Abstraction (training)

[Analeea]

Please help me to understand how inheritance works in the context of work with abstract classes and their heirs, to challenge the methods and designers of both, and to identify a specific example with me.

There's an abstract class. Animaland there's his descendant. Monkey♪

abstract public class Animal {
public int nLegs = 0;
public void numLegs(int nLegs) {
    this.nLegs = nLegs;
}


public abstract int getLegs();

//Constructor
public Animal (int nLegs) {
    numLegs(nLegs)
}

}

---

public class Monkey extends Animals {

public int getLegs(){
    return this.nLegs;
}
public monkey(){
    super(2);
}

public void saySome(){
    System.out.println("Hello, I am a monkey!");
}

}

♪ Animal Yes. переменная nLegs (legs) метод numLegs, the requisition of this variable, геттер getLegsto obtain its meaning and конструктор, rear value переменной nLegs by метода numLegs♪

♪ Monkey We are. реализуем геттер (exhaustive method to be implemented), создаем конструкторwhich causes конструктор Animal, assigning its local variable value 2I'm also writing an extra one. метод saySomethat puts the line in the console.

Is that right? Then let's go on.

In the main class, write:

public class MainClass{

 public static void main(String []args){

    Animal jay = new Monkey ();
    System.out.println(jay.nLegs);

   /* jay.saySome()*/
    ((Monkey) jay).saySome();

 }

}

Issues:

1. What does that mean when the object is jay? Monkey♪
   But how do I read the logic of that expression? Animal jay
= new Monkey (); like, Класса Animal объект jay = новый объект jay класса Monkey

2. Why is the method I've jay.saySome() doesn't work? Why does it work? ((Monkey) jay).saySome();?

[kalena]

The ghetter is not reappointed but is being implemented. ♪ Animal It's only declared but not implemented. ♪ Monkey You do it. Use the right terms, they don't just exist. Like the designer, he's not gonna work like a designer. Animal"and he's inside just calling the designer. Animal♪

Expression Animal jay = new Monkey(); means you're creating an object. Monkey and assign it to a variable type Animalbecause Monkey It's too. Animal♪ And then you work with the variable. jay c Animaland in the head knowing that it's actually true. Monkey♪

The commented method doesn't work exactly because jay - it's common. Animalwithout specifying. In class Animal No method saySome♪ But since you know that jay Yes Monkey, you clearly lead the type and cause the method from class. Monkey

UPD

Responses to questions from the commentary:

(2) For the object of the class, the descendant inherits all the fields of the ancestor class and the object of the class is both the object of the descendant class and the object of the ancestor class (I don't get Class(), you don't need to know this at the current stage). The class in the declaration of a variable can be understood as a type, yes.

(3) He cannot inherit both structures. He has one of his own, his own, as described in the class. Monkeyand he inherits the fields and methods of the class. Animal♪

(4) Monkey jay = new Monkey(); - that's why not. The question is what you want. For example, if you have another class, Elephant extends Animalit also implemented the method getLegsand you'll want to put all the animals on the same list in your code and then find out how many legs they have:

List<Animal> animals = fillList(); //Заполним список как-нибудь
for(Animal animal : animals){
    System.out.println(animal.getLegs());
} 

In this piece of code, you don't care where the elephant is, and where the monkey is, you just know what they have in common (all have legs) and you use it. In case within the cycle animal it'll be a elephant, it'll be a elephant. getLegs Class Elephantif animal It'll be a monkey, it'll be called. getLegs Class Monkey♪