java - How is the code flows in method overriding -

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for below code got output 

in base.foo()   in derived.bar()

code:

class base {     public static void foo(base bobj) {         system.out.println("in base.foo()");         bobj.bar();     }       public void bar() {         system.out.println("in base.bar()");     } } class derived extends base {     public static void foo(base bobj) {         system.out.println("in derived.foo()");         bobj.bar();     }     public void bar() {         system.out.println("in derived.bar()");     } } class overridetest {     public static void main(string []args) {         base bobj = new derived();         bobj.foo(bobj);     } }

how code flows? how getting output above. confused little bit , looking explanation.

static methods not overridden, there's no virtual invocation.

in derived class, static foo method hiding parent class' static foo method.

however, in main method, invoking foo on reference type base, hence base#foo gets invoked (so prints "in base.foo()").

in turn, static method invokes bar on passed object, is-a base, happens instance of derived.

since bar instance method , is overridden given signature matches, invocation resolves instance type, derived, hence "in derived.bar()" printed.

notes

  • a practice employ @override annotation on overridden methods. fail compile foo method it's static, pointing out there's no possible overriding there.
  • another practice (as pointed out jon skeet) avoid invoking static methods on instance objects (i.e. prefer explicit classname.staticmethod idiom), may cause confusion on scope in method invoked.

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