I came across the concept where a method cannot be overridden and overloaded at the same time. But when I used Arrays.toString(),
Correct me if I'm wrong, but the toString() method has been overridden and overloaded at the same time. How is that possible?
You are overriding when you have the exact same method signature in a subclass. You are overloading when you have two equal named methods with different type or number of parameters. You can override a method and then overload it but you can write a method and it can happend that when you overload it you end up overriding a super class method. One thing doesn't exclude the other.
Let's have a Super class like this:
public class Super {
public void testMethod() {}
// Overload
public void testMethod(String param) {}
}
And then extend it with a sub class like this:
public class Sub extends Super {
// Override only
#Override
public void testMethod() {}
// Overload only
public void testMethod(int param) {}
// Overload and Override
#Override
public void testMethod(String param) {}
}
As you can see, you can have only overload, only override or both in multiple ways. As said, one thing doesn't exclude the other.
Related
Some classes have many interfaces that they inherit from.
I want to make it explicit in my overrided methods what interface or class my method overrides so that when I look back at my code a month from now, I don't have to do a linear search over every interface being implemented and class being extended to find which interface or class the method is being overrode for.
Situation:
class Foo extends Bar implements Foo1, Foo2, Bar1, Bar2
{
public Foo(){}
#Override
void foo1() {}
#Override
void foo2() {}
#Override
void bar1() {}
#Override
void bar2() {}
}
I want to do something like:
class Foo extends Bar implements Foo1, Foo2, Bar1, Bar2
{
public Foo(){}
#Overrides(source="Foo2")
void foo1() {}
#Overrides(source="Bar1")
void foo2() {}
#Overrides(source="Bar2")
void bar1() {}
#Overrides(source="Foo1")
void bar2() {}
}
Is there a way to accomplish this?
Solution (thanks Nikem):
Can create a custom annotation (Overrides.java)
public #interface Overrides
{
String source() default "";
}
Override annotation is just a marker and compile time check to mark that -something- is being overridden without a care for where the method being overridden comes from. As user2357112 said, many IDEs can show you the interface/superclass a method comes from. If you need to have an IDE independent way to accomplish this then add a comment or javadocs saying which interface it comes from.
You can create your own annotation specifically for this purpose. Standard #Override annotation does not allow this.
Let's say I got the following, I would even call it pseudo-code
public class someClass {
public someClass{
example("Hello Stackoverflow");
}
#Override
public void example(){
System.out.println("Hello World");
}
public void example(String Hello){
System.out.println(Hello);
}
}
In this code the method public void example(String Hello) would be called instead of the public void example() method. How is the compiler working in this case ? The compiler has to decide which method to call in this case, because they got the same name. Is there something like an order e.g. first try #Override method, if that's not working go for the normal one. Or how does that work ?
No, what you've shown isn't overriding at all - it's overloading. (The use of the #Override annotation is obviously to do with overriding, but it's incorrect used here - there's no superclass method to override.)
Overriding is when a method signature is declared in a superclass, and then overridden in a subclass (with the same signature). The method implementation is chosen at execution time based on the execution-time type of the object you call it on.
Overloading is when more than one method is present with the same name, but different signatures. When invoking the method, the correct signature is picked at compile time based on the compile-time types of the arguments to the method.
For example:
public void foo(int x) {}
public void foo(String y) {}
public void foo(Object o) {}
public void foo() {}
foo(50); // Calls the first method
foo("hello"); // Calls the second method
// Calls the third method, because the compile-time type of the argument is
// Object, even though it's actually a reference to a string at execution time
foo((Object) "hello");
foo(); // Calls the fourth method
The #Override annotation tells the compiler: "Please fail to compile me unless I'm overriding an existing method defined in a parent class or an interface I implement."
However, you are doing method overloading - i.e. methods with the same name but different arguments.
Consequently, your code won't compile because your example() method doesn't override a method in a parent class or interface. Instead, it overloads another method in the same class. There is no annotation for that.
An example of a valid override would be adding a toString() method to your class, which would override the method declared in the Object class:
public class someClass {
public someClass{
example("Hello Stackoverflow");
}
public void example(){
System.out.println("Hello World");
}
public void example(String Hello){
System.out.println(Hello);
}
#Override
public String toString() {
return "Hello, World!";
}
}
Because you are calling a function with a parameter
example("Hello Stackoverflow");
that is function overloading study about overloading here
This has nothing at all to do with the #Override annotation.
This has nothing at all to do with the #Override annotation.
You can overload methods: having more than one method with the same name but different types and number of parameters.
The compiler will choose the method with the signature that matches best (the exact rules are a bit complicated since Java5, what with varargs and auto-boxing).
In your case, you call a method with a String parameter.
A method without parameters does not apply, will not be considered.
It's not an #Override since you do not override a parent method and code will not compile.
This is overload.
Here, you have 2 distinct methods. One take a parameter and the other one not.
So it simple:
when you call method with a parameter, it's the example(String hello) which will be called.
when you call method without a parameter, it's the example() method which will be called.
This is not an override.
Okay, I have never used generics heavily and was wondering if I could use them in the following way.
Lets suppose I have this class
public class ASuperClass {
public abstract void doSomething(String arg1, Integer arg2);
}
So, if I then extend the above class I would then have the following, where I would be forced to override doSomething with an argument list of String and Integer.
public class ASubClass extends ASuperClass{
#Override
public void doSomething(String arg1, Integer arg2){
some code.....
}
}
Now, lets suppose that in my subclass I am fine to override doSomething, but I need an additional String arg in my subclass. So, what I would like to do is have the below:
public class ASubClass extends ASuperClass{
#Override
public void doSomething(String arg1, Integer arg2, String arg3){
some code...
}
}
The above won't compile though because my subclass does not implement the abstract signature defined in the base. So, to get it to compile, I can of course do this:
public class ASubClass extends ASuperClass{
#Override
public void doSomething(String arg1, Integer arg2){}
public void doSomething(String arg1, Integer areg2, String arg3){
here is my code......
}
}
So, what exactly am I trying to do you are probably asking? What I am interested in is, is there a way to "force" any subclass of a class, using generics as the arg, to implement a base abstract method regardless of arg list. So that the following would compile extending the ASuperClass:
public class ASubClass extends ASuperClass{
#Override
public void doSomething(String arg1, Integer areg2, String arg3){
here is my code......
}
}
public class AnotherSubClass extends ASuperClass{
#Override
public void doSomething(String arg1, Integer areg2, String arg3, Integer arg4){
here is my code......
}
}
So, can you "generic-ize" the base class arg list so the above two classes would compile? Something like below? I know the syntax below is wrong, but can you make the arg list generic?
public class ASuperClass {
public abstract void doSomething(<?>);
}
Overall, my idea is to enforce consistency in an application where folks are extending some base functionality. Basically make sure that every sublass of ASuperClass has a doSomething() member function, to ensure consistency (method naming in particular) across subclasses, but it's arg list may be different per subclass.
Hopefully the above is not too confusing. Interested to know if this is possible.
Method parameters are a part of the method's "signature", which is used to decide overriding. In order for a subclass to override a method in its base class, to implement an abstract method of its abstract base class, or to implement an interface method, the entire signature, including the parameter types, must match.
If you would like the argument list to take a variable number of parameters, you can use this syntax:
public abstract void doSomething(Object ... args);
You can override a method with this signature in subclasses, and you can pass as many arguments as you wish. Some unfortunate consequence of this design decision are that
Your parameters are now untyped - each method needs to validate its list of parameters, including their type, before proceeding with the calculations, and
Parameters of primitive types need to be wrapped in Objects - primitives need to be boxed, or autoboxed.
public class ASuperClass
{
public abstract void doSomething(Object ... args);
}
public class ASubClass extends ASuperClass
{
#Override
public void doSomething(Object ... args)
{
here is my code......
}
}
This will allow both calls: doSomething(str1, int1) and doSomething(str1, int1, str2, int2);
If you define an abstract method like this it means that any object of a type that implements this abstract class is guaranteed to include this method.
For instance look at this code example.
List<ASuperClass> list = new ArrayList<ASuperClass>();
list.add(new ASubClass());
list.add(new AnotherSubClass());
So we create a list of objects all of type ASuperClass and then add two objects to this list. These objects are of different types but they both share this common parent. This means we can guarantee that they will respond to all of the method signatures on the ASuperClass type.
for(ASuperClass obj : list)
{
obj.doSomething("parm1", 1);
}
If you let the subtype determine the parameter list then this would break this. We would know that there was a method called doSomething but it would be useless to us because there would be no way of knowing what the parameters should be.
As others have suggested there is a hacky way of doing this.
public abstract void doSomething(Object ... params);
Whilst this is strictly true I'm sure nobody in their right mind would suggest actually doing this. Its only been suggested for completeness.
If on the other hand you wanted a variable number of parameters of the same type then you could use it more safely.
public abstract void doSomething(String parm1, Integer parm2, String ... params);
In this case you will always require a String, an Integer and then a variable number of strings. It's really equivalent to doing this.
public abstract void doSomething(String parm1, Integer parm2, String[] params);
Let's say I have a method called mymethod()
and this method overrides the method of the super class method.
What does it mean to override a method?
Does that mean mymethod() ignores everything that is in the method of the superclass, or does that means mymethod() also includes everything in the superclass method?
When overriding a method, can I only override the methods of the same name, or I can override methods of any name?
thanks.
An example:
public class Base {
public void saySomething() {
System.out.println("Hi, I'm a base class");
}
}
public class Child extends Base {
#Override
public void saySomething() {
System.out.println("Hi, I'm a child class");
}
}
Now assume we have a main function somewhere...
public static void main(String [] args) {
Base obj = new Child();
obj.saySomething();
}
When this runs, it will call Child's version of saySomething, because you overrode the parent's version by giving a new version of the function in Child.
The #Override annotation allows other developers (and you, when you forget) to know that this method overrides something in a base class/interface, and it also allows the compiler to yell at you if you're not actually overriding anything in a base class. For example, if you got the number of arguments wrong for a function, the compiler will give you an error saying your #Override is incorrect.
For example:
public class Child extends Base {
#Override
public void saySomething(int x) {
System.out.println("I'm a child and x is: " + x);
}
}
The compiler will yell at you because this version of saySomething takes an argument, but the parent's version doesn't have an argument, so you're #Override-ing something that's not in the parent.
On super
The Child version of saySomething will not invoke the Base version, you have to do it yourself with super.method().
For example:
public class Child extends Base {
#Override
public void saySomething() {
super.saySomething();
System.out.println("I'm also a child");
}
}
If you ran the main and used this Child class, it would print out I'm a base and I'm also a child.
Overriding means that when you call a method on your object, your object's method is called instead of the super class. The #Override annotation is something you use to make sure that you are overriding the correct method of the superclass. If you annotate a method that does not exist in the superclass, the Java compiler will give you an error. This way you can be sure that you are overriding the correct methods. This is especially useful in cases like this:
public class MyClass {
...
public boolean equals(MyClass myClass) {
...
}
}
There is a logic-bug in the code above. You haven't actually overridden the Object class's equals method. If you add the #Override annotation:
public class MyClass {
...
#Override
public boolean equals(MyClass myClass) {
...
}
}
The Java compiler will now complain because there is no corresponding method in the parent class. You'll then know that the correct solution is:
public class MyClass {
...
#Override
public boolean equals(Object o) {
...
}
}
To call the parent class's method, you can call super.overriddenMethod() where overriddenMethod is the name of the method you have overridden. So if you want to do something in addition to what the parent class already does, you can do something like this:
public class MyClass {
...
#Override
public void overriddenMethod() {
super.overriddenMethod();
/* whatever additional stuff you want to do */
}
}
If an inheriting class has on override method of the same name as the parent class it will be called instead of the one in the parent class. This only works if the names are the same, and of course if the signature of the method matches your call to the method.
What does it mean to override a method?
It means you replace the super class definition of the method with your own definition.
does that mean mymethod() ignores everything that is in the method of the super class?
or does that means mymethod() also includes everything in the superclass method?
You can choose whether to include the super class definition within your definition. To include it, you need to call super.mymethod() within mymethod().
and when overriding a method, can I only override the methods of the same name, or I can override methods of any name?
To override a method, you must supply a method in the sub class with the same signature (which means the same name, parameters and return type).
As a side note, the #Override annotation in your question does not actually cause your method to override another method. It causes a compile-time error if a method annotated with it does not have a signature matching a public or protected method of a super class (or interface as of 1.6).
I once had a student come to ask me why his code wasn't working. He had spent several days wondering why he could put something into a collection but was not able to find it. His code was something like:
public int hashcode()
instead of:
public int hashCode()
So the hashCode method never got called.
Adding #Overrides to a method makes it clear that you are overriding the method AND make sure that you really are overriding a method.
When you override a method of the super class, calling that method on your object calls its method instead of that of the super class.
You can call the super class's method (despite having overridden it) using super.methodName(). A common reason for this is when the overridden method would otherwise reimplement the super class method and add additional code specific to the extending class (public void methodName() { super.methodName(); /* more code */ }).
#Override annotation allows you to cause warning at compile time if the method isn't actually overriding anything. It isn't necessary, but these warning are a hit to you that you might have got the signature wrong in the extending class, forgot to implement the method at all in the super class, or some other silly mistake.
I have an abstract class that has a generic method and I want to override the generic method by substituting specific types for the generic parameter. So in pseudo-code I have the following:
public abstract class GetAndParse {
public SomeClass var;
public abstract <T extends AnotherClass> void getAndParse(T... args);
}
public class Implementor extends GetAndParse {
// some field declarations
// some method declarations
#Override
public <SpecificClass> void getAndParse(SpecificClass... args) {
// method body making use of args
}
}
But for some reason I'm not allowed to do this? Am I making some kind of syntax error or is this kind of inheritance and overriding not allowed? Specifically I'm getting an error about #Override because the eclipse IDE keeps reminding me to implement getAndParse.
Here's how I want the above code to work. Somewhere else in my code there is a method that expects instances of objects that implement GetAndParse which specifically means that they have a getAndParse method that I can use. When I call getAndParse on that instance the compiler checks to see whether I have used specific instances of T in the proper way, so in particular T should extend AnotherClass and it should be SpecificClass.
What we are having here is two different methods with individual type parameters each.
public abstract <T extends AnotherClass> void getAndParse(Args... args);
This is a method with a type parameter named T, and bounded by AnotherClass, meaning each subtype of AnotherClass is allowed as a type parameter.
public <SpecificClass> void getAndParse(Args... args)
This is a method with a type parameter named SpecificClass, bounded by Object (meaning each type is allowed as a type parameter). Do you really want this?
Is the type parameter used inside Args? I think the problem would be there.
The meaning of
public abstract <T extends AnotherClass> void getAndParse(T... args);
is that the caller of the method can decide with which type parameter he wants to call the method, as long as this is some subtype of AnotherClass. This means that in effect the method can be called with any objects of type AnotherClass.
Since the caller can decide the type parameter, you can't in a subclass narrow down the parameter type to SpecificClass - this would not be an implementation of the method, but another method with same name (overloading).
Maybe you want something like this:
public abstract class GetAndParse<T extends AnotherClass> {
public SomeClass var;
public abstract void getAndParse(T... args);
}
public class Implementor extends GetAndParse<SpecificClass> {
// some field declarations
// some method declarations
#Override
public void getAndParse(SpecificClass... args) {
// method body making use of args
}
}
Now the getAndParse method implements the parent class' method.
You are seeing this problem because of the concept called "Erasure" in Java Generics.
Java uses "erasure" to support backward compatibility. i.e Java code which did not use generics.
Erasure Procedure:
The compiler will first do a type checking and then it will remove(erase) all the type parameters as much as possible, and also insert TypeCasting where ever necessary.
example:
public abstract <T extends AnotherClass> void getAndParse(T paramAnotherClass);
will become
public abstract void getAndParse(AnotherClass paramAnotherClass);
In class "Implementor.java",
The code
public <SpecificClass> void getAndParse(T paramAnotherClass)
will become
public void getAndParse(SpecificClass paramAnotherClass){ }
the compiler will see that you have not implemented the abstract method correctly.
There is a type mismatch between the abstract method and the implemented method. This is why you are seeing the error.
More details can be found here.
http://today.java.net/pub/a/today/2003/12/02/explorations.html
You cannot override to specific type T because there is in fact (at the bytecode level if you wish) only one method getAndParse because of type erasure (see other answer):
public abstract void getAndParse(AnotherClass... args); // (1)
For every type of T, the same method is used.
You can overload it (I think):
public void getAndParse(SpecificClass... args); // (2)
but this will not a different method from (1) ant it will not be called by generic code:
T x = whatever;
object.getAndParse(x); // Calls (1) even if T is derived from SpecificClass
No, it's not valid. What would happen if someone with a GetAndParse reference called it with a different class extending AnotherClass?
That becomes a nonsense when someone has a reference to type GetAndParse and tries to call the getAndParse method. If Cat and Dog extend AnotherClass. I should expect to be able to call GetAndParse#getAndParse with either a Cat or a Dog. But the implementation has tried to restrict it and make it less compatible!
Static method can't override
class Vehicle{
static void park(int location){
System.out.println("Vehicle parking..");
}}
class Car extends Vehicle{
#Override //error
void park(int location) { //error
System.out.println("Car Parking..");
}}
Private method can't override
class Vehicle{
private void park(int location){
System.out.println("Vehicle parking..");
}
void callPark(){
park(100);
}}
class Car extends Vehicle{
//#Override
void park(int location) {
System.out.println("Car Parking..");
}}
class Demo {
public static void main(String[] args) {
Vehicle v1=new Car();
v1.callPark();
}}
Final method can't override
class Vehicle{
final void park(int location){
System.out.println("Vehicle parking..");
}}
class Car extends Vehicle{
//#Override
void park(int location) { //error
System.out.println("Car Parking..");
}}