Practice›Java Fundamentals›Module 8. Interfaces & Abstract Classes›8.3 Abstract classes

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8.3 Abstract classes — the abstract type in use

Lesson

An abstract class is still a type

Even though you cannot instantiate it, an abstract class is a type: you can declare Shape s = new Circle(1);, build a Shape[] of mixed subclasses, and take a Shape parameter. Calling area() or describe() dispatches to the actual object — the same polymorphism as before.

Worked example

Shape[] shapes = { new Circle(1), new Rect(2, 3) };
for (int i = 0; i < shapes.length; i++) {
    System.out.println(shapes[i].describe());
}
// Circle has area 3.14
// Rect has area 6.0

The array is typed Shape, but each element behaves as its real class.

▶ Try it: run this — one loop over an abstract type, two different areas.

abstract class Shape {
    private String name;
    public Shape(String name) { this.name = name; }
    public String getName() { return name; }
    public abstract double area();
    public String describe() { return name + " has area " + area(); }
}
class Circle extends Shape {
    private double r;
    public Circle(double r) { super("Circle"); this.r = r; }
    public double area() { return 3.14 * r * r; }
}
class Rect extends Shape {
    private double w, h;
    public Rect(double w, double h) { super("Rect"); this.w = w; this.h = h; }
    public double area() { return w * h; }
}
public class Main {
    public static void main(String[] args) {
        Shape[] shapes = { new Circle(1), new Rect(2, 3) };
        for (int i = 0; i < shapes.length; i++) {
            System.out.println(shapes[i].describe());
        }
    }
}

Can a variable have an abstract class type?

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A Shape[] where Shape is abstract can hold...

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Type exactly what this program prints.

abstract class Shape { public abstract double area(); }
class Sq extends Shape { public double area(){ return 4.0; } }
class Tri extends Shape { public double area(){ return 6.0; } }
public class Main { public static void main(String[] a){
    Shape[] arr = { new Sq(), new Tri() };
    for (int i = 0; i < arr.length; i++) System.out.println(arr[i].area());
}}

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Match each feature to where it belongs.

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True or false?

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