java polymorphism late binding rules

问题:

I am trying to do some polymorphism exercises and I cannot figure out how this kind of polymorphism works. I did not find any deep information about this kind of exercised. I hope you guys could give me some explanation.

exercise 1:

class Top {
    public void m( Middle p ) System.out.print("A ");
}

class Middle extends Top {
    public void m( Object p ) System.out.print("M ");
    public void m( Middle p ) System.out.print("L ");
}

class Bottom extends Middle {
    public void m( Object p ) System.out.print("V ");
    public void m( Middle p ) System.out.print("X ");
}

class Test {
    public static void run() {
        Top tm = new Middle();
        Middle mb = new Bottom();

        tm.m (mb);            -> L
        tm.m(new Bottom());   -> L   why?
        mb.m(mb);             -> X
        mb.m(tm);             -> V   why?
        mb.m(new Middle());   -> X
        new Bottom().m(tm);   -> V
    }
}

exercise 2:

class Top {
    public void gg( Top o ) System.out.print("A ");
    public void gg( Middle m ) System.out.print("B ");
}

class Middle extends Top {
    public void gg( Top o ) System.out.print("L ");
    public void gg( Bottom b ) System.out.print("M ");
}

class Bottom extends Middle {
    public void gg( Top o ) System.out.print("X ");
    public void gg( Middle m) System.out.print("Z ");
}

class Test {
    public static void run() {
        Top oo = new Top();
        Top ff = new Middle();
        Bottom uu = new Bottom();

        oo.gg(ff);      -> A
        oo.gg(uu);      -> A   why?
        ff.gg(ff);      -> L
        ff.gg(uu);      -> B   why?
        uu.gg(ff);      -> X
        uu.gg(uu);      -> X   why?
    }
}

Thank you in advance!

Greets

回答1:

In all of these cases, the methods that can be considered depend on the compile-time type of the variable, but the method that is actually called depends on the runtime type of the object. So for

Top ff = new Middle();

the methods of Middle are the ones that will be called - but these may be inherited from Top, and we can only call methods that are available in Top at compile time, because ff is declared as a Top.

To determine which of the overloaded methods is called, we look at the parameter type, and choose the most specific method. So if we have to choose between:

public void m( Object p ) System.out.print("M ");
public void m( Middle p ) System.out.print("L ");

and we are passing a Bottom, then the second method will be chosen. You can think of Bottom as being closer to Middle than Object in the class hierarchy:

Bottom -> Middle -> Top -> Object

Finally, some of your answers are just wrong (exercise 2) - I suggest you try running the code, which may need a bit of tweaking so it actually compiles.

oo.gg(uu); // -> A why?  -- actually produces B
uu.gg(uu); // -> X why?  -- actually produces M