That's not correct. Java method dispatch is always dynamic, so x.foo() will always call the foo() method on whatever concrete type x actually is. The declared t

That's not correct. Java method dispatch is always dynamic, so x.foo() will always call the foo() method on whatever concrete type x actually is. The declared type of the variable that holds the reference to that object doesn't matter, nor does casting (with a couple exceptions, none of which apply here). But don't trust me, try it! | Welcome to JShell -- Version 11.0.5 | For an introduction type: /help intro jshell> class Foo { void foo() { System.out.println("Super"); } } | created class Foo jshell> class Bar extends Foo { void foo() { System.out.println("Sub"); } } | created class Bar jshell> Foo f = new Bar(); f ==> Bar@58651fd0 jshell> Bar b = new Bar(); b ==> Bar@5419f379 jshell> f.foo(); Sub jshell> b.foo(); Sub jshell> ((Foo)b).foo(); Sub