Overrding

[Gautam Sewani]

public class SuperClass {
public static void main(String ar[])
{
System.out.println(new SubClass().g());
}
public int g()
{
return f();
}
private int f()
{
return 1;
}
}
class SubClass extends SuperClass{
public int f()
{
return 2;
}
}
I cannot understand why does this code produces an output of 1?
Can anyone explain?

[Arun Pai]

SubClass is not having method g(), it will implicitly call the SuperClass g(), HTH.
-Arun

[swapna sivaraju]

HI
Here the call new SubClass().g() invokes the inherited method g() from SuperClass (as g() is not present in SubClass) and then in the method
g() there is a call to f() in the statement return f()...now since f() is private in the SuperClass so the method f() of SuperClass is invoked not the one in SubClass( as u may think that u r working on the SubClass object)..Also,
u should know that private and static methods r never overridden...
In this code the method public int f() {}
is not overriding the private int f(){} in SubClass..
THats why the answer is 1.
Hope this helps
Swapna

[Steven Sun]

Another the to explain:
member variales are static-linked as well as the private methods,it is said that when using "new" to construct an instance,the member variables and preivate methods are definite.but other methods not,it is linked at runtime
in this case,method g() in Super already knows where to invoke the f(),because f() is private.it is already linked to the g() method as well as the instance.
so is clear