How the compiler chooses which method to invoke?

Question

  public class A {
        public void foo(Object o) {
            System.out.println("A.foo with an object");
  public class B extends A {
        public void foo(Object o) {
            System.out.println("B.foo with an object");   
        public void foo(String s) {
            System.out.println("B.foo with a string");   
public class C {
    public static void main(String[] args) {
        String s = "hello";
        A a = new B();
        a.foo(s);
}why the output is "B.foo with an object" and not "B.foo with a String".
or more generally: how the compiler choses the method to invoke?

Answer 1

or more generally: how the compiler choses the method
to invoke?Variable a is declared as a reference to an A. The compiler only knows this. It doesn't know that the object it happens to be pointing at is a B. So it only looks at A's methods.
A has only foo(Object), so it will call the method with the signature foo(Object).
Which signature to use is determined at compile time, and is determined by the type of the reference.
Now, as for whether A's foo or B's foo, that's where Java's polymorphism comes in. In this case, the VM looks at the object, not the reference, and calls the "deepest" (furthest from Object) class' method that matches that signature.
Which class' implementation of that signature to call is determined at runtime, and is determined by the object, not the reference.