Strange case of overriding a static method in java

Question

Everywhere they write that the static method cannot be overridden, but when I try to reduce the access specifier, from open to protected, it gives an error. eg

public class StaticOverrideFunda { public static void foo(){ System.out.println("Parent Foo"); } } public class B extends StaticOverrideFunda{ protected static void foo(){ System.out.println("Child Foo"); } /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub B.foo(); } } 

It says

Cannot reduce the visibility of a legacy method

So it depends on the basic rules, why do we say that foo is not overridden in class B? Why are we saying this is hiding / shading and not overriding?

Answer 1

It follows some of the same rules as redefinition, but that does not mean that it redefines. In this case, these are the rules in the section of 8.4.8.3 JLS , "Requirements for Overriding and Hiding":

The access modifier (§6.6) of the override or hide method must provide at least the same access as the overridden or hidden method, as follows: [...]

It still does not override, since the method will not be called polymorphically - you cannot write a call that sometimes ends with a call to StaticOverrideFunda.foo , and sometimes calls B.foo ; the target is fully determined at compile time.

It would be useful to consider the rest of section 8.4.8, which defines overriding as something that happens in instance methods.

Answer 2

You yourself posted the answer in your question, overriding a tool that has different code in the child class for the same method. Because static methods cannot be overridden, you cannot play with visibility by changing access specifiers.