Why can't I access static methods of an interface using an instance variable.
public class TestClass {
public static void main(String[] args) {
AWD a = new Car();
a.isRearWheelDrive(); //doesn't compile
}
}
interface AWD {
static boolean isRearWheelDrive() {
return false;
}
}
class Car implements AWD {
}
Static Interface Methods Aren't Inherited by Subclasses
You can't access static methods of interfaces through instances. You have to access them statically. This is a bit different from classes where accessing a static method through an instance is allowed, but often flagged as a code smell; static methods should be accessed statically.
That's because static methods of classes are inherited by subclasses, but static methods of interfaces aren't. That's stated in §8.4.8 of the specification:
8.4.8. Inheritance, Overriding, and Hiding
…
A class does not inherit static methods from its superinterfaces.
When you are looking up the accessible methods for the instance, the static method from the interface isn't among them.
Options for the code
So, as the code is now, you need to access the method statically:
AWD.isRearWheelDrive()
However, it seems like you want this to be an instance method, in which case you should probably be using a default method that returns false:
interface AWD {
default boolean isRearWheelDrive() {
return false;
}
}
Even that seems a little bit odd, though. It seems like you'd probably want that default method to be overriding some non-default method in a super-interface. That is, you probably want something like:
interface HasDriveWheels {
boolean isRearWheelDrive();
}
interface AllWheelDrive extends HasDriveWheels {
#Override
default boolean isRearWheelDrive() {
return false;
}
}
This is specified in The Java® Language Specification, §15.12.3. Compile-Time Step 3: Is the Chosen Method Appropriate?
If the form is ExpressionName . [TypeArguments] Identifier or Primary . [TypeArguments] Identifier, then the compile-time declaration must not be a static method declared in an interface, or a compile-time error occurs.
In retrospect, the ability to invoke a static method through an instance is of little use, even less than the inheritance of static methods. I’m quite sure that a lot of developers consider it a design mistake that is only kept due to compatibility reasons.
For the newer feature of static methods in an interface, there was no compatibility constraint that required repeating this mistake, hence, the rules for static methods in interfaces were designed differently. This is also the solution with the smallest impact on compatibility with old code.
Well this type of questions have been answered already but this confused me. I have spent lot of time going through lots of answer so please go through my question before making it duplicate.
case 1 -
static polymorphism - function overloading
dynamic polymorphism - function overriding
links -
https://stackoverflow.com/a/12894211/3181738
https://stackoverflow.com/a/20783390/3181738
and many others.
My confusion is that in all the examples dynamic polymorphism is shown using upcasting. What if I don't upcast.
class A{
public void show(){
System.out.print("A");
}
}
class B extends A{
public void show(){
System.out.print("B");
}
public static void main(String[] s){
A a = new B();
a.show(); // upcasting. It is dynamic polymorphism.
B b = new B();
b.show(); // Now java compiler can decide so is it still dynamic polymorphism?
}
}
case 2 -
static polymorphism is achieved via overloading if method is private , static or final.
So what about overloading of public and default methods?
Static methods can only be overloaded.
Non-static methods can be both overridden and overloaded.
public/protected/private doesn't affect this directly, except that since private methods can't be seen from subclasses, they can't be overridden either. The exact rules are defined here
And if you want to prevent the override of a non-static method, you can declare it final. (private methods are therefore implicitly final.)
Q1 upcasting: Upcasting and treating a subclass object as if it were a parent class object is the essence of polymorphism. So your question, "What if I don't upcast?" Then you are not using polymorphism
Q2 static polymorphism - method overridding: This is what it's saying:
static: static methods are overwritable, since they are associated with the class, not the object. So for instance, if you have a class with a main method, and then you subclass that, and add a main method to the subclass, when you execute the main method in the subclass, it is statically overridden
private: private methods are not inherited and cannot be called in the subclass, unless that are statically overridden
final: final methods are not overriddable. Therefore, redefining them in a subclass creates a statically overridden method
Case 1
At compile time, the compiler will identify the method signature that matches ie. it will be show() with no parameters in your example. Whether it is A's or B's method that gets invoked though is still runtime even though in your example you are sure it is a B.
Case 2
I think that the visibility rules of methods maybe causing some confusion for you.
The decision of which overloaded method that will be called at runtime will be decided at compile time ie. overloaded method is statically decided - this is regardless of whether it is public, package-private (default), or private. If the object, in which the method call occurs, does not have visibility of the method it fails at compile time.
As others have said - overriding does not apply to static methods since it is part of the class not the instance.
Why is it not possible to override static methods?
If possible, please use an example.
Overriding depends on having an instance of a class. The point of polymorphism is that you can subclass a class and the objects implementing those subclasses will have different behaviors for the same methods defined in the superclass (and overridden in the subclasses). A static method is not associated with any instance of a class so the concept is not applicable.
There were two considerations driving Java's design that impacted this. One was a concern with performance: there had been a lot of criticism of Smalltalk about it being too slow (garbage collection and polymorphic calls being part of that) and Java's creators were determined to avoid that. Another was the decision that the target audience for Java was C++ developers. Making static methods work the way they do had the benefit of familiarity for C++ programmers and was also very fast, because there's no need to wait until runtime to figure out which method to call.
Personally I think this is a flaw in the design of Java. Yes, yes, I understand that non-static methods are attached to an instance while static methods are attached to a class, etc etc. Still, consider the following code:
public class RegularEmployee {
private BigDecimal salary;
public void setSalary(BigDecimal salary) {
this.salary = salary;
}
public static BigDecimal getBonusMultiplier() {
return new BigDecimal(".02");
}
public BigDecimal calculateBonus() {
return salary.multiply(getBonusMultiplier());
}
/* ... presumably lots of other code ... */
}
public class SpecialEmployee extends RegularEmployee {
public static BigDecimal getBonusMultiplier() {
return new BigDecimal(".03");
}
}
This code will not work as you might expect. Namely, SpecialEmployee's get a 2% bonus just like regular employees. But if you remove the "static"s, then SpecialEmployee's get a 3% bonus.
(Admittedly, this example is poor coding style in that in real life you would likely want the bonus multiplier to be in a database somewhere rather than hard-coded. But that's just because I didn't want to bog down the example with a lot of code irrelevant to the point.)
It seems quite plausible to me that you might want to make getBonusMultiplier static. Perhaps you want to be able to display the bonus multiplier for all the categories of employees, without needing to have an instance of an employee in each category. What would be the point of searching for such example instances? What if we are creating a new category of employee and don't have any employees assigned to it yet? This is quite logically a static function.
But it doesn't work.
And yes, yes, I can think of any number of ways to rewrite the above code to make it work. My point is not that it creates an unsolvable problem, but that it creates a trap for the unwary programmer, because the language does not behave as I think a reasonable person would expect.
Perhaps if I tried to write a compiler for an OOP language, I would quickly see why implementing it so that static functions can be overriden would be difficult or impossible.
Or perhaps there is some good reason why Java behaves this way. Can anyone point out an advantage to this behavior, some category of problem that is made easier by this? I mean, don't just point me to the Java language spec and say "see, this is documented how it behaves". I know that. But is there a good reason why it SHOULD behave this way? (Besides the obvious "making it work right was too hard"...)
Update
#VicKirk: If you mean that this is "bad design" because it doesn't fit how Java handles statics, my reply is, "Well, duh, of course." As I said in my original post, it doesn't work. But if you mean that it is bad design in the sense that there would be something fundamentally wrong with a language where this worked, i.e. where statics could be overridden just like virtual functions, that this would somehow introduce an ambiguity or it would be impossible to implement efficiently or some such, I reply, "Why? What's wrong with the concept?"
I think the example I give is a very natural thing to want to do. I have a class that has a function that does not depend on any instance data, and which I might very reasonably want to call independent of an instance, as well as wanting to call from within an instance method. Why should this not work? I've run into this situation a fair number of times over the years. In practice I get around it by making the function virtual, and then creating a static method whose only purpose in life is to be a static method that passes the call on to the virtual method with a dummy instance. That seems like a very roundabout way to get there.
The short answer is: it is entirely possible, but Java doesn't do it.
Here is some code which illustrates the current state of affairs in Java:
File Base.java:
package sp.trial;
public class Base {
static void printValue() {
System.out.println(" Called static Base method.");
}
void nonStatPrintValue() {
System.out.println(" Called non-static Base method.");
}
void nonLocalIndirectStatMethod() {
System.out.println(" Non-static calls overridden(?) static:");
System.out.print(" ");
this.printValue();
}
}
File Child.java:
package sp.trial;
public class Child extends Base {
static void printValue() {
System.out.println(" Called static Child method.");
}
void nonStatPrintValue() {
System.out.println(" Called non-static Child method.");
}
void localIndirectStatMethod() {
System.out.println(" Non-static calls own static:");
System.out.print(" ");
printValue();
}
public static void main(String[] args) {
System.out.println("Object: static type Base; runtime type Child:");
Base base = new Child();
base.printValue();
base.nonStatPrintValue();
System.out.println("Object: static type Child; runtime type Child:");
Child child = new Child();
child.printValue();
child.nonStatPrintValue();
System.out.println("Class: Child static call:");
Child.printValue();
System.out.println("Class: Base static call:");
Base.printValue();
System.out.println("Object: static/runtime type Child -- call static from non-static method of Child:");
child.localIndirectStatMethod();
System.out.println("Object: static/runtime type Child -- call static from non-static method of Base:");
child.nonLocalIndirectStatMethod();
}
}
If you run this (I did it on a Mac, from Eclipse, using Java 1.6) you get:
Object: static type Base; runtime type Child.
Called static Base method.
Called non-static Child method.
Object: static type Child; runtime type Child.
Called static Child method.
Called non-static Child method.
Class: Child static call.
Called static Child method.
Class: Base static call.
Called static Base method.
Object: static/runtime type Child -- call static from non-static method of Child.
Non-static calls own static.
Called static Child method.
Object: static/runtime type Child -- call static from non-static method of Base.
Non-static calls overridden(?) static.
Called static Base method.
Here, the only cases which might be a surprise (and which the question is about) appear to be the first case:
"The run-time type is not used to determine which static methods are called, even when called with an object instance (obj.staticMethod())."
and the last case:
"When calling a static method from within an object method of a class, the static method chosen is the one accessible from the class itself and not from the class defining the run-time type of the object."
Calling with an object instance
The static call is resolved at compile-time, whereas a non-static method call is resolved at run-time. Notice that although static methods are inherited (from parent) they are not overridden (by child). This could be a surprise if you expected otherwise.
Calling from within an object method
Object method calls are resolved using the run-time type, but static (class) method calls are resolved using the compile-time (declared) type.
Changing the rules
To change these rules, so that the last call in the example called Child.printValue(), static calls would have to be provided with a type at run-time, rather than the compiler resolving the call at compile-time with the declared class of the object (or context). Static calls could then use the (dynamic) type hierarchy to resolve the call, just as object method calls do today.
This would easily be doable (if we changed Java :-O), and is not at all unreasonable, however, it has some interesting considerations.
The main consideration is that we need to decide which static method calls should do this.
At the moment, Java has this "quirk" in the language whereby obj.staticMethod() calls are replaced by ObjectClass.staticMethod() calls (normally with a warning). [Note: ObjectClass is the compile-time type of obj.] These would be good candidates for overriding in this way, taking the run-time type of obj.
If we did it would make method bodies harder to read: static calls in a parent class could potentially be dynamically "re-routed". To avoid this we would have to call the static method with a class name -- and this makes the calls more obviously resolved with the compile-time type hierarchy (as now).
The other ways of invoking a static method are more tricky: this.staticMethod() should mean the same as obj.staticMethod(), taking the run-time type of this. However, this might cause some headaches with existing programs, which call (apparently local) static methods without decoration (which is arguably equivalent to this.method()).
So what about unadorned calls staticMethod()? I suggest they do the same as today, and use the local class context to decide what to do. Otherwise great confusion would ensue. Of course it means that method() would mean this.method() if method was a non-static method, and ThisClass.method() if method were a static method. This is another source of confusion.
Other considerations
If we changed this behaviour (and made static calls potentially dynamically non-local), we would probably want to revisit the meaning of final, private and protected as qualifiers on static methods of a class. We would then all have to get used to the fact that private static and public final methods are not overridden, and can therefore be safely resolved at compile-time, and are "safe" to read as local references.
Actually we were wrong.
Despite Java doesn't allow you to override static methods by default, if you look thoroughly through documentation of Class and Method classes in Java, you can still find a way to emulate static methods overriding by following workaround:
import java.lang.reflect.InvocationTargetException;
import java.math.BigDecimal;
class RegularEmployee {
private BigDecimal salary = BigDecimal.ONE;
public void setSalary(BigDecimal salary) {
this.salary = salary;
}
public static BigDecimal getBonusMultiplier() {
return new BigDecimal(".02");
}
public BigDecimal calculateBonus() {
return salary.multiply(this.getBonusMultiplier());
}
public BigDecimal calculateOverridenBonus() {
try {
// System.out.println(this.getClass().getDeclaredMethod(
// "getBonusMultiplier").toString());
try {
return salary.multiply((BigDecimal) this.getClass()
.getDeclaredMethod("getBonusMultiplier").invoke(this));
} catch (IllegalAccessException e) {
e.printStackTrace();
} catch (IllegalArgumentException e) {
e.printStackTrace();
} catch (InvocationTargetException e) {
e.printStackTrace();
}
} catch (NoSuchMethodException e) {
e.printStackTrace();
} catch (SecurityException e) {
e.printStackTrace();
}
return null;
}
// ... presumably lots of other code ...
}
final class SpecialEmployee extends RegularEmployee {
public static BigDecimal getBonusMultiplier() {
return new BigDecimal(".03");
}
}
public class StaticTestCoolMain {
static public void main(String[] args) {
RegularEmployee Alan = new RegularEmployee();
System.out.println(Alan.calculateBonus());
System.out.println(Alan.calculateOverridenBonus());
SpecialEmployee Bob = new SpecialEmployee();
System.out.println(Bob.calculateBonus());
System.out.println(Bob.calculateOverridenBonus());
}
}
Resulting output:
0.02
0.02
0.02
0.03
what we were trying to achieve :)
Even if we declare third variable Carl as RegularEmployee and assign to it instance of SpecialEmployee, we will still have call of RegularEmployee method in first case and call of SpecialEmployee method in second case
RegularEmployee Carl = new SpecialEmployee();
System.out.println(Carl.calculateBonus());
System.out.println(Carl.calculateOverridenBonus());
just look at output console:
0.02
0.03
;)
Static methods are treated as global by the JVM, there are not bound to an object instance at all.
It could conceptually be possible if you could call static methods from class objects (like in languages like Smalltalk) but it's not the case in Java.
EDIT
You can overload static method, that's ok. But you can not override a static method, because class are no first-class object. You can use reflection to get the class of an object at run-time, but the object that you get does not parallel the class hierarchy.
class MyClass { ... }
class MySubClass extends MyClass { ... }
MyClass obj1 = new MyClass();
MySubClass obj2 = new MySubClass();
ob2 instanceof MyClass --> true
Class clazz1 = obj1.getClass();
Class clazz2 = obj2.getClass();
clazz2 instanceof clazz1 --> false
You can reflect over the classes, but it stops there. You don't invoke a static method by using clazz1.staticMethod(), but using MyClass.staticMethod(). A static method is not bound to an object and there is hence no notion of this nor super in a static method. A static method is a global function; as a consequence there is also no notion of polymorphism and, therefore, method overriding makes no sense.
But this could be possible if MyClass was an object at run-time on which you invoke a method, as in Smalltalk (or maybe JRuby as one comment suggest, but I know nothing of JRuby).
Oh yeah... one more thing. You can invoke a static method through an object obj1.staticMethod() but that really syntactic sugar for MyClass.staticMethod() and should be avoided. It usually raises a warning in modern IDE. I don't know why they ever allowed this shortcut.
Method overriding is made possible by dynamic dispatching, meaning that the declared type of an object doesn't determine its behavior, but rather its runtime type:
Animal lassie = new Dog();
lassie.speak(); // outputs "woof!"
Animal kermit = new Frog();
kermit.speak(); // outputs "ribbit!"
Even though both lassie and kermit are declared as objects of type Animal, their behavior (method .speak()) varies because dynamic dispatching will only bind the method call .speak() to an implementation at run time - not at compile time.
Now, here's where the static keyword starts to make sense: the word "static" is an antonym for "dynamic". So the reason why you can't override static methods is because there is no dynamic dispatching on static members - because static literally means "not dynamic". If they dispatched dynamically (and thus could be overriden) the static keyword just wouldn't make sense anymore.
Yes. Practically Java allows overriding static method, and No theoretically if you Override a static method in Java then it will compile and run smoothly but it will lose Polymorphism which is the basic property of Java. You will Read Everywhere that it is not possible to try yourself compiling and running. you will get your answer. e.g. If you Have Class Animal and a static method eat() and you Override that static method in its Subclass lets called it Dog. Then when wherever you Assign a Dog object to an Animal Reference and call eat() according to Java Dog's eat() should have been called but in static Overriding Animals' eat() will Be Called.
class Animal {
public static void eat() {
System.out.println("Animal Eating");
}
}
class Dog extends Animal{
public static void eat() {
System.out.println("Dog Eating");
}
}
class Test {
public static void main(String args[]) {
Animal obj= new Dog();//Dog object in animal
obj.eat(); //should call dog's eat but it didn't
}
}
Output Animal Eating
According to Polymorphism Principle of Java, the Output Should be Dog Eating.
But the result was different because to support Polymorphism Java uses Late Binding that means methods are called only at the run-time but not in the case of static methods. In static methods compiler calls methods at the compile time rather than the run-time, so we get methods according to the reference and not according to the object a reference a containing that's why You can say Practically it supports static overring but theoretically, it doesn't.
In Java (and many OOP languages, but I cannot speak for all; and some do not have static at all) all methods have a fixed signature - the parameters and types. In a virtual method, the first parameter is implied: a reference to the object itself and when called from within the object, the compiler automatically adds this.
There is no difference for static methods - they still have a fixed signature. However, by declaring the method static you have explicitly stated that the compiler must not include the implied object parameter at the beginning of that signature. Therefore, any other code that calls this must must not attempt to put a reference to an object on the stack. If it did do that, then the method execution would not work since the parameters would be in the wrong place - shifted by one - on the stack.
Because of this difference between the two; virtual methods always have a reference to the context object (i.e. this) so then it is possible to reference anything within the heap that belong to that instance of the object. But with static methods, since there is no reference passed, that method cannot access any object variables and methods since the context is not known.
If you wish that Java would change the definition so that a object context is passed in for every method, static or virtual, then you would in essence have only virtual methods.
As someone asked in a comment to the op - what is your reason and purpose for wanting this feature?
I do not know Ruby much, as this was mentioned by the OP, I did some research. I see that in Ruby classes are really a special kind of object and one can create (even dynamically) new methods. Classes are full class objects in Ruby, they are not in Java. This is just something you will have to accept when working with Java (or C#). These are not dynamic languages, though C# is adding some forms of dynamic. In reality, Ruby does not have "static" methods as far as I could find - in that case these are methods on the singleton class object. You can then override this singleton with a new class and the methods in the previous class object will call those defined in the new class (correct?). So if you called a method in the context of the original class it still would only execute the original statics, but calling a method in the derived class, would call methods either from the parent or sub-class. Interesting and I can see some value in that. It takes a different thought pattern.
Since you are working in Java, you will need to adjust to that way of doing things. Why they did this? Well, probably to improve performance at the time based on the technology and understanding that was available. Computer languages are constantly evolving. Go back far enough and there is no such thing as OOP. In the future, there will be other new ideas.
EDIT: One other comment. Now that I see the differences and as I Java/C# developer myself, I can understand why the answers you get from Java developers may be confusing if you are coming from a language like Ruby. Java static methods are not the same as Ruby class methods. Java developers will have a hard time understanding this, as will conversely those who work mostly with a language like Ruby/Smalltalk. I can see how this would also be greatly confusing by the fact that Java also uses "class method" as another way to talk about static methods but this same term is used differently by Ruby. Java does not have Ruby style class methods (sorry); Ruby does not have Java style static methods which are really just old procedural style functions, as found in C.
By the way - thanks for the question! I learned something new for me today about class methods (Ruby style).
Well... the answer is NO if you think from the perspective of how an overriden method should behave in Java. But, you don't get any compiler error if you try to override a static method. That means, if you try to override, Java doesn't stop you doing that; but you certainly don't get the same effect as you get for non-static methods. Overriding in Java simply means that the particular method would be called based on the run time type of the object and not on the compile time type of it (which is the case with overriden static methods). Okay... any guesses for the reason why do they behave strangely? Because they are class methods and hence access to them is always resolved during compile time only using the compile time type information. Accessing them using object references is just an extra liberty given by the designers of Java and we should certainly not think of stopping that practice only when they restrict it :-)
Example: let's try to see what happens if we try overriding a static method:-
class SuperClass {
// ......
public static void staticMethod() {
System.out.println("SuperClass: inside staticMethod");
}
// ......
}
public class SubClass extends SuperClass {
// ......
// overriding the static method
public static void staticMethod() {
System.out.println("SubClass: inside staticMethod");
}
// ......
public static void main(String[] args) {
// ......
SuperClass superClassWithSuperCons = new SuperClass();
SuperClass superClassWithSubCons = new SubClass();
SubClass subClassWithSubCons = new SubClass();
superClassWithSuperCons.staticMethod();
superClassWithSubCons.staticMethod();
subClassWithSubCons.staticMethod();
// ...
}
}
Output:-
SuperClass: inside staticMethod
SuperClass: inside staticMethod
SubClass: inside staticMethod
Notice the second line of the output. Had the staticMethod been overriden this line should have been identical to the third line as we're invoking the 'staticMethod()' on an object of Runtime Type as 'SubClass' and not as 'SuperClass'. This confirms that the static methods are always resolved using their compile time type information only.
I like and double Jay's comment (https://stackoverflow.com/a/2223803/1517187).
I agree that this is the bad design of Java.
Many other languages support overriding static methods, as we see in previous comments.
I feel Jay has also come to Java from Delphi like me.
Delphi (Object Pascal) was one of the languages implementing OOP before Java and one of the first languages used for commercial application development.
It is obvious that many people had experience with that language since it was in the past the only language to write commercial GUI products. And - yes, we could in Delphi override static methods. Actually, static methods in Delphi are called "class methods", while Delphi had the different concept of "Delphi static methods" which were methods with early binding. To override methods you had to use late binding, declare "virtual" directive. So it was very convenient and intuitive and I would expect this in Java.
In general it doesn't make sense to allow 'overriding' of static methods as there would be no good way to determine which one to call at runtime. Taking the Employee example, if we call RegularEmployee.getBonusMultiplier() - which method is supposed to be executed?
In the case of Java, one could imagine a language definition where it is possible to 'override' static methods as long as they are called through an object instance. However, all this would do is to re-implement regular class methods, adding redundancy to the language without really adding any benefit.
overriding is reserved for instance members to support polymorphic behaviour. static class members do not belong to a particular instance. instead, static members belong to the class and as a result overriding is not supported because subclasses only inherit protected and public instance members and not static members. You may want to define an inerface and research factory and/or strategy design patterns to evaluate an alternate approach.
By overriding we can create a polymorphic nature depending on the object type. Static method has no relation with object. So java can not support static method overriding.
By overriding, you achieve dynamic polymorphism.
When you say overriding static methods, the words you are trying to use are contradictory.
Static says - compile time, overriding is used for dynamic polymorphism.
Both are opposite in nature, and hence can't be used together.
Dynamic polymorphic behavior comes when a programmer uses an object and accessing an instance method. JRE will map different instance methods of different classes based on what kind of object you are using.
When you say overriding static methods, static methods we will access by using the class name, which will be linked at compile time, so there is no concept of linking methods at runtime with static methods. So the term "overriding" static methods itself doesn't make any meaning.
Note: even if you access a class method with an object, still java compiler is intelligent enough to find it out, and will do static linking.
Overriding in Java simply means that the particular method would be called based on the runtime type
of the object and not on the compile-time type of it (which is the case with overridden static methods). As static methods are class methods they are not instance methods so they have nothing to do with the fact which reference is pointing to which Object or instance, because due to the nature of static method it belongs to a specific class. You can redeclare it in the subclass but that subclass won't know anything about the parent class' static methods because, as I said, it is specific to only that class in which it has been declared. Accessing them using object references is just an extra liberty given by the designers of Java and we should certainly not think of stopping that practice only when they restrict it
more details and example
http://faisalbhagat.blogspot.com/2014/09/method-overriding-and-method-hiding.html
What good will it do to override static methods. You cannot call static methods through an instance.
MyClass.static1()
MySubClass.static1() // If you overrode, you have to call it through MySubClass anyway.
EDIT : It appears that through an unfortunate oversight in language design, you can call static methods through an instance. Generally nobody does that. My bad.
Answer of this question is simple, the method or variable marked as static belongs to the class only, So that static method cannot be inherited in the sub class because they belong to the super class only.
Easy solution: Use singleton instance. It will allow overrides and inheritance.
In my system, I have SingletonsRegistry class, which returns instance for passed Class. If instance is not found, it is created.
Haxe language class:
package rflib.common.utils;
import haxe.ds.ObjectMap;
class SingletonsRegistry
{
public static var instances:Map<Class<Dynamic>, Dynamic>;
static function __init__()
{
StaticsInitializer.addCallback(SingletonsRegistry, function()
{
instances = null;
});
}
public static function getInstance(cls:Class<Dynamic>, ?args:Array<Dynamic>)
{
if (instances == null) {
instances = untyped new ObjectMap<Dynamic, Dynamic>();
}
if (!instances.exists(cls))
{
if (args == null) args = [];
instances.set(cls, Type.createInstance(cls, args));
}
return instances.get(cls);
}
public static function validate(inst:Dynamic, cls:Class<Dynamic>)
{
if (instances == null) return;
var inst2 = instances[cls];
if (inst2 != null && inst != inst2) throw "Can\'t create multiple instances of " + Type.getClassName(cls) + " - it's singleton!";
}
}
A Static method, variable, block or nested class belongs to the entire class rather than an object.
A Method in Java is used to expose the behaviour of an Object / Class. Here, as the method is static (i.e, static method is used to represent the behaviour of a class only.) changing/ overriding the behaviour of entire class will violate the phenomenon of one of the fundamental pillar of Object oriented programming i.e, high cohesion. (remember a constructor is a special kind of method in Java.)
High Cohesion - One class should have only one role. For example: A car class should produce only car objects and not bike, trucks, planes etc. But the Car class may have some features(behaviour) that belongs to itself only.
Therefore, while designing the java programming language. The language designers thought to allow developers to keep some behaviours of a class to itself only by making a method static in nature.
The below piece code tries to override the static method, but will not encounter any compilation error.
public class Vehicle {
static int VIN;
public static int getVehileNumber() {
return VIN;
}}
class Car extends Vehicle {
static int carNumber;
public static int getVehileNumber() {
return carNumber;
}}
This is because, here we are not overriding a method but we are just re-declaring it. Java allows re-declaration of a method (static/non-static).
Removing the static keyword from getVehileNumber() method of Car class will result into compilation error, Since, we are trying to change the functionality of static method which belongs to Vehicle class only.
Also, If the getVehileNumber() is declared as final then the code will not compile, Since the final keyword restricts the programmer from re-declaring the method.
public static final int getVehileNumber() {
return VIN; }
Overall, this is upto software designers for where to use the static methods.
I personally prefer to use static methods to perform some actions without creating any instance of a class. Secondly, to hide the behaviour of a class from outside world.
Here is a simple explanation. A static method is associated with a class while an instance method is associated with a particular object. Overrides allow calling the different implementation of the overridden methods associated with the particular object. So it is counter-intuitive to override static method which is not even associated with objects but the class itself in the first place. So static methods cannot be overridden based on what object is calling it, it will always be associated with the class where it was created.
Now seeing above answers everyone knows that we can't override static methods, but one should not misunderstood about the concept of accessing static methods from subclass.
We can access static methods of super class with subclass reference if this static method has not been hidden by new static method defined in sub class.
For Example, see below code:-
public class StaticMethodsHiding {
public static void main(String[] args) {
SubClass.hello();
}
}
class SuperClass {
static void hello(){
System.out.println("SuperClass saying Hello");
}
}
class SubClass extends SuperClass {
// static void hello() {
// System.out.println("SubClass Hello");
// }
}
Output:-
SuperClass saying Hello
See Java oracle docs and search for What You Can Do in a Subclass for details about hiding of static methods in sub class.
Thanks
The following code shows that it is possible:
class OverridenStaticMeth {
static void printValue() {
System.out.println("Overriden Meth");
}
}
public class OverrideStaticMeth extends OverridenStaticMeth {
static void printValue() {
System.out.println("Overriding Meth");
}
public static void main(String[] args) {
OverridenStaticMeth osm = new OverrideStaticMeth();
osm.printValue();
System.out.println("now, from main");
printValue();
}
}
I know that a child cannot reduce the visibility of a non-static method and I understand why it is so.
I've read however that "static method can be hidden through its redeclaration". I however do not understand how this could be achieved in Java.
Is this really possible? If yes, how to do that (code example) and why was it introduced (it seems to contradict the principle of non-reducing the visibility of the interface)?
The short answer is: no, it is not possible. You have mixed up some terminology. Hiding has nothing to do with accessibility (which is what you are really asking about, not visibility, which is related to scope and shadowing and is discussed in Chapter 6 of the Java Language Specification (JLS)).
Now for the longer answer. The term overriding applies to instance methods, while the term hiding applies to class (static) methods. From the Java Tutorial topic Overriding and Hiding Methods:
The distinction between hiding a static method and overriding an instance method has important implications:
The version of the overridden instance method that gets invoked is the one in the subclass.
The version of the hidden static method that gets invoked depends on whether it is invoked from the superclass or the subclass.
Some of the other answers here provide incorrect examples about method hiding, so let's go back to the JLS, this time to §8.4.8:
Methods are overridden or hidden on a signature-by-signature basis.
That is, to override or hide a method in the parent class, the subclass must define a method with the same signature—basically, the same number and type of arguments (although generics and type erasure makes the rules a little more complicated than that). There are also rules about return types and throws clauses, but those seem irrelevant to this question.
Note that you can define a method in a subclass with the same name as a method in the parent class (or in an implemented interface) but with different number or type of arguments. In that case, you are overloading the method name and neither overriding nor hiding anything; the subclass method is a new method, pretty much independent of the inherited method(s). (There is an interaction when the compiler has to match methods to method calls, but that's about it.)
Now to your question: the terms accessibility and hiding (as well as visibility) are independent concepts in Java. There is, as you put it, a "principle" that there is simply no way for a subclass to reduce the accessibility of an inherited method. This applies regardless of whether you are overriding an instance method or hiding a class method. From the JLS §8.4.8.3:
The access modifier (§6.6) of an overriding or hiding method must provide at least as much access as the overridden or hidden method, as follows:
If the overridden or hidden method is public, then the overriding or hiding method must be public; otherwise, a compile-time error occurs.
If the overridden or hidden method is protected, then the overriding or hiding method must be protected or public; otherwise, a compile-time error occurs.
If the overridden or hidden method has default (package) access, then the overriding or hiding method must not be private; otherwise, a compile-time error occurs.
In summary, the fact that a static method can be hidden has nothing to do with changing the accessibility of the method.
Based on hagubear's valuable comments, it seems that the author of a statement meant hiding a method through overloading it with a method having the same declaration.
Quoting this link:
We can declare static methods with same signature in subclass, but it
is not considered overriding as there won’t be any run-time
polymorphism. (...) If a derived class defines a
static method with same signature as a static method in base class,
the method in the derived class hides the method in the base class.
Thus, defining a method in a child class having exact same declaration effectively hides the original method in child. However, as in case of fields, casting to the parent will restore the original access.
Sample code:
public class Test {
public static void main( String[] args ) {
B b = new B();
A a = b;
b.f(); // "Access somewhat denied"
a.f(); // "f()"
}
}
class A {
public static void f() {
System.out.println("f()");
}
}
class B extends A {
// *must* be public
public static void f() {
System.out.println("Access somewhat denied");
}
}
So I created a trivial test; IntelliJ indeed rejected it... and Yes, I know "it's a tool...but one I trust". In any case, I went to javac, which emitted the same ERROR:
Error:(...) java: ...Concrete.java:5: doSomethingStatic() in
...Concrete cannot override doSomethingStatic() in
...Base; attempting to assign weaker access privileges; was public
Based on this, and our skepticism in general, I suggest the error is in your documentation.
Below is my sample code, fairly definitive I think. It barfs at the protected.
public class Base
{
public static void doSomethingStatic(){}
}
public class Concrete extends Base
{
protected static void doSomethingStatic(){}
}
It can be hidden by an overloaded redeclaration in a derived class:
class Base
{
public static void doSomethingStatic(){}
}
class Derived extends Base
{
public static void doSomethingStatic(String arg){}
}
but only hidden to people who try to access it via the derived class.