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Unit 14: Polymorphism

Learning Objectives

Students should

  • understand polymorphism.
  • be aware of dynamic binding.
  • be aware of the equals method and the need to override it to customize the equality test.
  • understand when narrowing type conversion and type casting are allowed.

Taking on Many Forms

Method overriding enables polymorphism, the fourth and the last pillar of OOP. Arguably this is the most powerful pillar. It allows us to change how existing code behaves, without changing a single line of the existing code (or even having access to the code).

Consider the function say(Object) below: 

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void say(Object obj) {
  System.out.println("Hi, I am " + obj.toString());
}

Note that this method receives an Object instance. Since both Point <: Object and Circle <: Object, we can do the following (recap: "whenever a superclass is needed, a subclass can be given"): 

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Point p = new Point(0, 0);
say(p);
Circle c = new Circle(p, 4);
say(c);

When executed, say will first print Hi, I am (0.0, 0.0), followed by Hi, I am { center: (0.0, 0.0), radius: 4.0 }. We are invoking the overriding Point::toString() in the first call, and Circle::toString() in the second call. The same method invocation obj.toString() causes two different methods to be called in two separate invocations!

In biology, polymorphism means that an organism can have many different forms. Here, the variable obj can have many forms as well. Which method is invoked is decided during run-time, depending on the run-time type of the obj. This is called dynamic binding or late binding or dynamic dispatch.

Before we get into this in more detail, let's consider overriding Object::equals(Object).

The equals method

Object::equals(Object) compares if two object references refer to the same object. That is also the standard behavior of the == operator. In the case of == operator, we can say that it compares the value. So what is the value of a reference type? Well, it must be the reference (i.e., the address).

equals Method

While it is not necessary to know this, equals method should satisfy an equivalence relation on non-null object references:

  • It is reflexive: for any non-null reference value x, x.equals(x) should return true.
  • It is symmetric: for any non-null reference values x and y, x.equals(y) should return true if and only if y.equals(x) returns true.
  • It is transitive: for any non-null reference values x, y, and z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.
  • It is consistent: for any non-null reference values x and y, multiple invocations of x.equals(y) consistently return true or consistently return false, provided no information used in equals comparisons on the objects is modified.
  • For any non-null reference value x, x.equals(null) should return false.

One thing to be careful of when using equals method is that you must ensure in the method invocation obj.equals(arg), the value of obj is not null.

Suppose we have:

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Circle c0 = new Circle(new Point(0, 0), 10);
Circle c1 = new Circle(new Point(0, 0), 10);
Circle c2 = c1;

c2.equals(c1) returns true, but c0.equals(c1) returns false. Even though c0 and c1 are semantically the same, they refer to the two different objects.

To compare if two circles are semantically the same, we need to override this method1. After all, Java does not know what you meant by having two circle being equal. There are other possibilities including (but not limited to)

  • having equal radius
  • having equal center point
  • having both equal radius and center point
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// version 0.7
/**
 * A Circle object encapsulates a circle on a 2D plane.  
 */
class Circle {
  private Point c;   // the center
  private double r;  // the length of the radius

  /**
   * Create a circle centered on Point c with given radius r
   */
  public Circle(Point c, double r) {
    this.c = c;
    this.r = r;
  }

  /**
   * Return the area of the circle.
   */
  public double getArea() {
    return Math.PI * this.r * this.r;
  }

  /**
   * Return true if the given point p is within the circle.
   */
  public boolean contains(Point p) {
    return false;
    // TODO: Left as an exercise
  }

  /**
   * Return the string representation of this circle.
   */
  @Override
  public String toString() {
    return "{ center: " + this.c + ", radius: " + this.r + " }";
  }

  /**
   * Return true the object is the same circle (i.e., same center, same radius).
   */
  @Override
  public boolean equals(Object obj) {
    if (obj instanceof Circle) {
      Circle circle = (Circle) obj;
      return (circle.c.equals(this.c) && circle.r == this.r);
    }
    return false;
  }
}

This is more complicated than toString. There are a few new concepts involved here so let's go through them slowly. Click on the tabs below to find out more.

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public boolean equals(Object obj) {
  if (obj instanceof Circle) {
    Circle circle = (Circle) obj;
    return (circle.c.equals(this.c) && circle.r == this.r);
  }
  return false;
}

equals takes in a parameter of compile-time type Object (Line 1). It only makes sense if we compare (during run-time) a circle with another circle. So, we first check if the run-time type of obj is a subtype of Circle. This is done using the instanceof operator. The operator returns true if obj has a run-time type that is a subtype of Circle.

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public boolean equals(Object obj) {
  if (obj instanceof Circle) {
    Circle circle = (Circle) obj;
    return (circle.c.equals(this.c) && circle.r == this.r);
  }
  return false;
}

To compare this circle with the given circle, we have to access the center c and radius r. But if we access obj.c or obj.r, the compiler will complain. As far as the compiler is concerned, obj has the compile-time type Object, and there is no such fields c and r in the class Object! This is why, after assuring that the run-time type of obj is a subtype of Circle, we assign obj to another variable circle that has the compile-time type Circle. We finally check if the two centers are equal (again, Point::equals is left as an exercise) and the two radii are equal2.

Another important point is that circle.c and circle.r are allowed even when the fields c and r are declared private because we are inside the class Circle. So even if the fields belong to another object, as long as it is in the same class, it is allowed.

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public boolean equals(Object obj) {
  if (obj instanceof Circle) {
    Circle circle = (Circle) obj;
    return (circle.c.equals(this.c) && circle.r == this.r);
  }
  return false;
}

The statement that assigns obj to circle involves type casting. We mentioned before that Java is strongly typed and so it is very strict about type conversion. Here, Java allows type casting from type \(T\) to \(S\) if \(S <: T\)3. By allow, we meant that Java compiler will not give compilation error.

This is called narrowing type conversion. Unlike widening type conversion, which is always allowed and always correct, a narrowing type conversion requires explicit typecasting and validation during run-time. If we do not ensure that obj has the correct run-time type, casting can lead to a run-time error (which if you recall, is bad).

The syntax for explicit type conversion is (Type) expr. Java will attempt to convert the result of the expression expr into the type Type. While you may also add explicit type casting for widening type conversion, such explicit type casting is not necessary and can be omitted.

All these complications would go away, however, if we define Circle::equals to take in a Circle as a parameter, like this:

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class Circle {
    :
  /**
   * Return true the object is the same circle (i.e., same center, same radius).
   */
  @Override
  public boolean equals(Circle circle) {
    return (circle.c.equals(this.c) && circle.r == this.r);
  }
}

This version of equals however, does not override Object::equals(Object). Since we hinted to the compiler that we meant this to be an overriding method, using @Override, the compiler will give us an error. This is not treated as method overriding, since the signature for Circle::equals(Circle) is different from Object::equals(Object).

Why then is overriding important? Why not just leave out the line @Override and live with the non-overriding, one-line, equals method above? Keep this question in mind as this will be answered when we discuss dynamic binding.

Why is it called Narrowing?

Because the subclass inherits from superclass, it may seem like the subclass has "more" properties. So it seems counter-intuitive why such conversion is called narrowing. Take, for instance, Circle and ColoredCircle. ColoredCircle has an additional property called color. So why converting from Circle to ColoredCircle is called narrowing?

The reason here is that a ColoredCircle is a Circle but the opposite is not true (i.e., Circle is not ColoredCircle). Consider adding another class SpinningCircle such that SpinningCircle <: Circle. Then, Circle contains all possible Circle + all possible ColoredCircle + all possible SpinningCircle. On the other hand, SpinningCircle only contains all possible SpinningCircle.

We can see that the possible values for Circle is larger than the possible values for SpinningCircler. So converting from Circle to SpinningCircle will indeed narrow down our possible value.

The Power of Polymorphism

Let's consider the following example. Suppose we have a general contains method that takes in an array of objects. The array can store any type of objects: Circle, Square, Rectangle, Point, String, etc. The method contains also takes in a target obj to search for, and returns true if there is an object in array that equals to obj.

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// version 0.1 (with polymorphism)
boolean contains(Object[] array, Object obj) {
  for (Object curr : array) {
    if (curr.equals(obj)) {
      return true;
    }
  }
  return false;
}

With overriding and polymorphism, the magic happens in Line 4 -- depending on the run-time type of curr, the corresponding, customized version of equals is called to compare against obj. So if the run-time type of curr is Circle, then we will invoke Circle::equals(Object) and if the run-time type of curr is Point, then we will invoke Point::equals(Object). This, of course, assumes that Object::equals(Object) is overridden in both classes.

Now, if Circle::equals takes in a Circle as the parameter (i.e., Circle::equals(Circle)), the call to equals inside the method contains would not invoke Circle::equals(Circle). It would invoke Object::equals(Object) instead due to the matching method signature, and we can't search for Circle based on semantic equality.

But why are we searching for equals(Object) in the first place? Look closely at how the method is invoked: curr.equals(obj). Here, we can see that the parameter we are passing is obj. The compile-time type of obj is Object as seen from the parameter declaration at Line 2. So at compile-time, we only know that its type is Object.

To have a generic contains method without polymorphism and overriding, we will have to do something like this:

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// version 0.2 (without polymorphism)
boolean contains(Object[] array, Object obj) {
  for (Object curr : array) {
    if (obj instanceof Circle) {
      if (curr.equals((Circle)obj)) {
        return true;
      }
    } else if (obj instanceof Square) {
      if (curr.equals((Square)obj)) {
        return true;
      }
    } else if (obj instanceof Point) {
      if (curr.equals((Point)obj)) {
        return true;
      }
    }
     :
  }
  return false;
}

which is not scalable since every time we add a new class, we have to come back to this method and add a new branch to the if-else statement!

As this example has shown, polymorphism allows us to write succinct code that is future proof. By dynamically deciding which method implementation to execute during run-time, the implementer can write short yet very general code that works for existing classes as well as new classes that might be added in the future by the client, without even the need to re-compile!

Java Array

Note that there are two ways to declare an array in Java:

  1. Type[] var
  2. Type var[] (i.e., C-style array)

Please follow the first style as it will be less confusing. That is also the preferred style based on our style guide.

The next point will be important later when we talk about variance of types. When we declare a class in Java like the class A below:

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class A { .. }

Another class is also created corresponding to the type of array of type A. The name given by Java internally is [LA;. Notice how the name is not a valid name in Java, so you cannot create such a class by yourself. This class has a field called length which is the number of element the array instance may contain. This should hopefully explains why to get the number of elements in an array, we use arr.length and not arr.length() because it is a field and not a method.

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class [LA; { .. } // but not a legal name

The more interesting thing is when you declare a subtype of class A such as the class B below:

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class B extends A { .. }

The type of array of type B (i.e., [LB;) is automatically created such that [LB; is a subtype of [LA;. In other words,

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class [LB; extends [LA; { .. } // still not a legal name

Adding Class

So to recap, one of the main benefit of polymorphism is the ability to extend your current program with new classes without without modifying what you have written before. To make the example more explicit, imagine that the following classes are already written.

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class Animal {
  public String name() {
    return "Animal";
  }

  public String sound() {
    return "Grrrr";
  }
}

class Cow extends Animal {
  @Override
  public String name() {
    return "Cow";
  }

  @Override
  public String sound() {
    return "Moo";
  }
}

class Duck extends Animal {
  @Override
  public String name() {
    return "Duck";
  }

  @Override
  public String sound() {
    return "Quack";
  }
}

Consider the following method to sing Old MacDonald.

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void sing(Animal[] animals) {
  for(Animal animal : animals) {
    System.out.println("Old MacDonald had a farm E-I-E-I-O");
    System.out.println("And on that farm he had a " + animal.name() + " E-I-E-I-O");
    System.out.println("With a " + animal.sound() + " " + animal.sound() + " here");
    System.out.println("and a " + animal.sound() + " " + animal.sound() + " there");
    System.out.println("Here a " + animal.sound() + " there a " + animal.sound());
    System.out.println("Everywhere a " + animal.sound() + " " + animal.sound());
    System.out.println("Old MacDonald had a farm E-I-E-I-O");
  }
}

Now if we want to add a new animal, we simply have to create a new class that extends the Animal class and implements the two methods name and sound. We do not have to modify the all the codes we have written before.

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class Dog extends Animal {
  @Override
  public String name() {
    return "Dog";
  }

  @Override
  public String sound() {
    return "Woof";
  }
}

Notice how the method sing(Animal[]) works without modification. Additionally, we do not have to modify all the other classes as well. That is the power of polymorphism.

Adding Functionality

There is a "dual" problem to adding a class and that's adding functionality. This problem is difficult in Java. Consider the animals above including Animal, Cow, Duck, and Dog. Now since all animals can move but they move differently, let's add a method called move(). Where should we add this method to?

Firstly, since all animals can move, we have to add this method to the Animal class. Secondly, since all animals move differently, we also have to override this method in Cow, Duck, Dog. Therefore, the act of adding a single functionality requires us to modify all the relevant classes!

In this case, we only have 4 classes, but if we have more classes, there are more places we have to changed. This breaks the principle of abstraction. In fact, this breaks abstraction barrier if we are not the implementer of one of the animal classes. To make matters worse, if we forgot to override this method in one of the subclasses, we will not even get compilation error.

Design

Hopefully you understand the power of polymorphism as well as its limitation. In particular, when you design your classes according to OOP principle, you should try to design it in such a way that the changes you expect are about adding classes rather than adding functionality.

  1. If we override equals(Object), we should generally override hashCode() as well, but let's leave that for another lesson on another day. ↩

  2. The right way to compare two floating-point numbers is to take their absolute difference and check if the difference is small enough. We are sloppy here to keep the already complicated code a bit simpler. You shouldn't do this in your code. ↩

  3. This is not the only condition where type casting is allowed. We will look at other conditions in later units. ↩