Is there a work around that will allow me to cast an object of the base class to an object of the derived class?
something like the following
B extends A
A a = new A();
B b = (B)a
Is there a trick that will achieve this?
No, absolutely not. What would you expect the values of any fields declared in B but not in A to be? For example, what would you expect this to do:
Object x = new Object();
String text = (String) x;
System.out.println(text);
An Object has no text data... so what would it mean to cast it as a string?
You can only cast a reference to a type which is appropriate for the actual type of the object.
The desire to do this usually indicates a design problem somewhere... or it might mean that you want something like:
public class A {
public A() {
// Whatever
}
public A(A a) {
// Use the existing values in "a" to initialize this object
}
}
public class B extends A {
/** Creates a new B from the values in an A, with suitable defaults. */
public B(A a) {
super(a);
// Now initialize any fields in B with appropriate values
}
}
Then:
A a = new A();
B b = new B(a);
That will create two objects, unlike a cast... but it would at least be valid.
How is that even possible? Think about it. It is like saying if you have a class FourWheeler, you can simply cast it into a Ferrari and make it a Ferrari!
No, this isn't possible. When B extends A it inherits the behavior of A, but on the same time, there is nothing stopping you from defining new behavior for B (where those new behaviors won't be part of A)
For example say A has a single method called 'methodA'. Now when B extends A it inherits 'methodA' but it also declares another method called 'methodB'. So under such circumstance you will get a runtime 'ClassCastException' when you try to call the 'methodB' over an instance of Object A.
Related
This question already has answers here:
A Base Class pointer can point to a derived class object. Why is the vice-versa not true?
(13 answers)
Closed 7 years ago.
This is rather basic question. But I can't understand well the concept of inheritance.
Suppose I have two classes, A and B with both have a test() method that returned 1 and 2 respectively, and B inherited A class. In main method I declare the instance as such;
A a1 = new B();
and call the method a1.test(), it will return 2. This is the concept of polymorphism. But when I have a method test2() in just subclass, I can't call the method using the same instance declaration as above. Why is that happen?
I can't call the method using the same instance declaration as above. Why is that happen?
Because the type of the variable is A, and class A does not have a method test2(). The Java compiler only looks at the type of the variable to check if you can call a method, it does not look at the actual object (which is in this case a B).
This is all easier to understand if you use more concrete and meaningful names for your classes, instead of abstract names such as A and B. Let's call them Animal and Bear instead:
class Animal {
}
class Bear extends Animal {
public void growl() { ... }
}
class Cat extends Animal {
public void meow() { ... }
}
Animal a1 = new Bear();
Animal a2 = new Cat();
// Doesn't work, because not every Animal is a Bear, and not all
// animals can growl.
a1.growl();
// You wouldn't expect this to work, because a2 is a Cat.
a2.growl();
Because variable type is A, and class A does not have a method test2():
Rather you can use:
A a1 = new B(); // upcasting
B b1 = (B)a1; // Downcasting a1 to type B
b1.test2(); // now you can call test2 function
Because, the left side of your condition determines which method's you can call, and right side determines which methods will be called. So in this case class A does't have test2() method.
Imagine A = "TV" and B = "HD_TV".
You can say
TV tv = new HD_TV() // HD TV
and
TV tv = new TV() // ordinary TV
because an HD_TV is a TV.
You can say:
tv.show(movie)
It will show what is on TV, but you will get a better picture with the HDTV.
You cannot say:
tv.showHD(hdMovie) // Compiler error !!!
because in declaring tv as TV, you are saying it might not be an HD TV. Even though you can see that in this case it is, the compiler still respects your declaration that it is just a TV and you can only use methods supported for a TV.
That is because you are declaring the instance a1 as an A. Because B inherits A, you can call all the functions declared in A and they might have a different meaning if they are overloaded in B, but you do not have any access to B-only things.
You can see the first A as some kind of a header file, if you are familiar with that. It declares what A contains, without looking at how the functions are implemented or what the default vars are of everything in A. As a direct consequence, you can only access everything that is declared to literally be in A.
The left-hand side - A in this case - is the declared type, and it doesn't know about anything specific to child classes. The right-hand side - ´B´ in this case - is the actual type, and this provides the behaviour.
So, this will work because the declared type B knows about methods available in the class B.
B b1 = new B();
b1.test2();
If it was possible to have a1.test2(), that would mean every class would have to know about every child it has - including those in other libraries, when projects are assembled!
When B inherits A class and the reference of A is created with object of B like A a1 = new B();.
On Compile time java compiler looks for method availability in class A.
So it allows calling method test() but not the test2().
As test() method is available in class A but test2() is not available in class A.
You can type cast the object created like ((B)a1).test2().
This will work.
Here a1 object is of type A. a1 is pointing to an object of type B.
a1 is a reference of type A to an object of type B.
since a1 is of type A it know only test() which is declared in its class definition already. In case you want to access test2 declared in class B you need to type cast the a1 object back to B
like
B b1 = (B)a1
b1.test2() will be accessible.
This happens because you declare A variable and use B class which is an A. The compiler know it's an A but doesn't know it's a B later in the code. It's easier to use real life objects as example.
For example you have:
class Pet() {
method feed();
}
And a
class Dog() extends Pet {
method feed();
method bark()
}
If you have a code in another class:
So if you have code :
Pet dogPet=new Dog();
You know it's a dog here because you create the instance and you can use:
((Dog)dogPet).bark(); or just declare the variable as a dog instead of pet.
But if you have a method in another class:
void someMethod(Pet aPet){
// Here you don't know if the Pet is a dog or not. So you know only that it
//can be fed but you don't know if it barks. Even if a Dog is supplied to the method
}
In a1 = new B(), the actual type of the object created is B but you reference it as its supertype so you can call a method that accepts A (polymorphism).
So if a method is overridden in subclass, a1.test() is executing subclass's test().
In order to execute test2() you have to do that: ((B) a1).test2();
There is a concept called Up casting and Down casting.Up-casting is casting to a supertype, while downcasting is casting to a subtype. Supercasting is always allowed, but subcasting involves a type check and can throw a ClassCastException.,See the Example Code:
class A{
public int test(){
return 1;
}
}
class B extends A{
public int test(){
return 2;
}
public int test2(){
return 3;
}
}
and
A a1 = new B();
a1.test2();//not possible
Here you can't invoke methods of class B.
class A {
int super_var = 1;
}
class B extends A {
int sub_var = 2;
}
public class Demo{
public static void main(String []args){
A a = new B();
System.out.print(a.sub_var); //compile error
}
}
why this will end with a compile error ? reference (a) referencing to an Object of B it has sub_var so why is it restricted ? why reference (a) can access only the fields in A ?
Let's say you have these classes:
public class Animal() {
// ...
}
public class Fish extends Animal() {
public void swim() {...}
}
If you declared an Animal:
Animal x = new Fish();
and you called the swim() method
x.swim();
Would you expect it to work? I don't think so, because not every animal can swim. That's why you have to explicitly specify that the animal x is a Fish:
((Fish) x).swim();
In your case, if you wanted to call that method, you should specify (technically, it's called cast) the type:
System.out.print(((B)a).sub_var);
Note:
This works similar for methods and variables. I used a method in the example since it's more illustrative.
Edit:
Let's see this example:
Animal x;
if (some_condition)
x = new Fish();
else
x = new Cat();
x.swim();
This restriction exists, because Java won't know if, at execution time, the object assigned to x will have the method swim(). So to avoid this, you have to cast to the respective type to call a method that doesn't exist in superclass.
At first it does sound like it should work. (And in some languages it probably does.) But think about this example:
public class Demo {
public static void main(String []args){
A a = new B();
print( a );
}
public static void print( A arg ) {
System.out.print(arg.sub_var); //compile error
}
}
This functionally does the same thing but the print is in another method. If your version worked, this one could be expected to work too.
But what if someone then does this?
Demo.print( new A() );
This should fail because A doesn't have a sub_var. It would have to throw some kind of runtime error instead.
So the design decision in Java was not to allow this and if you declare a local variable/field/method parameter as type A, then you can only access things that every object that is either A or a subclass is guaranteed to have.
If you want to access more, you need to cast it to the subclass, which will throw an exception if you try it on an object that doesn't fit.
A a = new A();
System.out.print(((B)a).sub_var); //ClassCastException is thrown here
You can not access B members with the reference of Parent object A.
Instead change your println statement like below to access,
System.out.print(((B)a).sub_var);
Is there a variable called sub_var in the parent class ? No. That is why you get the error -
sub_var cannot be resolved or is not a field
See this
System.out.print(a.super_var); //okay
System.out.print(a.sub_var); //compile error
you create an object of type B and assign it to a variable of type A. The type A does not declare sub_var. This field is declared only in type B. the compiler only sees what is declared in type A, although the variable is instantiated to an object of type B.
If you want to access sub_var you would have to cast a to B.
System.out.println( ((B)a).sub_var);
The sub_var is in class B, so you can only access through a reference of class B. To the compiler A a = new B(); means a is an instance of class A.
Ok.. So,
When you have a hierarchy of classes such as
public class A {...}
and,
public class B extends A {...}
...When you create objects, what is the difference between:
A object = new A();
A object = new B();
B object = new B();
Thank you for your time.
public class A
{
public void methodA(){}
}
public class B extends A
{
public void methodB(){}
}
I hope this can demonstrate the difference.
A obj = new A();
a.methodA(); //works
A obj = new B();
obj.methodA(); //works
obj.methodB(); //doesn't work
((B)obj).methodB(); //works
B obj = new B();
obj.methodA(); //works
obj.methodB(); //works
A object = new A();
You are creating an A instance in a reference of type A. You may can access only A methods/properties and parents methods/properties.
A object = new B();
You are creating B instance in a reference of type A. In this way object could behave in a polymorphic way, for example if you make object.method() and method is overriden in B then it will call this override method. You have to take care in not to break the Liskov Substitution Principle. You may can access only A methods/properties and parents methods/properties. This is the preferred way when you only need supertype contract.
B object = new B();
You are creating a B instance in a reference variable of type B. You may can access only B methods/properties and parents methods/properties.
A line like
A var = new B();
is kind of a shorthand for two separate steps.
A var; // (1) Make a variable of TYPE A.
var = new B(); // (2) Make an object of CLASS B, that from now on may be
// referred to by the variable var.
So a variable has a TYPE, and an object has a CLASS. Often they match up. The type of a variable is often actually a class, although not necessarily. It's important to understand the difference between the type of a variable, and the class of the object that the variable refers to.
An object typically belongs to more than one class. If class B extends class A, that means that all objects of class B are also objects of class A. And all objects of any class at all are also objects of class Object. In other words, when we say that an object is a B, that's more specific than saying it's an A. Just like when we say that Yogi is a bear, that's more specific than saying Yogi is an animal, because all bears are animals.
So a variable of type A can indeed refer to an object of class B, if A is a class that B extends. But if you've got a variable of type A, you can't use it to do things that are specific to objects of type B. For example, suppose class A has a method called display() and class B has a method called explain(). The compiler will let you call display() on a variable of type A, but it won't let you call explain(). If it did, it would be risking trying to call explain() on an object that's not actually a B, which would fail.
So whenever there are methods that class B defines, you'll need a variable of type B in order to be able to call them. Of course, you can also use that same variable to call the methods that are defined in class A. In a sense then, if class B extends class A, then a variable of type B is more powerful than a variable of type A - you can do more stuff with it.
So the question arises - why would I ever want to write
A var = new B();
when a variable of type B would be more powerful than var in this example?
The short answer is that it communicates to people looking at the code. It says, "yes, I know this variable refers to a B, but I actually only intend to use the methods provided by class A. This can actually be helpful to someone trying to understand your code, or to maintain it.
There are also cases where it can make a real difference to method calls involving that variable. Suppose there's another class C, which has two methods with the same name but slightly different signatures, like this.
public class C {
public void process(A arg){
// Do some stuff
}
public void process(B arg){
// Do some other stuff
}
}
In this particular case, the version of process that gets called depends on the type of the variable, not the class of the object. So if you write
C processor = new C();
A var = new B();
processor.process(var);
this will call the first version of process - the one with A in the signature. Because of the type of the variable. But if you write
C processor = new C();
B var = new B();
processor.process(var);
this will call the second version of process - the one with B in the signature.
A object = new A();
object of type A (you can access fields or method from A)
A object = new B();
object of type A (you cannot access fields or method from B, only from A)
B object = new B();
object of type B (you can access fields or method from A and B)
A object1 = new A();
A object2 = new B();
B object3 = new B();
object1 is declared as a reference to an A object. Since class B extends class A, it could be set to either or (new A() or new B() would be valid).
object2 is declared as a reference to an A object, but is actually a B object. Say the B class has a method called eatFood(). If you tried to access that method with object2.eatFood(), the compiler would throw an error because the eatFood method is only in the B class. Even though the object is actually a B object, the compiler thinks it is an A object due to the type declaration. To access the eatFood method, you would have to typecast it: ((B)object2).eatFood().
object3 is simply a reference to a B object, and in reality IS a B object. It could access A methods as well as B methods.
A object = new B();
This declares that object will refer to an object of class A or any of its subclasses (when it isn't null). The compiler will treat it as an object of type A, so you can only access methods and fields that are declared for A (or one of its superclasses). It also means that you can later assign it to any other object that is of class A or a subclass:
A object1 = new B();
B object2 = new B();
// reassign later
object1 = new A(); // legal
object2 = new A(); // ILLEGAL
class C extends A { ... }
object1 = new C(); // legal
object2 = new C(); // ILLEGAL
So the initial declaration declares object as having type A. But its initial value is an object of type B, which is OK because B is a subclass of A.
That should explain the difference between your second and third examples. The difference between the first and second is simply that (at run time) the first creates a new object of type A and the second creates a new object of type B.
I need to link an object of type B to any instance of type A (circular dependencies). I could also declare another method, which must be called after the constructor of A and link a new B to an A-instance. What I want to achieve is not having to call such a method manually. This is the sample code:
public Class A{
B b;
public A(){
b = new B(this); // this does not work,
// as this references an object that has not been created yet
}
}
public Class B{
A a;
public B(A a){
this.a = a; //or something else
}
}
I commented the problematic line. I also understand why it can't work. What I need to know is, if there is a well-known design pattern to avoid this problematic? Or should I just redesign my class model, putting anything in B to A? Any suggestions?
It does work. It's problematic in that it exposes an object before it's been fully initialized (so if the B constructor calls methods on the parameter, for example, that could be a bad thing), but it does work. The reference B.a will be a reference to the instance of A that's been/being constructed.
I would recommend trying to avoid the cyclic reference where possible, but where the alternative is even worse, the code you've given will work.
Using this approach is not recommended and possibly create runtime exception because the object is not fully initialized. We can take simple scenario as:
public class A {
B b;
String s;
public A(){
b = new B(this); // this does not work, as this references an object that has not been created yet
s = "print me";
}
}
public class B {
A a;
public B(A a){
this.a = a; //or something else
System.out.println(this.a.s); // will same as a.s;
}
}
It will output null because a partially initialized reference is copied to constructor. This code compiles because compiler doesn't find anything missing in code as all appropriate references and variable are there.
I have two classes A and B while B is a subtype of A:
public class A {
private String stringVar;
public A() {
stringVar = "";
}
public String getStringVar() {
return stringVar;
}
public void setStringVar(String str) {
this.stringVar = str;
}
#Override
public String toString() {
return getStringVar();
}
}
Class B:
public class B extends A {
private int intVar;
public B() {
intVar = 0;
}
public int getIntVar() {
return intVar;
}
public void setIntVar(int intVar) {
this.intVar = intVar;
}
#Override
public String toString() {
return super.toString() + " " + getIntVar();
}
}
As you can see in the following main method I assign the b to a. Now "a" can't invoke b's methods which is clear, because I'm using an instance of type A now. But it behaves like a B when toString is invoked. Curious, I would have expected toString of a. Why is this so?
public class Main {
public static void main(String[] args) {
A a = new A();
B b = new B();
b.setIntVar(200);
b.setStringVar("foo");
a = b;
System.out.println(a);
}
}
Because a points to the implementation of B.
And is declared as A.
So behavior of B. And methods visible of A.
To use B methods do like this
((B) a).getIntVar();
Think of it like this
Object o = new FancyObject();
When compiling this only Objects methods will be accepted even though it's a FancyObjcet with lots of methods.
To use the methods of FancyObject on o do like this.
Object o = new FancyObject();
(FancyObject o).fancyMethod();
Quote "because I'm using an instance of type A now" you are still using an instance of type B. You can see it like you have upcasted b but it's the same instance.
Picture cross linked from another site with credits in the picture, if this is against the rules then somebody is free to edit this part of my answer.
This is nature of inheritance / polymorphism and overriding methods.
Overrided methods will be determined in runtime based on objects real type and not based on reference type.
Therefore a.toString() is actually b.toString() because it is determined in runtime.
http://download.oracle.com/javase/tutorial/java/IandI/override.html
The concept you need to understand is the difference between References and Objects.
a is a reference (a local variable in this case) that points first to an Object of type A and then to an Object of type B.
The compiler knows that it must be of type A (or a subtype thereof), so it can safely call all methods A defines, but they will be called on the actual Object, not on the original Type of a.
This is polymorphism: The object that a holds has static type A, but it is still an Object of dynamic type B. Dynamic dispatch therefore chooses the overridden toString() defined in B.
That's exactly how Java's runtime polymorphism works. All that matters is the actual type at runtime. What you have done is take a reference to an A and point it at an instance of B. You have changed the type of the thing that a points to.
Try
a = (A)b;
No, B Overrides the toString method of A, so if an object is an instance of B, when you call its toString method, you get whatever method that instance has. In general, if you have an object and call its methods, the method called is the one that is in the instance, not in the variable type. The only exception is static methods.
In C++, this is not the case. The method called is the one of the variable type, if one exists, unless you explicitly select the above described behavior by making a method virtual.
That is called runtime polymorphism in OOP.