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Java in a Nutshell

Previous Chapter 3
Classes and Objects in Java

3.7 Overriding Methods

When a class defines a method using the same name, return type, and arguments as a method in its superclass, the method in the class overrides the method in the superclass. When the method is invoked for an object of the class, it is the new definition of the method that is called, not the superclass' old definition.

Method overriding is an important and useful technique in object-oriented programming. Suppose we define a subclass Ellipse of our Circle class. [9] Then it would be important for Ellipse to override the area() and circumference() methods of Circle. Ellipse would have to implement new versions of these functions because the formulas that apply to circles don't work for ellipses.

[9] This is admittedly a strange thing to do, since, mathematically, a circle is a kind of ellipse, and it is customary to derive a more specific class from a more general one. Nevertheless, it is a useful example here.

Before we go any further with the discussion of method overriding, be sure that you understand the difference between method overriding and method overloading, which we discussed earlier. As you probably recall, method overloading refers to the practice of defining multiple methods (in the same class) with the same name but with differing argument lists. This is very different from method overriding, and it is important not to get them confused!

Overriding Is Not Shadowing

Although Java treats the variables and methods of a class analogously in many ways, method overriding is not like variable shadowing at all: You can refer to shadowed variables simply by casting an object to the appropriate type. You cannot invoke overridden methods with this technique, however. Example 3.12 illustrates this crucial difference.

Example 3.12: Method Overriding versus Variable Shadowing

class A {
    int i = 1;
    int f() { return i; }
class B extends A {
    int i = 2;                          // Shadows variable i in class A.
    int f() { return -i; }              // Overrides method f in class A.
public class override_test {
    public static void main(String args[]) {
        B b = new B();
        System.out.println(b.i);     // Refers to B.i; prints 2.
        System.out.println(b.f());   // Refers to B.f(); prints -2.
        A a = (A) b;                 // Cast b to an instance of class A.
        System.out.println(a.i);     // Now refers to A.i; prints 1;
        System.out.println(a.f());   // Still refers to B.f(); prints -2;

While this difference between method overriding and variable shadowing may seem surprising at first, a little thought makes the purpose clear. Suppose we have a bunch of Circle and Ellipse (a subclass of Circle) objects that we are manipulating. To keep track of the circles and ellipses, we store them in an array of type Circle[], casting all the Ellipse objects to Circle objects before we store them. Then, when we loop through the elements of this array, we don't have to know or care whether the element is actually a Circle or an Ellipse. What we do care very much about, however, is that the correct value is computed when we invoke the area() method of any element of the array. That is, we don't want to use the formula for the area of a circle when the object is actually an ellipse!

Seen in this context, it is not surprising at all that method overriding is handled differently by Java than variable shadowing.

final Methods

If a method is declared final, it means that the method declaration is the "final" one--that it cannot be overridden. static methods and private methods (which we haven't learned about yet) cannot be overridden either, nor can the methods of a final class. If a method cannot be overridden, the compiler may perform certain optimizations on it, as we'll see below.

Dynamic Method Lookup

If we have an array of Circle and Ellipse objects, how does the compiler know to call the Circle area() method or the Ellipse area() method for any given item in the array? The compiler does not know this; it can't. The compiler knows that it does not know, however, and produces code that uses "dynamic method lookup" at run-time. When the interpreter runs the code, it looks up the appropriate area() method to call for each of the objects. That is, when the interpreter interprets the expression s.area(), it dynamically looks for an area() method associated with the particular object referred to by the variable s. It does not simply use the area() method that is statically associated with the type of the variable s. [10]

[10] C++ programmers should note that dynamic method lookup is what C++ does for virtual functions. An important difference between Java and C++ is that Java does not have a virtual keyword; methods in Java are "virtual" by default.

Dynamic method lookup is fast, but it is not as fast as invoking a method directly. Fortunately, there are a number of cases in which Java does not need to use dynamic method lookup. static methods cannot be overridden, so they are always invoked directly. private methods (which we haven't learned about yet) are not inherited by subclasses and so cannot be overridden by subclasses; this means the Java compiler can safely invoke them without dynamic method lookup as well. final methods are invoked directly for the same reason: they cannot be overridden. Finally, when a method of a final class is invoked through an instance of the class or a subclass of it, then it, too, can be invoked without the overhead of dynamic lookup. These static, final, and private methods that can be invoked directly are also candidates for inlining--i.e., if the methods are short, the compiler may simply insert the method body into the code rather than inserting a call to the method.

Invoking an Overridden Method

We've seen the important differences between method overriding and variable shadowing. Nevertheless, the Java syntax for invoking an overridden method is very similar to the syntax for accessing a shadowed variable: both use the super keyword. Example 3.13 illustrates this.

Example 3.13: Invoking an Overridden Method

class A {
    int i = 1;
    int f() { return i; }        // A very simple method.
class B extends A {
    int i;                       // This variable shadows i in A.
    int f() {                    // This method overrides f() in A.
        i = super.i + 1;         // It retrieves A.i this way.
        return super.f() + i;    // And it invokes A.f() this way.

Recall that when you use super to refer to a shadowed variable, it is the same as casting this to the superclass type and accessing the variable through that. On the other hand, using super to invoke an overridden method is not the same as casting this. In this case, super has the special purpose of turning off dynamic method lookup and invoking the specific method that the superclass defines or inherits.

In Example 3.13 we use super to invoke an overridden method that is actually defined in the immediate superclass. super also works perfectly well to invoke overridden methods that are defined further up the class hierarchy. This is because the overridden method is inherited by the immediate superclass, and so the super syntax does in fact refer to the correct method.

To make this more concrete, suppose class A defines method f, and that B is a subclass of A, and that C is a subclass of B that overrides method f. Then you can still use:


to invoke the overridden method from within class C. This is because class B inherits method f from class A. If classes A, B, and C all define method f, however, then calling super.f() in class C invokes class B's definition of the method. In this case, there is no way to invoke A.f() from within class C. super.super.f() is not legal Java syntax!

It is important to note that super can only be used to invoke overridden methods from within the class that does the overriding. With our Circle and Ellipse classes, for example, there is no way to write a program (with or without super) that invokes the Circle area() method on an object of type Ellipse. The only way to do this is to use super in a method within the Ellipse class.

Finally, note that this form of super does not have to occur in the first statement in a method, as it does when used to invoke a superclass constructor method.

Finalizer Chaining Revisited

Now that we've discussed method overriding and how to invoke an overridden method, we can return to the issue of the finalizer method that we left dangling earlier on.

In Java, constructor methods are automatically chained, but finalizer methods are not. If you define a finalize() method to free resources allocated by your class, you may be overriding a finalize() method in a superclass that frees resources allocated by that class. If your finalizer method does not invoke the superclass finalizer, the superclass finalizer never gets called, and resources are not deallocated when they should be.

To prevent this, you should be sure to invoke the superclass finalize() method. The best time to do this is usually after your finalize() method has done all of its deallocation. It is a good idea to add the following call:


as the last line of all your finalizer methods. You should do this even if you know that none of your class's superclasses have finalizer methods, because future implementations of the class may include one.

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