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| 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:

    super.f()

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:

    super.finalize();

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.

Subclasses and Inheritance |  | Data Hiding and Encapsulation