Re: No late binding for static methods

Question

Only the sale class has a showAdvertisement() method.
When the compiler compiles that class it decides at compile time which exact method is called for each static method call (that's the big difference to non-static method calls: the exact method to be called is only decided at runtime in those cases).
There's only one method showAdvertisement() and that is called in both cases in your code. Since that single method always calls the same static announcement() method, you will see the result you posted.

Answer 1

JoachimSauer wrote:
When the compiler compiles that class it decides at compile time which exact method is called for each static method call (that's the big difference to non-static method calls: the exact method to be called is only decided at runtime in those cases).I thought static methods were bound at classload-time. Demo:
public class A {
    public static void f() {
        System.out.println("A.f");
public class B extends A  {
    public static void f() {
        System.out.println("B.f");
public class C  {
    public static void main(String[] args) {
        B.f();
}If you run the above code, "A.f" is printed, as you would expect -- there is only one static method f defined.
But if you uncomment the code in class B and recompile only it (not C!), then run C, you will get the output "B.f".
If you run javap -c C you will see:
public static void main(java.lang.String[]);
  Code:
   0:   invokestatic    #2; //Method B.f:()V
   3:   returnSo the byte code isn't binding the call to A.f at compile time. The choice is made at classload-time. That is why there are similar restrictions on hiding static methods as for overriding instance methods: [http://java.sun.com/docs/books/jls/third_edition/html/classes.html#8.4.8.3]