Why does the line marked with //1 print 57 instead of 39?

Question

class X {
    protected int v = 0;

    public X() {
        v += 10;
    }

    public void proc(X p) {
        System.out.println(43);
    }
}

class Y extends X {
    public Y() {
        v += 5;
    }

    public void proc(X p) {
        System.out.println(57);
    }

    public int getV() {
        return v;
    }
}

class Z extends Y {
    public Z() {
        v += 9;
    }

    public void proc(Z p) {
        System.out.println(39);
    }
}

class Main {
    public static void main(String[] args) {
        X x = new Z();
        Y y = new Z();
        Z z = new Z();
        x.proc(z);// 1
        System.out.println(y.getV());
    }
}

From what I can understand,the method proc() is called on an object of type X that "holds" a type Z and at runtime JVM checks the object's type and overrides the method with the proc() method from Y.But the method parameter is of type Z,why doesn't it call the overloaded method from the Z class?

Answer 1

It happens because you don't override the method 'proc' in Z class. When you overriding the method, you can't use argument, which has a child class of original argument class. If you add @Override on Z.proc(Z p), your code will be not compiled.

Let's imagine that it is possible, then you can use some method from Z class during executing Z.proc(Z p).

class Z extends Y {
    public Z() {
        v += 9;
    }

    public void proc(Z p) {
        someActions();
        System.out.println(39);
    }

    private void someActions() {
        System.out.println("Some actions");
    }

}

Now when you executing

X x = new Z();
x.proc(new X());

What should happens? There is no 'someActions' method in the X class. How it should works? That's why Z.proc(Z p) doesn't override X.proc(X p). Class Z, has two different methods: Z.proc(Z p) and Y.proc(X p).

When you calling

X x = new Z();
x.proc(new Z());

JVM looks for closest overrided or original method with signature 'proc(X)' to Z class (because X class has 'proc(X)' method) finds it in Y class and executing Y.proc(xp). That's why you see '57' in output.

Answer 2

Because you're passing an X to proc , so the proc for class Y prevails.

If you passed an actual Z declared as a Z , it would have printed 39.

Z x = new Z();
X y = new Z();
Z z = new Z();


z.proc(x); // prints 39
z.proc(y); // prints 57

Answer 3

class X {
    protected int v = 0;

    public X() {
        v += 10;
    }

    public void proc(X p) {
        System.out.println(43);
    }
}

class Y extends X {
    public Y() {
        v += 5;
    }

    public void proc(X p) {
        System.out.println(57);
    }

    public int getV() {
        return v;
    }
}

class Z extends Y {
    public Z() {
        v += 9;
    }

    public void proc(Z p) {
        System.out.println(39);
    }
}

class Main {
    public static void main(String[] args) {
        X x = new Z();
        Y y = new Z();
        Z z = new Z();
        x.proc(z);// 1
        System.out.println(y.getV());
    }
}

From what I can understand,the method proc() is called on an object of type X that "holds" a type Z and at runtime JVM checks the object's type and overrides the method with the proc() method from Y.But the method parameter is of type Z,why doesn't it call the overloaded method from the Z class?

Answer 4

Just to elaborate over Federico answer. The "x" object is, in fact, a "X" variable that point to a "Z" instance in memory. "Z" instance has only two methods:

void proc(Z p)
void proc(X p)

But the runtime only is aware of the second method that is implemented at "X" class and its overridden by the method implemented in "Y" class. So, when you call:

x.proc(z)

The runtime only knows about of this second method and call it, executing the overridden one.