Java中的“超级”关键字

Question

I have an issue regarding the super keyword in java. Follow the example below:

public class Circle {
private double radius;
private double area; 

public void setRadius(double radius){
    this.radius = 1; 

}
public double getRadius(){
    return this.radius;
}
public void setArea(double radius){
    this.area = area;
}
public double getArea(){
    return this.area = Math.PI * radius * radius;
}
}


public class Cylinder extends Circle {
    private double height;
    public Cylinder(){
        super();
        height = 1;
    }
    public Cylinder(double height){
        super();
        this.height = height;
    }
    public Cylinder(double radius, double height){
        super();
        this.height = height;
        public double getHeight(){
            return height;
        }
    }
public double getVolume(){
    return getArea()*height;
}
}

My point is that, when I use super() in the subclass, how java will know which constructor to call in my subclass ? Since in my superclass, I have two constructors with no arguments; public double getRadius() and public double getArea()

Answer 1

There is only one constructor in your super class, the default no argument constructor which is not explicitly defined within the class. Each of the subclass's constructors will invoke this constructor in the super class.

getRadius() and getArea() are just methods within the super class, they are not constructors. Remember that constructors always take the form of:

[access modifier] [Class Name](/* arguments optional*/){}

So a constructor for the Circle class would look like:

public Circle(/*Arguments could go here */){

}

If there were more than one constructor in the super class, let's say:

public Circle(int radius){
  //....
}

public Circle(double radius){
  //....
}

The compiler would use the arguments to determine which constructor to call.

Answer 2

Constructor name will be the class name. You are talking about the methods there.

class classname{

   classname(){// constructor name as class name.
   }

}