This is my code
package alpha ;
class A1
{
static class A11
{
private
final // WHAT IS THE EFFECT OF THIS MODIFIER?
void fun ( String caller )
{
System . out . println ( "A11:\t" + caller ) ;
}
}
static class A12 extends A11
{
private void fun ( String caller )
{
super . fun ( caller + caller ) ;
}
}
public static void main ( String [ ] args )
{
A12 a12 = new A12 ( ) ;
a12 . fun ( "Hello" ) ;
}
}
I have found that with or without the final mdifer in A1.A11 the program compiles and runs.
I can understand that without the final modifier, A1.A12 can see and thus override the fun method. It is private but they are in the same class so there is no visibility issue.
I can not understand why it works with the final modifier. Should not the overriding in A1.A12 be prohibited?
This is the output of the program with the final modifer in place
java alpha/A1
A11: HelloHello
If it was simply ignoring the other fun method then
Your methods are private.
Change their visibility to protected to see expected behavior, that is, only when the method is protected, public or default visibility, the concept of overriding even exists.
Doing something like this --
class A1
{
static class A11
{
public
final // WHAT IS THE EFFECT OF THIS MODIFIER?
void fun ( String caller )
{
System . out . println ( "A11:\t" + caller ) ;
}
}
static class A12 extends A11
{
public void fun ( String caller )
{
super . fun ( caller + caller ) ;
}
}
public static void main ( String [ ] args )
{
A12 a12 = new A12 ( ) ;
a12 . fun ( "Hello" ) ;
}
}
will now throw the compile-time exception
fun(java.lang.String) in A1.A12 cannot override fun(java.lang.String) in A1.A11; overridden method is final