually done.
//3.- An Animal may be a Cat, but a Dog or a Rhinoceros too.
//All of them have their own noises and own functionalities.
//Uncomment below and read the error in the console:
// __Type_____tag________object
//Cat catAnimal = new Animal();
}
}
upcasting means casting the object to a supertype, while downcasting means casting to a subtype.
In java, upcasting is not necessary as it's done automatically. And it's usually referred as implicit casting. You can specify it to make it clear to others.
Thus, writing
Animal a = (Animal)d;
or
Animal a = d;
leads to exactly the same point and in both cases will be executed the callme() from Dog.
Downcasting is instead necessary because you defined a as object of Animal. Currently you know it's a Dog, but java has no guarantees it's. Actually at runtime it could be different and java will throw a ClassCastException, would that happen. Of course it's not the case of your very sample example. If you wouldn't cast a to Animal, java couldn't even compile the application because Animal doesn't have method callme2().
In your example you cannot reach the code of callme() of Animal from UseAnimlas (because Dog overwrite it) unless the method would be as follow:
class Dog extends Animal
{
public void callme()
{
super.callme();
System.out.println("In callme of Dog");
}
...
}
We can create object to Downcasting. In this type also. : calling the base class methods
Animal a=new Dog();
a.callme();
((Dog)a).callme2();
Just confused on how to following answer is correct.
class Cat {
public void isClawedBy(Cat c){
System.out.println("Clawed by a cat");
}
}
class Kitten extends Cat{
public void isClawedBy(Kitten c){
System.out.println("Clawed by a Kit");
}
}
If the following is called
Cat g = new Cat();
Cat s = new Kitten();
Kitten t = new Kitten();
g.isClawedBy(t);
s.isClawedBy(t);
t.isClawedBy(t);
How is the answer:
Clawed by Cat
Clawed by Cat
Clawed by Kitten
I'm confused on why s.isClawedBy(t) = Clawed by Cat.
Since the dynamic type of s is a kitten, and t is a kitten.
Is it because the arguments are different, so it doesn't override it?
Another part I am confused on. //Note the arguments have been swapped.
class Cat {
public void isClawedBy(Kitten c){
System.out.println("Clawed by a cat");
}
}
class Kitten extends Cat{
public void isClawedBy(Cat c){
System.out.println("Clawed by a Kit");
}
}
If the following is called
Cat g = new Cat();
Cat s = new Kitten();
Kitten t = new Kitten();
g.isClawedBy(t);
s.isClawedBy(t);
t.isClawedBy(t);
The output is:
Clawed by Cat
Clawed by Cat
Clawed by Cat
How does it work for when t is called?
About the second query : t.isClawedBy(t) giving the output of Clawed by Cat.
Since t is a Kitten and the argument passed in the method t.isClawedBy(t) is also Kitten , the method from the superclass Cat will be called because it matches the arguments perfectly.
Class Kitten does not override isClawedBy(Cat c). It adds a new method isClawedBy(Kitten c). The runtime sees s referenced as a Cat at the time s.isClawedBy(t) is called, and it ends up calling the Cat method.
If you change Kitten to:
class Kitten extends Cat{
#Override
public void isClawedBy(Cat c){
System.out.println("Clawed by a Kit");
}
}
Then you will see the output you desire. Even more interesting, you can do:
((Kitten) s).isClawedBy(t); and you will see the proper method called.
I'm confused on why s.isClawedBy(t) = Clawed by Cat. Since the dynamic type of s is a kitten, and t is a kitten.
s has reference type Cat but holds a Kitten object. t has reference type Kitten and holds an object Kitten. When the method is run at run-time it is first checked if the reference type has such a method and then the most specific version of the method is called. Since the subclass doesn't override the method (different types in the parameter) the method in the reference type is called.
For your second part it is the exact same thing happening, there is no overload and a Kitten can indeed be passed as a Cat to the method in the reference type so again this method is the most specific one in the hierarchy and it is called.
As for the first question;
Overriding resolution is done at runtime, but overloading resolution is done at compile time.
Since the signature of your methods are not identical (different parameter types. It doesn't matter that one's the subclass of another), they're overloading.
Since resolved at compile time, the compiler does not know what the instance type is; only the declared type.
To the compiler, s.isClawedBy(t) is the method isClawedBy(Kitten) of the declared type Cat.
The compiler says "yep, Cat can accept a Kitten in its method, that's what this method is".
So, at runtime, which method that will call has ALREADY been chosen by the compiler. The lookup is NOT performed at runtime.
Thus, at runtime, despite s actually being a Kitten object, the Cat method is called.
I am trying to understand Java's polymorphism, and I have one question about downcasting an object.
Let's say for this example I have two subclasses Dog and Cat that inherit from a superclass Animal
From what I understood, the only way to downcast an object is if this Object is already of the good type, like this:
Animal a = new Dog();
Dog d = (Dog) a;
This works right?
But what if I want to create a regular animal without knowing what it would be, and then cast it when I know, how can I do that?
Animal a = new Animal();
Dog d = (Dog) a;
This will throw a ClassCastException at runtime right?
The only way I found to do that is to create a new Dog constructor that creates a dog from a regular animal:
Animal a = new Animal();
Dog d = new Dog(a);
with
public Class Dog extends Animal{
public Dog(Animal a){
super(a);
}
}
So my question is, how am I supposed to do this?
Am I doing it the best way?
Am I not supposed to do this at all, if I have to it means my program is not well conceived?
Is there a better way I missed?
Thanks a lot!
nbarraille
If you want to create an instance of a type that may vary depending upon non-local conditions, use an Abstract Factory (as described in the Design Patterns book).
In it's simplest form:
interface AnimalFactory {
Animal createAnimal();
}
class DogFactory implements AnimalFactory {
public Dog createAnimal() {
return new Dog();
}
}
Note also there is a difference between the static type of a reference and the dynamic type of the object. Even though you have an Animal reference, if the original object is a Dog, it still behaves like a Dog.
You should only cast to a class that the object really is, so if you have a Dog that extends Animal you can cast it to an Animal (because it is one) but you shouldn't cast an Animal to a Dog because not all Animals are Dogs. The Dog class may well have extra fields that are not implemented by the Animal class and so the cast doesn't make sense (what do you initialise those values to?).
Java is a strongly typed language, and that means you can only cast an object to a type it extends from (either a superclass or an interface).
Even if you "fake it", e.g. copy all a classes methods and fields, you just can't cast an object to a type it doesn't extend.
public class Foo{
public String phleem;
public void bar(){
}
}
public class Bar{
public String phleem;
}
public interface Baz{
public void bar();
}
Given the above code, you can't cast a Foo object to either a Bar or a Baz, although the class structure seems to imply that you could. There is no inheritance involved, so a ClassCastException is thrown.
Here you are talking about downcasting, so in this scenario always superclass should be used as a reference and child object should be pointed by that.
This usd basically in factory patter.