为什么class A中的initialize（）方法为protected的时候，在执行public A(int f1) 里的 initialize()方法时，它调用的是class B里面的 initialize()方法。而把classA里面的 initialize()改为private时，它执行的就是class A 里面的 initialize()方法。有人能为我解答一下么

public static class A {
    private int f1 = 7;

    public A(int f1) {
        this.f1 = f1;
        initialize();
    }
    //what a mean s...
    // so what's happening here...
    // when creating the B object, first the A constructor is called and A::f1 is set to 6
    // this is just a smoke candle! cause now initialize is called. Cause we have a B::object polymorphism is beaming us to the
    // B::initialize method (not the A::initialize as we might expect due to the code layout). B::initialize is printing f1
    // and f1 of course here refers to the B::f1 field... and that has not been initialized yet, it is still 0
    // So if we stop dynamic binding from invoking B::initialize we're all set. And that's pretty easy to do if we stop overriding
    // 2 possibilities: a) making the base initialize final or b) making it private. Both stops overriding as the child class
    // won't have access to the method in question anymore. Then  B::initialize is
    // an independent method. We can verify this with annotating it with @Override. The compiler should warn us that
    // B::initialize doesn't override A::initialize (after applying mentioned change)

    // oh, I see final does not work... the compiler warns about initialize not being able to override, but private works...???
    // and ahhh, of course it is logical that it does not work with final... final should prevent overriding and exactly this
    // is happening while private just defines visibility. And that it is doing perfectly.
    // And the @Override annotation works as well... hope I got it and it is really as I imagine it to be.
    private void initialize() {
        System.out.println(f1);
    }
}