How private method of super class are resolved?

Question

class A{ 
  private void sayA(){
     System.out.println("Private method of A");
  }
  public static void main(String args[]){
      A instanceA=new B();
      instanceA.sayA();
  }
}

class B extends A{
}

I was expecting it to throw a run time exception as at compile-time the compiler checks if sayA() can be called on a reference of A and at run-time it'll check if sayA() can be called on B's object. But it instead printed "Private method of A".

Answer 1

Accessibility is a compile time concept (reflected in Reflection APIs).

From the Java Language Specification

Note that accessibility is a static property that can be determined at compile time; it depends only on types and declaration modifiers.

That is, the compiler doesn't care what the runtime type of the instance referenced by your variable named instanceA will be

A instanceA = new B();

It only cares that you invoked a method on a reference of static type A. That method is private and since you are within the body of the class which declares it, it is visible, and therefore usable.

Otherwise, the member or constructor is declared private, and access is permitted if and only if it occurs within the body of the top level class (§7.6) that encloses the declaration of the member or constructor.

---

For spiderman in comments, consider the following

class A {
    private void privateMethod () {
        System.out.println("private method");
    }
    public void publicMethod() {
        privateMethod();
    }
}

class B extends A {}

class Example {
    public static void main(String[] args) {
        new B().publicMethod();
    }
}

Answer 2

private means that only the class that declares the field can see it. Because you're calling instanceA.sayA(); from within class A the method is visible and the code both compiles and runs. If you were to try to call that method from within class B or any other class you would get the compile warning that The method sayA() from the type A is not visible