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Java: Inheritance, Dynamic Binding or what? [duplicate]

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 does java compiler choose correct methods and variables in inheritance

I am a bit confused when inheritance and type casting is mixed. I want to understand the rules which java compiler follows when choosing correct methods and variables in inheritance.
I have read something like
Variables are bound at compile time and methods are bound at run time.
The second one is from stackoverflow (#John Skeet)
Overload resolution (which method signature is called) is determined at compile-time, based on the compile-time types of both the method target and the argument expressions
The implementation of that method signature (overriding) is based on the actual type of the target object at execution time.
But the problem is that they are explanations of specific situations and do not give the general process to be followed when other factors (like exception handling) are taken into consideration.
This may be bad practice but assume that both methods and variables are overridden(hidden in case of static variables).
Now if java compiler has to choose at compile time which method/variable needs to be invoked then what algorithm will it follow? Similarly at run time what is the algorithm that java compiler will use(based on the actual type of the object whose reference is being used)?

All method signatures and variables are verified during compile-time but the actual method calls are done / resolved during run-time.
For example :
class A {
int i=5;
public void doSomething(){
//print "in A"
}
}
class B extends A{
int i=10;
public void doSomething(){
// print "in B"
}
public static void main(String[] args){
A a = new B();
a.doSomething();
}
}
Now, when you call a.doSomething(); , during compilation, the compiler just checks whether doSomething() is defined for class A (lhs reference). It is not even bothered about whether the method is also defined for class B. Even if the method were not present in B, the program would compile fine.
Next, during runtime, the JVM dynamically decides which method to call based on type of Object (B in your case.).
So, "in B" is printed.
Now, coming back to fields. Access to fields are resolved during compile time. So, if a field doesn't exist during compile-time, then the compilation fails. The fields are called based on the reference type. So, a.i will print 5 (A's value of i) because the field was resolved during compile-time. Thus, the difference is, method calls are resolved during runtime, their signatures are needed / checked during compile-time where-as fields are checked / resolved during compile-time.

Does Java support dynamic method invocation?

class A { void F() { System.out.println("a"); }}
class B extends A { void F() { System.out.println("b"); }}
public class X {
public static void main(String[] args) {
A objA = new B();
objA.F();
}
}
Here, F() is being invoked dynamically, isn't it?
This article says:
... the Java bytecode doesn’t support
dynamic method invocation. There are
three supported invocations modes :
invokestatic, invokespecial,
invokeinterface or invokevirtual.
These modes allows to call methods
with known signature. We talk of
strongly typed language. This allows
to to make some checks directly at
compile time.
On the other side, the dynamic
languages use dynamic types. So we can
call a method unknown at the compile
time, but that’s completely impossible
with the Java bytecode.
What am I missing?

You are confusing dynamic invocation with dynamic binding..
The first one allows the type checker to accept programs in which you are not sure if a method will be present on an object at run-time, while dynamic binding just chooses the right implementation according to the runtime type of the object but maintaining the statically type checking.
What does it mean?
It means that in your example, Java will call the implementation on object B because the runtime type of the objA variable is B; and it will compile because it knows that a B is a A so the method invocation won't fail at runtime (objA will have a F implementation for sure).
With dynamic invocation instead it won't check at compile time that the type of the object on which you are calling F contains that method, of course it will raise an exception if during execution the method won't be available on specified object.
Just for trivia: the invokedynamic feature will be added with Java7 because many scripting languages have been written to work on top of JVM and the lack of a dynamic invocation feature forced the developers of these languages to add a middle layer between the script and the real JVM that cares about dynamic invocation using the reflection. Of course this approach causes a lot of overhead (think about Grovvy's MetaClass), that's why Sun decided to give them a help..

In your example the correct method is called because polymorphically the instance of B appears like an instance of A. The method can be located by examining the runtime type of the object; that is, B; as opposed to the compile-time type of the object reference, A. The other important part is the signature of method - these must always match (polymorphically of course).
This differs from dynamic languages because in those there is essentially no compile-time for the object - and everything must be resolved at runtime.

In fact, what you're missing is that this is the part of 'invokevirtual' which is explained in the article.
You're simply overriding the method and that uses a virtual method table to invoke the correct method.

I wouldn't call your example "dynamic", rather virtual. Because at compile time the method name and signature is known (and its existence is checked by the compiler). The only that is resolved at runtime is the concrete implementation to be used for that method.
A more proper example of "dynamic" method invocation would involve reflection, (see the Method class ). In that way, methods whose existence are unkown at compile type can be invoked in runtime (this is extensively used by frameworks, not so much by application code).
The article you mention seems a little misleading in that respect. But it's true that the signature of the methods you explicitly call must be known/checked in compile time, and so, in that sense, Java is not dynamic.

You can make functional interfaces.
class Logger {
private BiConsumer<Object, Integer> logger = null;
// ...
private Logger(Object logger) {
this.akkaLogger = (LoggingAdapter) logger;
this.logger = (message, level) -> {
switch (level) {
case INFO: akkaInfo(message);
break;
case DEBUG: akkaDebug(message);
break;
case ERROR: akkaError(message);
break;
case WARN: akkaWarn(message);
break;
}
};
}
private Logger() {
this.logger = (message, level) -> System.out.println(message);
}
// ...
}

Why does Java's invokevirtual need to resolve the called method's compile-time class?

Consider this simple Java class:
class MyClass {
public void bar(MyClass c) {
c.foo();
}
}
I want to discuss what happens on the line c.foo().
Original, Misleading Question
Note: Not all of this actually happens with each individual invokevirtual opcode. Hint: If you want to understand Java method invocation, don't read just the documentation for invokevirtual!
At the bytecode level, the meat of c.foo() will be the invokevirtual opcode, and, according to the documentation for invokevirtual, more or less the following will happen:
Look up the foo method defined in compile-time class MyClass. (This involves first resolving MyClass.)
Do some checks, including: Verify that c is not an initialization method, and verify that calling MyClass.foo wouldn't violate any protected modifiers.
Figure out which method to actually call. In particular, look up c's runtime type. If that type has foo(), call that method and return. If not, look up c's runtime type's superclass; if that type has foo, call that method and return. If not, look up c's runtime type's superclass's superclass; if that type has foo, call that method and return. Etc.. If no suitable method can be found, then error.
Step #3 alone seems adequate for figuring out which method to call and verifying that said method has the correct argument/return types. So my question is why step #1 gets performed in the first place. Possible answers seem to be:
You don't have enough information to perform step #3 until step #1 is complete. (This seems implausible at first glance, so please explain.)
The linking or access modifier checks done in #1 and #2 are essential to prevent certain bad things from happening, and those checks must be performed based on the compile-time type, rather than the run-time type hierarchy. (Please explain.)
Revised Question
The core of the javac compiler output for the line c.foo() will be an instruction like this:
invokevirtual i
where i is an index to MyClass' runtime constant pool. That constant pool entry will be of type CONSTANT_Methodref_info, and will indicate (maybe indirectly) A) the name of the method called (i.e. foo), B) the method signature, and C) the name of compile time class that the method is called on (i.e. MyClass).
The question is, why is the reference to the compile-time type (MyClass) needed? Since invokevirtual is going to do dynamic dispatch on the runtime type of c, isn't it redundant to store the reference to the compile-time class?

It is all about performance. When by figuring out the compile-time type (aka: static type) the JVM can compute the index of the invoked method in the virtual function table of the runtime type (aka: dynamic type). Using this index step 3 simply becomes an access into an array which can be accomplished in constant time. No looping is needed.
Example:
class A {
void foo() { }
void bar() { }
}
class B extends A {
void foo() { } // Overrides A.foo()
}
By default, A extends Object which defines these methods (final methods omitted as they are invoked via invokespecial):
class Object {
public int hashCode() { ... }
public boolean equals(Object o) { ... }
public String toString() { ... }
protected void finalize() { ... }
protected Object clone() { ... }
}
Now, consider this invocation:
A x = ...;
x.foo();
By figuring out that x's static type is A the JVM can also figure out the list of methods that are available at this call site: hashCode, equals, toString, finalize, clone, foo, bar. In this list, foo is the 6th entry (hashCode is 1st, equals is 2nd, etc.). This calculation of the index is performed once - when the JVM loads the classfile.
After that, whenever the JVM processes x.foo() is just needs to access the 6th entry in the list of methods that x offers, equivalent to x.getClass().getMethods[5], (which points at A.foo() if x's dynamic type is A) and invoke that method. No need to exhaustively search this array of methods.
Note that the method's index, remains the same regardless of the dynamic type of x. That is: even if x points to an instance of B, the 6th methods is still foo (although this time it will point at B.foo()).
Update
[In light of your update]: You're right. In order to perform a virtual method dispatch all the JVM needs is the name+signature of the method (or the offset within the vtable). However, the JVM does not execute things blindly. It first checks that the cassfiles loaded into it are correct in a process called verification (see also here).
Verification expresses one of the design principles of the JVM: It does not rely on the compiler to produce correct code. It checks the code itself before it allows it to be executed. In particular, the verifier checks that every invoked virtual method is actually defined by the static type of the receiver object. Obviously, the static type of the receiver is needed to perform such a check.

That's not the way I understand it after reading the documentation. I think you have steps 2 and 3 transposed, which would make the whole series of events more logical.

Presumably, #1 and #2 have already happened by the compiler. I suspect that at least part of the purpose is to make sure that the they still hold with the version of the class in the runtime environment, which may be different from the version the code was compiled against.
I haven't digested the invokevirtual documentation to verify your summary, though, so Rob Heiser could be right.

I'm guessing answer "B".
The linking or access modifier checks done in #1 and #2 are essential to prevent certain bad things from happening, and those checks must be performed based on the compile-time type, rather than the run-time type hierarchy. (Please explain.)
#1 is described by 5.4.3.3 Method Resolution, which makes some important checks. For example, #1 checks the accessibility of the method in the compile-time type and may return an IllegalAccessError if it is not:
...Otherwise, if the referenced method is not accessible (§5.4.4) to D, method resolution throws an IllegalAccessError. ...
If you only checked the run-time type (via #3), then the run-time type could illegally widen the accessibility of the overridden method (a.k.a. a "bad thing"). Its true that the compiler should prevent such a case, but the JVM is nevertheless protecting itself from rogue code (e.g. manually-constructed malevolent code).

To totally understand this stuff, you need to understand how method resolution works in Java. If you're looking for an in-depth explanation, I suggest looking at the book, "Inside the Java Virtual Machine". The following sections from Chapter 8, "The Linking Model", are available online and seem particularly relevant:
Chapter introduction
Resolution of CONSTANT_Methodref_info Entries
Direct References
(CONSTANT_Methodref_info entries are entries in the class file header that describe the methods called by that class.)
Thanks to Itay for inspiring me to do the Googling required to find this.

Overriding private methods in Java

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.