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.
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.
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.
I know it is a pretty beaten topic here but there is something i need to clarify, so bear with me for a minute.
Static methods are inherited just as any other method and follow the same inheritance rules about access modifiers (private methods are not inherited etc.)
Static methods are not over-ridden they are redefined. If a subclass defines a static method with the same signature as one in the superclass, it is said to be shadowing or hiding the superclass's version not over-riding it since they are not polymorphic as instance methods.
The redefined static methods still seem to be following some of (if not all) of the over-riding rules.
Firstly, the re-defined static method cannot be more access restricted than the superclass's static method. Why??
Secondly, the return types must also be compatible in both the superclass's and subclass's method. For example:
class Test2 {
static void show() {
System.out.println("Test2 static show");
}
}
public class StaticTest extends Test2 {
static StaticTest show() {
System.out.println("StaticTest static show");
return new StaticTest();
}
public static void main(String[] args) {
}
}
In eclipse it shows an error at line at:
The return type is incompatible with Test2.show()
Why??
And Thirdly, are there any other rules that are followed in re-defining static methods which are same as the rules for over-riding, and what is the reason for those rules?
Thanx in advance!!
The requirements for hiding of static methods is spelled out in detail in §8.4.8.3 of the Java Language Specification. By and large, it's the same as for instance methods:
The return type of the hiding method (in the subclass) must be assignment-compatible with the return type of the hidden method (in the superclass).
The access modifier of the hiding method must be no more restrictive than that of the hidden method.
It is an error for a method m in class T to have the same signature after erasure as another method n that is accessible in T unless the signature of m before erasure is a subsignature of method n.
There are restrictions on the throws clauses of methods that hide, override, or implement other methods that are declared to throw checked exceptions. (Basically, the hiding method cannot be declared to throw checked exceptions that are not declared in the hidden/overridden/implemented method.)
I think that's it, but see the JLS for more details. The JLS does not explain the rationale for these rules, but most of them seem intended to prevent problems with polymorphism. You want a subclass to be usable wherever a parent class is being used.
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.
How is method overriding implemented in Java? In C++ we have the concept of vtable.. how is this implemented internally in Java?
To answer the question, which is specifically how overriding is implemented in the virtual machine, there's a write up available in Programming for the Java Virtual Machine (Google Books link).
The VM will look for an appropriate method definition in the referenced class, and then work its way up through the inheritance stack. Obviously at some stage various optimisations will apply.
See here for a description of the relevant bytecode instruction invokevirtual:
invokevirtual looks at the descriptor
given in , and determines
how many arguments the method takes
(this may be zero). It pops these
arguments off the operand stack. Next
it pops objectref off the stack.
objectref is a reference to the object
whose method is being called.
invokevirtual retrieves the Java class
for objectref, and searches the list
of methods defined by that class and
then its superclasses, looking for a
method called methodname, whose
descriptor is descriptor.
As gustafc has highlighted below, various optimisations can apply, and no doubt the JIT will introduce further.
Method overriding in Java is a concept based on polymorphism OOPS
concept which allows programmer to create two methods with same name
and method signature on interface and its various implementation and
actual method is called at runtime depending upon type of object at
runtime. Method overriding allows you to write flexible and extensible
code in Java because you can introduce new functionality with minimal
code change.
There are few rules which needs to be followed while overriding any method in Java, failure to follow these rules result in compile time error in Java.
First and most important rule regarding method overriding in Java is that you can only override method in sub class. You can not override method in same class.
Second important rule of method overriding in Java that name and signature of method must be same in Super class and Sub class or in interface and its implementation.
Third rule to override method in Java is that overriding method can not reduce accessibility of overridden method in Java. For example if overridden method is public than overriding method can not be protected, private or package-private; But opposite is true overriding method can increase accessibility of method in Java, i.e. if overridden method is protected than overriding method can be protected or public.
Another worth noting rule of method overriding in Java is that overriding method can not throw checked Exception which is higher in hierarchy than overridden method. Which means if overridden method throws IOException than overriding method can not throw java.lang.Exception in its throws clause because java.lang.Exception comes higher than IOException in Exception hierarchy. This rule doesn't apply to RuntimeException in Java, which is not even need to be declared in throws clause in Java.
You can not override private, static and final method in Java. private and static method are bonded during compile time using static binding in Java and doesn't resolve during runtime. overriding final method in Java is compile time error. Though private and static method can be hidden if you declare another method with same and signature in sub class.
Overridden method is called using dynamic binding in Java at runtime based upon type of Object.
If you are extending abstract class or implementing interface than you need to override all abstract method unless your class is not abstract. abstract method can only be used by using method overriding.
Always use #Override annotation while overriding method in Java. Though this is not rule but its one of the best Java coding practice to follow. From Java 6 you can use #Override annotation on method inherited from interface as well.
--Defining the same method with the same method signature(i.e. same number/type of arguments and same return type/s.)in base and derived class.
--Which method is to be called is decided at runtime so, it is also called runtime polymorphism/late binding.
--we should override the method defined in the superclass.
--when the method is called the method defined in the subclass is called and executed instead of the one in superclass.
--we overcome this by use of 'super' keyword.
//program
class A
{
void disp(){
System.out.println("class A");
}
}
class B extends A
{
public void disp(){
System.out.println("class B");
}
}
public class ExampleIdea{
public static void main(String[] args){
A a = new B(); //Parent reference but B class object (we can achieve polymorphism when parent class reference is used to refer a child class object)
B b = new B(); //B class reference and B class object
A c = new A();
a.disp(); //runs the method in B class
b.disp(); //runs the method in B class
c.disp(); //runs then method in A class
}
}//end of program
when we run this output will be class B.
In order to access the function of class A we have to use super keyword
as:
class B extends A
{
public void disp(){
System.out.println("class B");
super.disp();
}
Maybe this comparison of C++ Vtables and Java method invocation tables is of interest.
Method overriding
It means one method is available in supper class which is displaying some string, but you want to extend this class, and also you want to print your own method at that time you need to overriding this method into your local class.
This is a basic introduction to the method overriding.
Example:
class Animal
{
public void displayMessage()
{
System.out.println("Animal is eating");
}
}
class Dog extends Animal
{
public void displayMessage()
{
System.out.println("Dog is eating");
}
public static void main(String arg[])
{
Dog d=new Dog();
d.displayMessage();
}
}
OUTPUT:
Dog is eating
Advantages of Method Overriding is that the class ca give its own specific implementation to an inherited method without any modifying the parent class method.
The rules of Method overriding are:
The argument list : the argument list must be the same.
Access Modifier : if you're overriding the method you must give the same modifier of super class method suppose super class have the public method you cannot give the protected or private vice versa.
As long as the function (method) you wish to override is not marked as final you just simply override the method by extending the class which holds that method and then provide a method with same signature but different body.