You can, as Objective C method dispatch is all dynamic. Just call it with [self methodOfChild], which will probably generate a compiler warning (which you can silence by casting self to id). But, for the love of goodness, don’t do it. Parents are supposed to provide for their children, not the children for their parents. … Read more
Any constructor for any class as you know creates an object. So, the constructor should contain proper initialization code for its class. But if you have some class which extends another one (lets call it “parent”) then constructor for the class cannot contain all the code needed for the initialization by definition (for example, you … Read more
If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass. If the super class does not have a no-argument constructor, you will get a compile-time error. Object does have such a constructor, so if Object is the only superclass, there is … Read more
Constructors are not inherited, you must create a new, identically prototyped constructor in the subclass that maps to its matching constructor in the superclass. Here is an example of how this works: class Foo { Foo(String str) { } } class Bar extends Foo { Bar(String str) { // Here I am explicitly calling the … Read more
Use an interface if you want to define a contract. I.e. X must take Y and return Z. It doesn’t care how the code is doing that. A class can implement multiple interfaces. Use an abstract class if you want to define default behaviour in non-abstract methods so that the endusers can reuse it without … Read more
That is how Java works. The constructors of the parent classes are called, all the way up the class hierarchy through Object, before the child class’s constructor is called. Quoting from the docs: With super(), the superclass no-argument constructor is called. With super(parameter list), the superclass constructor with a matching parameter list is called. Note: … Read more
You can do: super.alphaMethod1(); Note, that super is a reference to the parent class, but super() is its constructor.
? as a type parameter can only be used in methods. eg: printAll(MyList<? extends Serializable>) I cannot define classes with ? as type parameter. A wildcard (?) isn’t a formal type parameter, but rather can be used as a type argument. In the example you give, ? extends Serializable is given as a type argument … Read more
When creating an Employee you’re creating a Person at the same time. To make sure the Person is properly constructed, the compiler adds an implicit call to super() in the Employee constructor: class Employee extends Person { Employee(int id) { super(); // implicitly added by the compiler. } } Since Person does not have a … Read more
No, the private member are not inherited because the scope of a private member is only limited to the class in which it is defined. Only the public and protected member are inherited. From the Java Documentation, Private Members in a Superclass A subclass does not inherit the private members of its parent class. However, … Read more