Related
Eg
class Abc{
public void fun()
{
System.out.println("public method") ;
}
#Override
public String toString()
{
// "has to be public method but the access is default because of the class access";
return super.toString();
}
}
In the above Eg - fun method access is public but the class access is default hence what is the use of having the method public rather than default because it cannot be access without creating object.
In the same way the toString method has to be public since (overriding cannot restrict the access). But it is already getting restricted since the class access is default
My Basic question is
What is the use of having a less restricted method in a more restricted class?
My Basic question is What is the use of having a less restricted method in a more restricted class?
There are several purposes related to inheritance and polymorphism. The biggest of those is probably that methods that override superclass methods or implement interface methods cannot be more restricted than the method they override. In this regard, I note that you express a possible misconception when you say:
the toString method has to be public since (overriding cannot restrict the access). But it is already getting restricted since the class access is default
That class Abc has default access does not prevent any other class from having references to instances of Abc or from invoking the public methods of that class. The declared type of any such reference will be a supertype of class Abc, but the implementation invoked will be that provided by Abc.
Additionally, there is the question of signaling intent. It is helpful to view the access specified for class members to be qualified by the access level of the class. Thus, declaring fun() to be public says that it can be accessed by everyone who can access an instance of Abc. This is somewhat useful in its own right, but it also turns out to be especially useful if the access level of the class is ever changed. If the class's designer chose member access levels according to this principle, then those do not need to be revisited under these circumstances.
On its face?
Absolutely no purpose whatsoever. It's measurable (you can use reflection to fetch a java.lang.reflect.Method object that represents that method, and you can ask it if it is public, and the answer would be 'yes', but it doesn't actually change how accessible that method is.
However, taking a step back and thinking about the task of programming in its appropriate light, which includes the idea that code is usually a living, breathing idea that is continually updated: Hey, your class is package private today. Maybe someone goes in and makes it public tomorrow, and you intended for this method to be just as public as the type itself if ever that happens.
Specifically in this case, you're overriding a method. Java is always dynamic dispatch. That means that if some code has gotten a hold of an instance of this class (Created with new Abc()), and that code is not in this package, that they can still invoke this method.
Let's see it in action:
package abc;
class Abc {
#Override public String toString() { return "Hello"; }
}
public class AbcMaker {
public Object make() { return new Abc(); }
}
Note that here AbcMaker is public and make() can be invoked just fine from outside code; they know what Object is, and Abc is an Object. This lets code from outside the abc package invoke and obtain Abc instances, even though Abc is package private. This is fine.
They can then do:
package someOtherPackage;
class Test {
public void foo() {
System.out.println(new AbcMaker().make().toString());
}
}
And that would print Hello, as expected - that ends up invoking the toString defined in your package private class, invoked from outside the package!
Thus, we come to a crucial conclusion:
That toString() method in your package private class is ENTIRELY PUBLIC.
It just is. You can't override a method and make it less accessible than it is in your parent type, because java is covariant, meaning any X is a valid standin for Y, if X is declared as X extends Y. if Y has public toString, then X's can't take that away, as all Xs must be valid Ys (and Ys have public toString methods, therefore Xs must also have this).
Thus, the compiler is forcing you here. The banal reason is 'cuz the spec says so', but the reason the spec says this is to make sure you, the programmer, are not confused, and that what you type and write matches with reality: That method is public, it has to be, so the compiler will refuse to continue unless you are on the same page as it and also say so.
I am not asking this -> Why is there no multiple inheritance in Java, but implementing multiple interfaces is allowed?
In Java, multiple inheritance isn't allowed, but, after Java 8, Interfaces can have default methods (can implement methods itself), just like abstract classes. Within this context, it multiple inheritance should also be allowed.
interface TestInterface
{
// abstract method
public void square(int a);
// default method
default void show()
{
System.out.println("Default Method Executed");
}
}
Things are not so simple.
If a class implements multiple interfaces that defines default methods with the same signature the compiler will force you to override this method for the class.
For example with these two interfaces :
public interface Foo {
default void doThat() {
// ...
}
}
public interface Bar {
default void doThat() {
// ...
}
}
It will not compile :
public class FooBar implements Foo, Bar{
}
You should define/override the method to remove the ambiguity.
You could for example delegate to the Bar implementation such as :
public class FooBar implements Foo, Bar{
#Override
public void doThat() {
Bar.super.doThat();
}
}
or delegate to the Foo implementation such as : :
public class FooBar implements Foo, Bar {
#Override
public void doThat() {
Foo.super.doThat();
}
}
or still define another behavior :
public class FooBar implements Foo, Bar {
#Override
public void doThat() {
// ...
}
}
That constraint shows that Java doesn't allow multiple inheritancy even for interface default methods.
I think that we cannot apply the same logic for multiple inheritances because multiples issues could occur which the main are :
overriding/removing the ambiguity for a method in both inherited classes could introduce side effects and change the overall behavior of the inherited classes if they rely on this method internally. With default interfaces this risk is also around but it should be much less rare since default methods are not designed to introduce complex processings such as multiple internal invocations inside the class or to be stateful (indeed interfaces cannot host instance field).
how to inherit multiple fields ? And even if the language allowed it you would have exactly the same issue as this previously quoted : side effect in the behavior of the inherited class : a int foo field defined in a A and B class that you want to subclass doesn't have the same meaning and intention.
The language designers already thought about that, so these things are enforced by the compiler. So if you define:
interface First {
default void go() {
}
}
interface Second {
default void go() {
}
}
And you implement a class for both interfaces:
static class Impl implements First, Second {
}
you will get a compilation error; and you would need to override go to not create the ambiguity around it.
But you could be thinking that you can trick the compiler here, by doing:
interface First {
public default void go() {
}
}
static abstract class Second {
abstract void go();
}
static class Impl extends Second implements First {
}
You could think that First::go already provides an implementation for Second::go and it should be fine. This is too taken care of, thus this does not compile either.
JLS 9.4.1.3 : Similarly, when an abstract and a default method with matching signatures are inherited, we produce an error. In this case, it would be possible to give priority to one or the other - perhaps we would assume that the default method provides a reasonable implementation for the abstract method, too. But this is risky, since other than the coincidental name and signature, we have no reason to believe that the default method behaves consistently with the abstract method's contract - the default method may not have even existed when the subinterface was originally developed. It is safer in this situation to ask the user to actively assert that the default implementation is appropriate (via an overriding declaration).
The last point I would bring in, to solidify that multiple inheritance is not allowed even with new additions in java, is that static methods from interfaces are not inherited. static methods are inherited by default:
static class Bug {
static void printIt() {
System.out.println("Bug...");
}
}
static class Spectre extends Bug {
static void test() {
printIt(); // this will work just fine
}
}
But if we change that for an interface (and you can implement multiple interfaces, unlike classes):
interface Bug {
static void printIt() {
System.out.println("Bug...");
}
}
static class Spectre implements Bug {
static void test() {
printIt(); // this will not compile
}
}
Now, this is prohibited by the compiler and JLS too:
JLS 8.4.8 : A class does not inherit static methods from its superinterfaces.
Java doesn't allow multiple inheritance for fields. This would be difficult to support in the JVM as you can only have references to the start of an object where the header is, not arbitrary memory locations.
In Oracle/Openjdk, objects have a header followed by the fields of the most super class, then the next most super class, etc. It would be a significant change to allow the fields of a class to appear at different offsets relative to the header of an object for different subclasses. Most likely object references would have to become a reference to the object header and a reference to the fields to support this.
default methods in interfaces pose a problem that :
If both of the implemented interfaces define a default method with
same method signature, then the implementation class does not know
which default method to use.
The implementation class should define explicitly specify which default method to use or define it's own one.
Thus default methods in Java-8 do not facilitate multiple inheritance. The main motivation behind default methods is that if at some point we need to add a method to an existing interface, we can add a method without changing the existing implementation classes. In this way, the interface is still compatible with older versions. However, we should remember the motivation of using Default Methods and should keep the separation of interface and implementation.
The main issues with multiple inheritance are ordering (for overriding and calls to super), fields and constructors; interfaces don't have fields or constructors, so they don't cause problems.
If you look at other languages they usually fall in two broad categories:
Languages with multiple inheritance plus a few features to disambiguate special cases: virtual inheritance [C++], direct calls to all superconstructors in the most-derived class [C++], linearization of superclasses [Python], complex rules for super [Python], etc.
Languages with a differente concept, usually called interfaces, traits, mixins, modules, etc. that impose some limitations such as: no constructors [Java] or no constructors with parameters [Scala until very recently], no mutable fields [Java], specific rules for overriding (e.g. mixins take precedence over base classes [Ruby] so you can include them when you need a bunch of utility methods), etc. Java has become a language like these.
Why just by disallowing fields and constructors you solve many issues related to multiple inheritance?
You can't have duplicated fields in duplicated base classes.
The main class hierarchy is still linear.
You can't construct your base objects the wrong way.
Imagine if Object had public/protected fields and all subclasses had constructors setting those fields. When you inherit from more than one class (all of them derived from Object), which one gets to set the fields? The last class? They become siblings in the hierarchy, so they know nothing about each other. Should you have multiple copies of Object to avoid this? Would all classes interoperate correctly?
Remember that fields in Java are not virtual (overridable), they are simply data storage.
You could make a language where fields behave like methods and could be overridden (the actual storage would be always private), but that would be a much bigger change and problably wouldn't be called Java anymore.
Interfaces can't be instantiated by themselves.
You should always combine them with a concrete class. That eliminates the need for constructors and makes the programmer's intent clearer too (that is, what is meant to be a concrete class and what's an accessory interface/mixin). This also provides a well-defined place to solve all ambiguities: the concrete class.
That is mostly related to "diamonds problem" i think. Right now if you implement multiple interfaces with the same method, compiler forces you to override method the one you want to implement, because it don't know which on to use. I guess Java creators wanted to remove this problem back when interfaces couldn't use default methods. Now they came up with idea, that is good to be able to have methods with implementation in interfaces, as you can still use those as functional interfaces in streams / lambda expressions and utilize their default methods in processing. You cannot do that with classes but diamond problem still exist there. That is my guess :)
class A{
void m1(){
System.out.println("m1-A");
}
}
class B{
void m1(){
System.out.println("m1-B");
}
}
class C extends A, B{ // this will give an error
// inheritance means making all variables and/or methods available to the child class, here child class will get confused as which m1() method to inherit, hence an error
}
JAVA DOES SUPPORT MULTIPLE INHERITANCE.
If you make a OVERALL COMPARISON OF THE PROGRAMMING LANGUAGE,JAVA,THEN YOU COME TO KNOW THAT I AM TRUE.
Java's topclass or the root class in the Ancestor Hierarchy is the Object class.
This class is a Superclass of all other classes. Hence, each class in Java that we declare or is predefined in the API itself inherits this Object class.
Moreover, Java provides us to inherit one more class of our choice.
Hence, we can say that we are performing INTERLOCKED BUT MULTIPLE INHERITANCE.
2ND Way
Java supports Multiple Inheritance of Interfaces. So you can use as many interface implementations you want. But note, implementing an interface does not define IS A relationship as in case of Inheritance of Classes is possible.
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.
For example a class has public do, packaged do, protected do, and private do. If an instance of the class calls do it gets the private one, if a subclass calls it, then it gets protected, if a same package calls it, then it gets packaged and if anything else calls it, it gets public?
No.
It is a compile-time error to declare
two methods with override-equivalent
signatures (defined below) in a class.
Two methods have the same signature if
they have the same name and argument
types.
(From JLS §8.4.2)
class A
{
public Object do() { ... }
protected Object do() { ... }
Object do() { ... }
private Object do() { ... }
}
No. This will not compile with or without a subclass. Nor should it be expected too. How would the compiler have any idea which method to invoke? It's impossible.
To maybe make it a little clearer, a more distinctive overload--one that returns some other type than Object is not even an acceptable overload because the compiler would still have trouble determining which method to call. A weaker form of overloading would be even less acceptable.
No. This is an element that comprises its identity (just as you can't overload the name or return type of the method). You can only overload which variables are passed to the method.
If you tried to make a method public whose super method was private, you would get a compiler error, and your program would not run.
No, because this methods all have the same method signature:
Definition: Two of the components of a method declaration comprise the method
signature — the method's name and the parameter types.
Compiler can only distinguish between methods based on their signature.
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.