Your code as written will not successfully compile. You cannot say "A ac = new C();" because C does not extend A. If you claim to have run this and gotten output, you must have mis-copied something from your running code into this post.
If, for the sake of argument, your code really said "A ac = new A();", then your code still wouldn't run, because A.someMethod takes an A, not a C. The statement ac.someMethod(c) executes the function A.someMethod. ac is of type A, so executing any function against ac will get the function of that name from class A. Trying to pass a parameter of type C to a function declared to take a parameter of type A will not "switch" you to using a function from a different class that does take such a parameter. Overloading only works within a class.
Perhaps what you're thinking of is an example more like this:
class A {
public void someMethod(A a) {
System.out.println("A");
}
public void someMethod(B b) {
System.out.println("B");
}
}
class Dmd {
public static void main(String[] args) {
A a = new A();
B b = new B();
a.someMethod(b);
}
}
This will output "B".
The difference here is that the class A has two versions of the function someMethod. One of them takes an A, one takes a B. When we call it with a B, we get the B version.
Do you see how this is different from your example? You are declaring three classes each with a function someMethod. Here we have one class with two functions both named someMethod. That's very different in Java.
