This should demonstrate how to solve it: import java.lang.reflect.Field; class Super { private int i = 5; } public class B extends Super { public static void main(String[] args) throws Exception { B b = new B(); Field f = b.getClass().getSuperclass().getDeclaredField(“i”); f.setAccessible(true); System.out.println(f.get(b)); } } (Or Class.getDeclaredFields for an array of all fields.) Output: 5
The rules for auto are, for the most part, the same as for template type deduction. The example posted works for the same reason you can pass objects of private types to template functions: template <typename T> void fun(T t) {} int main() { Foo f; fun(f.Baz()); // ok } And why can we pass … Read more
I think one reason it works this way is because access modifiers work at compile time. As such, determining whether or not a given object is also the current object isn’t easy to do. For example, consider this code: public class Foo { private int bar; public void Baz(Foo other) { other.bar = 2; } … Read more
The inner class is just a way to cleanly separate some functionality that really belongs to the original outer class. They are intended to be used when you have 2 requirements: Some piece of functionality in your outer class would be most clear if it was implemented in a separate class. Even though it’s in … Read more
No, there’s no way to do it. That would essentially be scoping in reverse. Methods defined inside the constructor have access to private variables because all functions have access to the scope in which they were defined. Methods defined on a prototype are not defined within the scope of the constructor, and will not have … Read more
Update: See others answer, this is outdated. Short answer, no, there is no native support for private properties with ES6 classes. But you could mimic that behaviour by not attaching the new properties to the object, but keeping them inside a class constructor, and use getters and setters to reach the hidden properties. Note that … Read more