Shadowing static method

Question

I was reading about shadowing static methods, and I can't understand one thing, how compiler chose method to run. Suppose I have 3 classes.

public static class A {
        public static int test() {
            return 1;
        }
    }

    public static class B extends A {
        public static int test() {
            return 2;
        }
    }

    public static class C extends B {
        public static int test() {
            return 3;
        }
    }

Obviously when I call method A.test() B.test() C.test() i will have results 1 2 3

But if i call it from instance (I know bad practice), then When this code:

C c = new C();
System.out.println(c.test());

will print out 3 as i expect, but when I do

B b = c;
System.out.println(b.test());

then my output will be 2. Which is surprising for me, as b was instantiate as object of class C. Any reason why it was implemented like that?

Answer 1

then my output will be 2. Which is surprising for me, as b was instantiate as object of class C. Any reason why it was implemented like that?

Static method's bind to Type rather than instance. So the right side part doesn't matter in case of static method.

Inshort, static members belongs to class rather than instance.

Refer the JLS : 15.12.4. Run-Time Evaluation of Method Invocation

Answer 2

The decision is made by the compiler, always, and is based on the declared type of the variable. The runtime type of the object -- which is what matters for instance methods -- isn't considered at all. In fact, you can call static methods through a reference variable which is set to null, and it will work just fine.

Answer 3

First and foremost accessing static methods through instances is bad & compiler should have given you a warning. You should heed it. Then when you invoke b.test the type of b is B not C as you are expecting. So B.test is invoked.