Related
As succinctly described here, overriding private methods in Java is invalid because a parent class's private methods are "automatically final, and hidden from the derived class". My question is largely academic.
How is it not a violation of encapsulation to not allow a parent's private method to be "overridden" (ie, implemented independently, with the same signature, in a child class)? A parent's private method cannot be accessed or inherited by a child class, in line with principles of encapsulation. It is hidden.
So, why should the child class be restricted from implementing its own method with the same name/signature? Is there a good theoretical foundation for this, or is this just a pragmatic solution of some sort? Do other languages (C++ or C#) have different rules on this?
You can't override a private method, but you can introduce one in a derived class without a problem. This compiles fine:
class Base
{
private void foo()
{
}
}
class Child extends Base
{
private void foo()
{
}
}
Note that if you try to apply the #Override annotation to Child.foo() you'll get a compile-time error. So long as you have your compiler/IDE set to give you warnings or errors if you're missing an #Override annotation, all should be well. Admittedly I prefer the C# approach of override being a keyword, but it was obviously too late to do that in Java.
As for C#'s handling of "overriding" a private method - a private method can't be virtual in the first place, but you can certainly introduce a new private method with the same name as a private method in the base class.
Well, allowing private methods to be overwritten will either cause a leak of encapsulation or a security risk. If we assume that it were possible, then we’d get the following situation:
Let's say that there's a private method boolean hasCredentials() then an extended class could simply override it like this:
boolean hasCredentials() { return true; }
thus breaking the security check.
The only way for the original class to prevent this would be to declare its method final. But now, this is leaks implementation information through the encapsulation, because a derived class now cannot create a method hasCredentials any more – it would clash with the one defined in the base class.
That’s bad: lets say this method doesn’t exist at first in Base. Now, an implementor can legitimately derive a class Derived and give it a method hasCredentials which works as expected.
But now, a new version of the original Base class is released. Its public interface doesn’t change (and neither do its invariants) so we must expect that it doesn’t break existing code. Only it does, because now there’s a name clash with a method in a derived class.
I think the question stems from a misunderstanding:
How is it /not/ a violation of encapsulation to not allow a parent's private method to be "overridden" (ie, implemented independently, with the same signature, in a child class)
The text inside the parentheses is the opposite of the text before it. Java does allow you to “independently implement [a private method], with the same signature, in a child class”. Not allowing this would violate encapsulation, as I’ve explained above.
But “to not allow a parent's private method to be "overridden"” is something different, and necessary to ensure encapsulation.
"Do other languages (C++ or C#) have different rules on this?"
Well, C++ has different rules: the static or dynamic member function binding process and the access privileges enforcements are orthogonal.
Giving a member function the private access privilege modifier means that this function can only be called by its declaring class, not by others (not even the derived classes). When you declare a private member function as virtual, even pure virtual (virtual void foo() = 0;), you allow the base class to benefit from specialization while still enforcing the access privileges.
When it comes to virtual member functions, access privileges tells you what you are supposed to do:
private virtual means that you are allowed to specialize the behavior but the invocation of the member function is made by the base class, surely in a controlled fashion
protected virtual means that you should / must invoke the upper class version of the member function when overriding it
So, in C++, access privilege and virtualness are independent of each other. Determining whether the function is to be statically or dynamically bound is the last step in resolving a function call.
Finally, the Template Method design pattern should be preferred over public virtual member functions.
Reference: Conversations: Virtually Yours
The article gives a practical use of a private virtual member function.
ISO/IEC 14882-2003 §3.4.1
Name lookup may associate more than one declaration with a name if it finds the name to be a function name; the declarations are said to form a set of overloaded functions (13.1). Overload resolution (13.3) takes place after name lookup has succeeded. The access rules (clause 11) are considered only once name lookup and function overload resolution (if applicable) have succeeded. Only after name lookup, function overload resolution (if applicable) and access checking have succeeded are the attributes introduced by the name’s declaration used further in expression processing (clause 5).
ISO/IEC 14882-2003 §5.2.2
The function called in a member function call is normally selected according to the static type of the object expression (clause 10), but if that function isvirtualand is not specified using aqualified-idthen the function actually called will be the final overrider (10.3) of the selected function in the dynamic type of the object expression [Note: the dynamic type is the type of the object pointed or referred to by the current value of the object expression.
A parent's private method cannot be accessed or inherited by a child class, inline with principles of encapsulation. It is hidden.
So, why should the child class be
restricted from implementing its own
method with the same name/signature?
There is no such restriction. You can do that without any problems, it's just not called "overriding".
Overridden methods are subject to dynamic dispatch, i.e. the method that is actually called is selected at runtime depending on the actual type of the object it's called on. With private method, that does not happen (and should not, as per your first statement). And that's what is meant by the statement "private methods can't be overridden".
I think you're misinterpreting what that post says. It's not saying that the child class is "restricted from implementing its own method with the same name/signature."
Here's the code, slightly edited:
public class PrivateOverride {
private static Test monitor = new Test();
private void f() {
System.out.println("private f()");
}
public static void main(String[] args) {
PrivateOverride po = new Derived();
po.f();
});
}
}
class Derived extends PrivateOverride {
public void f() {
System.out.println("public f()");
}
}
And the quote:
You might reasonably expect the output to be “public f( )”,
The reason for that quote is that the variable po actually holds an instance of Derived. However, since the method is defined as private, the compiler actually looks at the type of the variable, rather than the type of the object. And it translates the method call into invokespecial (I think that's the right opcode, haven't checked JVM spec) rather than invokeinstance.
It seems to be a matter of choice and definition. The reason you can't do this in java is because the specification says so, but the question were more why the specification says so.
The fact that C++ allows this (even if we use virtual keyword to force dynamic dispatch) shows that there is no inherent reason why you couldn't allow this.
However it seem to be perfectly legal to replace the method:
class B {
private int foo()
{
return 42;
}
public int bar()
{
return foo();
}
}
class D extends B {
private int foo()
{
return 43;
}
public int frob()
{
return foo();
}
}
Seems to compile OK (on my compiler), but the D.foo is not related to B.foo (ie it doesn't override it) - bar() always return 42 (by calling B.foo) and frob() always returns 43 (by calling D.foo) no matter whether called on a B or D instance.
One reason that Java does not allow override the method would be that they didn't like to allow the method to be changed as in Konrad Rudolph's example. Note that C++ differs here as you need to use the "virtual" keyword in order to get dynamic dispatch - by default it hasn't so you can't modify code in base class that relies on the hasCredentials method. The above example also protects against this as the D.foo does not replace calls to foo from B.
When the method is private, it's not visible to its child. So there is no meaning of overriding it.
I apologize for using the term override incorrectly and inconsistent with my description. My description describes the scenario. The following code extends Jon Skeet's example to portray my scenario:
class Base {
public void callFoo() {
foo();
}
private void foo() {
}
}
class Child extends Base {
private void foo() {
}
}
Usage is like the following:
Child c = new Child();
c.callFoo();
The issue I experienced is that the parent foo() method was being called even though, as the code shows, I was calling callFoo() on the child instance variable. I thought I was defining a new private method foo() in Child() which the inherited callFoo() method would call, but I think some of what kdgregory has said may apply to my scenario - possibly due to the way the derived class constructor is calling super(), or perhaps not.
There was no compiler warning in Eclipse and the code did compile. The result was unexpected.
Beyond anything said before, there's a very semantic reason for not allowing private methods to be overridden...THEY'RE PRIVATE!!!
If I write a class, and I indicate that a method is 'private', it should be completely unseeable by the outside world. Nobody should be able access it, override it, or anything else. I simply ought to be able to know that it is MY method exclusively and that nobody else is going to muck with it or depend on it. It could not be considered private if someone could muck with it. I believe that it's that simple really.
A class is defined by what methods it makes available and how they behave. Not how those are implemented internally (e.g. via calls to private methods).
Because encapsulation has to do with behavior and not implementation details, private methods have nothing to do with the idea encapsulation. In a sense, your question makes no sense. It's like asking "How is putting cream in coffee not a violation of encapsulation?"
Presumably the private method is used by something that is public. You can override that. In doing so, you've changed behavior.
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();
}
}
This question already has answers here:
why java polymorphism not work in my example
(3 answers)
Closed 6 years ago.
Generally Overriding is the concept of Re-defining the meaning of the member in the sub class.Why variables are not behaving like methods while Overriding in java ?
For instance:
class Base {
int a = 10;
void display() {
System.out.println("Inside Base :");
}
}
class Derived extends Base {
int a = 99;
#Override
// method overriding
void display() {
System.out.println("Inside Derived :");
}
}
public class NewClass {
public static void main(String... a) {
Derived d = new Derived();
Base b = d;
b.display(); // Dynamic method dispatch
System.out.println("a=" + b.a);
}
}
Since data member a is package access specified, it is also available to the Derived class. But generally while calling the overridden method using the base class reference, the method that is redefined in derived class is called (Dynamic method dispatch)..but it is not the same for the variable..why.?
EXPECTED OUTPUT
Inside Derived :
a=99
OBTAINED OUTPUT:
Inside Derived :
a=10
Prints 10 - why the variable does not behave similar to method in the derived class?
Why the variables are not allowed to be overridden in the sub class?
You typed b as an instance of Base. So when the compiler needs to resolve b.a, it looks to the definition of Base for the meaning of b.a. There is no polymorphism for instance fields.
Because the only thing that polymorphism ever applies to in Java is instance method.
Hence, you can neither override static members, nor the instance member fields. By, having these members in a derived class with the same names you're simply hiding them with a new definition.
System.out.println("a="+b.a);
Although, Base b may point to a sub-class object (at runtime) the a above has already been bound to Base class at compile time (static binding). Hence, it prints 10.
Variables behave like that because they lack behavior. In other words, variables are passive.
There is nothing about a variable's definition that a derived class can reasonably change by overriding:
It cannot change its type, because doing so may break methods of the base class;
It cannot reduce its visibility, because that would break the substitution principle.
It cannot make it final without making it useless to the base class.
Therefore, member variables declared in derived classes hide variables from the base class.
There is no way to override a class variable. You do not override class variables in Java you hide them. Overriding is for instance methods.
In this case, it might be a good idea to write a getter method:
public int getA(){
return 99;
}
Now you can override it in a derived class.
First, we don't override any class variable. Methods only.
Second, if you would like to see that the variable value has been updated or replaced, you should rather declare it as "static int" instead of "int". In this way, it will work as everybody is sharing the same variable, and the new value will be put on it.
Third, if you would like to see that the variable value being assigned and used differently, you could design it as passing a parameter in constructor, or something similar, to make it work accordingly as you desire.
The answer to this has to do with variable scoping, not polymorphism. In other words, you're overriding that variable in the class scope. So, d.a will return the variable in Derived's class scope, but b.a will return the variable in Base's class scope.
In OOP (Object Oriented Programming) the idea is to hide the data in the object and let object only communicate with invoking methods. That's why variables cannot be overloaded, in fact they are "scoped"/"attached" to a specific class.
Also the derived class should not define a again, it is already defined in the base class, so simply set a on the object to the desired value, e.g:
class Base {
private int a = 10;
public int getA() { return a; }
public void setA(inta) { this.a = a; }
}
class Derived extends Base {
// adding new variables, override methods, ...
}
// then later:
Derived d = new Derived();
d.setA(99); // override the default value 10
What would happen if variables could override other variables? Suddenly your class has to be aware of what variables the parent class is using, lest you accidentally override one and break whatever was using it in the parent class. The whole point of encapsulation is to avoid having that kind of intimate knowledge of another object's internal state. So instead, variables shadow same-named other variables, and which one you see depends on what type you're trying to reach the variable through.
There's hope, though. If all you want is to override the value, you don't have to redeclare the variable. Just change the value in an init block. If the base class is harmed by you doing that, then it chose the wrong visibility for that variable.
class Base {
int a = 10;
}
class Derived extends Base {
{ a = 99; }
}
Of course, this doesn't work very well for final variables.
we don't override any class variable. Methods only.
If you would like to see that the variable value has been updated or
replaced, you should rather declare it as "static int" instead of
"int". In this way, it will work as everybody is sharing the same
variable, and the new value will be put on it.
If you would like to see that the variable value being assigned and
used differently, you could design it as passing a parameter in
constructor, or something similar, to make it work accordingly as
you desire.
Moreover, if variables are overridden then what is left with a parent class of its own,it breaches the class security if java would give the access to change the value of variable of parent class.
As succinctly described here, overriding private methods in Java is invalid because a parent class's private methods are "automatically final, and hidden from the derived class". My question is largely academic.
How is it not a violation of encapsulation to not allow a parent's private method to be "overridden" (ie, implemented independently, with the same signature, in a child class)? A parent's private method cannot be accessed or inherited by a child class, in line with principles of encapsulation. It is hidden.
So, why should the child class be restricted from implementing its own method with the same name/signature? Is there a good theoretical foundation for this, or is this just a pragmatic solution of some sort? Do other languages (C++ or C#) have different rules on this?
You can't override a private method, but you can introduce one in a derived class without a problem. This compiles fine:
class Base
{
private void foo()
{
}
}
class Child extends Base
{
private void foo()
{
}
}
Note that if you try to apply the #Override annotation to Child.foo() you'll get a compile-time error. So long as you have your compiler/IDE set to give you warnings or errors if you're missing an #Override annotation, all should be well. Admittedly I prefer the C# approach of override being a keyword, but it was obviously too late to do that in Java.
As for C#'s handling of "overriding" a private method - a private method can't be virtual in the first place, but you can certainly introduce a new private method with the same name as a private method in the base class.
Well, allowing private methods to be overwritten will either cause a leak of encapsulation or a security risk. If we assume that it were possible, then we’d get the following situation:
Let's say that there's a private method boolean hasCredentials() then an extended class could simply override it like this:
boolean hasCredentials() { return true; }
thus breaking the security check.
The only way for the original class to prevent this would be to declare its method final. But now, this is leaks implementation information through the encapsulation, because a derived class now cannot create a method hasCredentials any more – it would clash with the one defined in the base class.
That’s bad: lets say this method doesn’t exist at first in Base. Now, an implementor can legitimately derive a class Derived and give it a method hasCredentials which works as expected.
But now, a new version of the original Base class is released. Its public interface doesn’t change (and neither do its invariants) so we must expect that it doesn’t break existing code. Only it does, because now there’s a name clash with a method in a derived class.
I think the question stems from a misunderstanding:
How is it /not/ a violation of encapsulation to not allow a parent's private method to be "overridden" (ie, implemented independently, with the same signature, in a child class)
The text inside the parentheses is the opposite of the text before it. Java does allow you to “independently implement [a private method], with the same signature, in a child class”. Not allowing this would violate encapsulation, as I’ve explained above.
But “to not allow a parent's private method to be "overridden"” is something different, and necessary to ensure encapsulation.
"Do other languages (C++ or C#) have different rules on this?"
Well, C++ has different rules: the static or dynamic member function binding process and the access privileges enforcements are orthogonal.
Giving a member function the private access privilege modifier means that this function can only be called by its declaring class, not by others (not even the derived classes). When you declare a private member function as virtual, even pure virtual (virtual void foo() = 0;), you allow the base class to benefit from specialization while still enforcing the access privileges.
When it comes to virtual member functions, access privileges tells you what you are supposed to do:
private virtual means that you are allowed to specialize the behavior but the invocation of the member function is made by the base class, surely in a controlled fashion
protected virtual means that you should / must invoke the upper class version of the member function when overriding it
So, in C++, access privilege and virtualness are independent of each other. Determining whether the function is to be statically or dynamically bound is the last step in resolving a function call.
Finally, the Template Method design pattern should be preferred over public virtual member functions.
Reference: Conversations: Virtually Yours
The article gives a practical use of a private virtual member function.
ISO/IEC 14882-2003 §3.4.1
Name lookup may associate more than one declaration with a name if it finds the name to be a function name; the declarations are said to form a set of overloaded functions (13.1). Overload resolution (13.3) takes place after name lookup has succeeded. The access rules (clause 11) are considered only once name lookup and function overload resolution (if applicable) have succeeded. Only after name lookup, function overload resolution (if applicable) and access checking have succeeded are the attributes introduced by the name’s declaration used further in expression processing (clause 5).
ISO/IEC 14882-2003 §5.2.2
The function called in a member function call is normally selected according to the static type of the object expression (clause 10), but if that function isvirtualand is not specified using aqualified-idthen the function actually called will be the final overrider (10.3) of the selected function in the dynamic type of the object expression [Note: the dynamic type is the type of the object pointed or referred to by the current value of the object expression.
A parent's private method cannot be accessed or inherited by a child class, inline with principles of encapsulation. It is hidden.
So, why should the child class be
restricted from implementing its own
method with the same name/signature?
There is no such restriction. You can do that without any problems, it's just not called "overriding".
Overridden methods are subject to dynamic dispatch, i.e. the method that is actually called is selected at runtime depending on the actual type of the object it's called on. With private method, that does not happen (and should not, as per your first statement). And that's what is meant by the statement "private methods can't be overridden".
I think you're misinterpreting what that post says. It's not saying that the child class is "restricted from implementing its own method with the same name/signature."
Here's the code, slightly edited:
public class PrivateOverride {
private static Test monitor = new Test();
private void f() {
System.out.println("private f()");
}
public static void main(String[] args) {
PrivateOverride po = new Derived();
po.f();
});
}
}
class Derived extends PrivateOverride {
public void f() {
System.out.println("public f()");
}
}
And the quote:
You might reasonably expect the output to be “public f( )”,
The reason for that quote is that the variable po actually holds an instance of Derived. However, since the method is defined as private, the compiler actually looks at the type of the variable, rather than the type of the object. And it translates the method call into invokespecial (I think that's the right opcode, haven't checked JVM spec) rather than invokeinstance.
It seems to be a matter of choice and definition. The reason you can't do this in java is because the specification says so, but the question were more why the specification says so.
The fact that C++ allows this (even if we use virtual keyword to force dynamic dispatch) shows that there is no inherent reason why you couldn't allow this.
However it seem to be perfectly legal to replace the method:
class B {
private int foo()
{
return 42;
}
public int bar()
{
return foo();
}
}
class D extends B {
private int foo()
{
return 43;
}
public int frob()
{
return foo();
}
}
Seems to compile OK (on my compiler), but the D.foo is not related to B.foo (ie it doesn't override it) - bar() always return 42 (by calling B.foo) and frob() always returns 43 (by calling D.foo) no matter whether called on a B or D instance.
One reason that Java does not allow override the method would be that they didn't like to allow the method to be changed as in Konrad Rudolph's example. Note that C++ differs here as you need to use the "virtual" keyword in order to get dynamic dispatch - by default it hasn't so you can't modify code in base class that relies on the hasCredentials method. The above example also protects against this as the D.foo does not replace calls to foo from B.
When the method is private, it's not visible to its child. So there is no meaning of overriding it.
I apologize for using the term override incorrectly and inconsistent with my description. My description describes the scenario. The following code extends Jon Skeet's example to portray my scenario:
class Base {
public void callFoo() {
foo();
}
private void foo() {
}
}
class Child extends Base {
private void foo() {
}
}
Usage is like the following:
Child c = new Child();
c.callFoo();
The issue I experienced is that the parent foo() method was being called even though, as the code shows, I was calling callFoo() on the child instance variable. I thought I was defining a new private method foo() in Child() which the inherited callFoo() method would call, but I think some of what kdgregory has said may apply to my scenario - possibly due to the way the derived class constructor is calling super(), or perhaps not.
There was no compiler warning in Eclipse and the code did compile. The result was unexpected.
Beyond anything said before, there's a very semantic reason for not allowing private methods to be overridden...THEY'RE PRIVATE!!!
If I write a class, and I indicate that a method is 'private', it should be completely unseeable by the outside world. Nobody should be able access it, override it, or anything else. I simply ought to be able to know that it is MY method exclusively and that nobody else is going to muck with it or depend on it. It could not be considered private if someone could muck with it. I believe that it's that simple really.
A class is defined by what methods it makes available and how they behave. Not how those are implemented internally (e.g. via calls to private methods).
Because encapsulation has to do with behavior and not implementation details, private methods have nothing to do with the idea encapsulation. In a sense, your question makes no sense. It's like asking "How is putting cream in coffee not a violation of encapsulation?"
Presumably the private method is used by something that is public. You can override that. In doing so, you've changed behavior.
class A
{
int i=10;
void show()
{
System.out.println("class A");
}
}
class B extends A
{
int i=5;
public void show()
{
System.out.println("class B");
}
}
class M
{
public static void main(String s[])
{
A a=new B();
a.show();
System.out.println(a.i);
}
}
OUTPUT= class B
10
If class A method is overridden by class B method then why not the variable 'i'?
Because variables are not virtual, only methods are.
It is not overwritten, but hidden. In your output you specifically requested the value of a.i, not ((B)a).i.
This is a "feature" of the implementation. In memory, this looks like so:
a:
pointer to class A
int i
b:
pointer to class B
int i (from A)
int i (from B)
When you access i in an instance of B, Java needs to know which variable you mean. It must allocate both since methods from class A will want to access their own field i while methods from B will want their own i (since you chose to create a new field i in B instead of making A.i visible in B). This means there are two i and the standard visibility rules apply: Whichever is closer will win.
Now you say A a=new B(); and that's a bit tricky because it tells Java "treat the result from the right hand side as if it were an instance of A".
When you call a method, Java follows the pointer to the class (first thing in the object in memory). There, it finds a list of methods. Methods overwrite each other, so when it looks for the method show(), it will find the one defined in B. This makes method access fast: You can simply merge all visible methods in the (internal) method list of class B and each call will mean a single access to that list. You don't need to search all classes upwards for a match.
Field access is similar. Java doesn't like searching. So when you say B b = new B();, b.i is obviously from B. But you said A a = new B() telling Java that you prefer to treat the new instance as something of type A. Java, lazy as it is, looks into A, finds a field i, checks that you can see that field and doesn't even bother to look at the real type of a anymore (because that would a) be slow and b) would effectively prevent you from accessing both i fields by casting).
So in the end, this is because Java optimizes the field and method lookup.
Why no field overrides in Java though?
Well, because instance field lookups in Java happen at compile time: Java simply gives you the value of the field at a given offset in object's memory (based on the type information at hand during compilation: in this case a is declared to be of type A).
void foo() {
A a = new B();
int val = a.i; // compiler uses type A to compute the field offset
}
One may ask "Why didn't compiler use type B since it knows that a is in fact an instance of B? Isn't it obvious from the assignment just above?". Of course, in the case above, it's relatively obvious and compiler may try to be smarter and figure it out.
But that's compiler design "rat hole", what if a "trickier" piece of code is encountered, like so:
void foo(A a) {
int val = a.i;
}
If compiler were "smarter", it would become its job to look at all invocations of foo() and see what real type was used, which is an impossible job since compiler can not predict what other crazy things may be passed to foo() by unknown or yet unwritten callers.
It's a design decision by the developers of Java, and is documented in the Java Language Specification.
A method with the same method signature as a method in its parent class overrides the method in its parent class.
A variable with the same name as a variable in its parent class hides the parent's variable.
The difference is that hidden values can be accessed by casting the variable to its parent type, while overridden methods will always execute the child class's method.
As others have noted, in C++ and C#, to get the same override behavior as Java, the methods need to be declared virtual.
a is an instance of A. You call the constructor B(). But it is still a A class.
That is why i equals 10;
The override from the method will be succeded.
Note a class starts not with
public class A()
but with;
public class A { ... }
Tip: You can use setters and getters to make sure of what data-members you use.
Or: You simply can set the values at the constructor instead of the class declaration.
Because by default the variables are private. You must declare it as "protected", then will be properly inherited.