2

Object Orientation

CERTIFICATION OBJECTIVES

•     Describe Encapsulation

•     Implement Inheritance

•     Use IS-A and HAS-A Relationships (OCP)

•     Use Polymorphism

•     Use Overriding and Overloading

•     Understand Casting

•     Use Interfaces

•     Understand and Use Return Types

•     Develop Constructors

•     Use static Members

   Two-Minute Drill

Q&A Self Test

Being an Oracle Certified Associate (OCA) 8 means you must be at one with the object-oriented aspects of Java. You must dream of inheritance hierarchies; the power of polymorphism must flow through you; and encapsulation must become second nature to you. (Coupling, cohesion, composition, and design patterns will become your bread and butter when you’re an Oracle Certified Professional [OCP] 8.) This chapter will prepare you for all the object-oriented objectives and questions you’ll encounter on the exam. We have heard of many experienced Java programmers who haven’t really become fluent with the object-oriented tools that Java provides, so we’ll start at the beginning.

CERTIFICATION OBJECTIVE

Encapsulation (OCA Objectives 6.1 and 6.5)

6.1 Create methods with arguments and return values; including overloaded methods.

6.5 Apply encapsulation principles to a class.

Imagine you wrote the code for a class and another dozen programmers from your company all wrote programs that used your class. Now imagine that later on, you didn’t like the way the class behaved, because some of its instance variables were being set (by the other programmers from within their code) to values you hadn’t anticipated. Their code brought out errors in your code. (Relax, this is just hypothetical.) Well, it is a Java program, so you should be able to ship out a newer version of the class, which they could replace in their programs without changing any of their own code.

This scenario highlights two of the promises/benefits of an object-oriented (OO) language: flexibility and maintainability. But those benefits don’t come automatically. You have to do something. You have to write your classes and code in a way that supports flexibility and maintainability. So what if Java supports OO? It can’t design your code for you. For example, imagine you made your class with public instance variables, and those other programmers were setting the instance variables directly, as the following code demonstrates:

And now you’re in trouble. How are you going to change the class in a way that lets you handle the issues that come up when somebody changes the size variable to a value that causes problems? Your only choice is to go back in and write method code to adjust size (a setSize(int a) method, for example) and then insulate the size variable with, say, a private access modifier. But as soon as you make that change to your code, you break everyone else’s!

The ability to make changes in your implementation code without breaking the code of others who use your code is a key benefit of encapsulation. You want to hide implementation details behind a public programming interface. By interface, we mean the set of accessible methods your code makes available for other code to call—in other words, your code’s API. By hiding implementation details, you can rework your method code (perhaps also altering the way variables are used by your class) without forcing a change in the code that calls your changed method.

If you want maintainability, flexibility, and extensibility (and, of course, you do), your design must include encapsulation. How do you do that?

   Keep instance variables hidden (with an access modifier, often private).

   Make public accessor methods, and force calling code to use those methods rather than directly accessing the instance variable. These so-called accessor methods allow users of your class to set a variable’s value or get a variable’s value.

   For these accessor methods, use the most common naming convention of set<SomeProperty> and get<SomeProperty>.

Figure 2-1 illustrates the idea that encapsulation forces callers of our code to go through methods rather than accessing variables directly.

FIGURE 2-1   The nature of encapsulation

We call the access methods getters and setters, although some prefer the fancier terms accessors and mutators. (Personally, we don’t like the word “mutate.”) Regardless of what you call them, they’re methods that other programmers must go through in order to access your instance variables. They look simple, and you’ve probably been using them forever:

Wait a minute. How useful is the previous code? It doesn’t even do any validation or processing. What benefit can there be from having getters and setters that add no functionality? The point is, you can change your mind later and add more code to your methods without breaking your API. Even if today you don’t think you really need validation or processing of the data, good OO design dictates that you plan for the future. To be safe, force calling code to go through your methods rather than going directly to instance variables. Always. Then you’re free to rework your method implementations later, without risking the wrath of those dozen programmers who know where you live.

Note: In Chapter 6 we’ll revisit the topic of encapsulation as it applies to instance variables that are also reference variables. It’s trickier than you might think, so stay tuned! (Also, we’ll wait until Chapter 6 to challenge you with encapsulation-themed mock questions.)

CERTIFICATION OBJECTIVE

Inheritance and Polymorphism (OCA Objectives 7.1 and 7.2)

7.1 Describe inheritance and its benefits.

7.2 Develop code that demonstrates the use of polymorphism; including overriding and object type versus reference type (sic).

Inheritance is everywhere in Java. It’s safe to say that it’s almost (almost?) impossible to write even the tiniest Java program without using inheritance. To explore this topic, we’re going to use the instanceof operator, which we’ll discuss in more detail in Chapter 4. For now, just remember that instanceof returns true if the reference variable being tested is of the type being compared to. This code

produces this output:

they’re not equal
t1’s an Object

Where did that equals method come from? The reference variable t1 is of type Test, and there’s no equals method in the Test class. Or is there? The second if test asks whether t1 is an instance of class Object, and because it is (more on that soon), the if test succeeds.

Hold on…how can t1 be an instance of type Object, when we just said it was of type Test? I’m sure you’re way ahead of us here, but it turns out that every class in Java is a subclass of class Object (except, of course, class Object itself). In other words, every class you’ll ever use or ever write will inherit from class Object. You’ll always have an equals method, a clone method, notify, wait, and others available to use. Whenever you create a class, you automatically inherit all of class Object’s methods.

Why? Let’s look at that equals method for instance. Java’s creators correctly assumed that it would be very common for Java programmers to want to compare instances of their classes to check for equality. If class Object didn’t have an equals method, you’d have to write one yourself—you and every other Java programmer. That one equals method has been inherited billions of times. (To be fair, equals has also been overridden billions of times, but we’re getting ahead of ourselves.)

The Evolution of Inheritance

Up until Java 8, when the topic of inheritance was discussed, it usually revolved around subclasses inheriting methods from their superclasses. While this simplification was never perfectly correct, it became less correct with the new features available in Java 8. As the following table shows, it’s now possible to inherit concrete methods from interfaces. This is a big change. For the rest of the chapter, when we talk about inheritance generally, we will tend to use the terms “subtypes” and “supertypes” to acknowledge that both classes and interfaces need to be accounted for. We will tend to use the terms “subclass” and “superclass” when we’re discussing a specific example that’s under discussion. Inheritance is a key aspect of most of the topics we’ll be discussing in this chapter, so be prepared for LOTS of discussion about the interactions between supertypes and subtypes!

As you study the following table, you’ll notice that as of Java 8 interfaces can contain two types of concrete methods, static and default. We’ll discuss these important additions later in this chapter.

Table 2-1 summarizes the elements of classes and interfaces relative to inheritance.

TABLE 2-1   Inheritable Elements of Classes and Interfaces

For the exam, you’ll need to know that you can create inheritance relationships in Java by extending a class or by implementing an interface. It’s also important to understand that the two most common reasons to use inheritance are

   To promote code reuse

   To use polymorphism

Let’s start with reuse. A common design approach is to create a fairly generic version of a class with the intention of creating more specialized subclasses that inherit from it. For example:

outputs:

displaying shape
moving game piece

Notice that the PlayerPiece class inherits the generic displayShape() method from the less-specialized class GameShape and also adds its own method, movePiece(). Code reuse through inheritance means that methods with generic functionality—such as displayShape(), which could apply to a wide range of different kinds of shapes in a game—don’t have to be reimplemented. That means all specialized subclasses of GameShape are guaranteed to have the capabilities of the more general superclass. You don’t want to have to rewrite the displayShape() code in each of your specialized components of an online game.

But you knew that. You’ve experienced the pain of duplicate code when you make a change in one place and have to track down all the other places where that same (or very similar) code exists.

The second (and related) use of inheritance is to allow your classes to be accessed polymorphically—a capability provided by interfaces as well, but we’ll get to that in a minute. Let’s say that you have a GameLauncher class that wants to loop through a list of different kinds of GameShape objects and invoke displayShape() on each of them. At the time you write this class, you don’t know every possible kind of GameShape subclass that anyone else will ever write. And you sure don’t want to have to redo your code just because somebody decided to build a dice shape six months later.

The beautiful thing about polymorphism (“many forms”) is that you can treat any subclass of GameShape as a GameShape. In other words, you can write code in your GameLauncher class that says, “I don’t care what kind of object you are as long as you inherit from (extend) GameShape. And as far as I’m concerned, if you extend GameShape, then you’ve definitely got a displayShape() method, so I know I can call it.”

Imagine we now have two specialized subclasses that extend the more generic GameShape class, PlayerPiece and TilePiece:

Now imagine a test class has a method with a declared argument type of GameShape, which means it can take any kind of GameShape. In other words, any subclass of GameShape can be passed to a method with an argument of type GameShape. This code

outputs:

displaying shape
displaying shape

The key point is that the doShapes() method is declared with a GameShape argument but can be passed any subtype (in this example, a subclass) of GameShape. The method can then invoke any method of GameShape, without any concern for the actual runtime class type of the object passed to the method. There are implications, though. The doShapes() method knows only that the objects are a type of GameShape since that’s how the parameter is declared. And using a reference variable declared as type GameShape—regardless of whether the variable is a method parameter, local variable, or instance variable—means that only the methods of GameShape can be invoked on it. The methods you can call on a reference are totally dependent on the declared type of the variable, no matter what the actual object is, that the reference is referring to. That means you can’t use a GameShape variable to call, say, the getAdjacent() method even if the object passed in is of type TilePiece. (We’ll see this again when we look at interfaces.)

IS-A and HAS-A Relationships

Note: As of Winter 2017, the OCA 8 exam won’t ask you directly about IS-A and HAS-A relationships. But understanding IS-A and HAS-A relationships will help OCA 8 candidates with many of the questions on the exam.

IS-A

In OO, the concept of IS-A is based on inheritance (or interface implementation). IS-A is a way of saying, “This thing is a type of that thing.” For example, a Mustang is a type of Horse, so in OO terms we can say, “Mustang IS-A Horse.” Subaru IS-A Car. Broccoli IS-A Vegetable (not a very fun one, but it still counts). You express the IS-A relationship in Java through the keywords extends (for class inheritance) and implements (for interface implementation).

A Car is a type of Vehicle, so the inheritance tree might start from the Vehicle class as follows:

In OO terms, you can say the following:

Vehicle is the superclass of Car.
Car is the subclass of Vehicle.
Car is the superclass of Subaru.
Subaru is the subclass of Vehicle.
Car inherits from Vehicle.
Subaru inherits from both Vehicle and Car.
Subaru is derived from Car.
Car is derived from Vehicle.
Subaru is derived from Vehicle.
Subaru is a subtype of both Vehicle and Car.

Returning to our IS-A relationship, the following statements are true:

“Car extends Vehicle” means “Car IS-A Vehicle.”
“Subaru extends Car” means “Subaru IS-A Car.”

And we can also say:

“Subaru IS-A Vehicle”

because a class is said to be “a type of” anything further up in its inheritance tree. If the expression (Foo instanceof Bar) is true, then class Foo IS-A Bar, even if Foo doesn’t directly extend Bar, but instead extends some other class that is a subclass of Bar. Figure 2-2 illustrates the inheritance tree for Vehicle, Car, and Subaru. The arrows move from the subclass to the superclass. In other words, a class’s arrow points toward the class from which it extends.

FIGURE 2-2   Inheritance tree for Vehicle, Car, Subaru

HAS-A

HAS-A relationships are based on use, rather than inheritance. In other words, class A HAS-A B if code in class A has a reference to an instance of class B. For example, you can say the following:

A Horse IS-A Animal. A Horse HAS-A Halter.

The code might look like this:

In this code, the Horse class has an instance variable of type Halter (a halter is a piece of gear you might have if you have a horse), so you can say that a “Horse HAS-A Halter.” In other words, Horse has a reference to a Halter. Horse code can use that Halter reference to invoke methods on the Halter and get Halter behavior without having Halter-related code (methods) in the Horse class itself. Figure 2-3 illustrates the HAS-A relationship between Horse and Halter.

FIGURE 2-3   HAS-A relationship between Horse and Halter

HAS-A relationships allow you to design classes that follow good OO practices by not having monolithic classes that do a gazillion different things. Classes (and their resulting objects) should be specialists. As our friend Andrew says, “Specialized classes can actually help reduce bugs.” The more specialized the class, the more likely it is that you can reuse the class in other applications. If you put all the Halter-related code directly into the Horse class, you’ll end up duplicating code in the Cow class, UnpaidIntern class, and any other class that might need Halter behavior. By keeping the Halter code in a separate, specialized Halter class, you have the chance to reuse the Halter class in multiple applications.

Users of the Horse class (that is, code that calls methods on a Horse instance) think that the Horse class has Halter behavior. The Horse class might have a tie(LeadRope rope) method, for example. Users of the Horse class should never have to know that when they invoke the tie() method, the Horse object turns around and delegates the call to its Halter class by invoking myHalter.tie(rope). The scenario just described might look like this:

In OO, we don’t want callers to worry about which class or object is actually doing the real work. To make that happen, the Horse class hides implementation details from Horse users. Horse users ask the Horse object to do things (in this case, tie itself up), and the Horse will either do it or, as in this example, ask something else (like perhaps an inherited Animal class method) to do it. To the caller, though, it always appears that the Horse object takes care of itself. Users of a Horse should not even need to know that there is such a thing as a Halter class.

CERTIFICATION OBJECTIVE

Polymorphism (OCA Objective 7.2)

7.2 Develop code that demonstrates the use of polymorphism; including overriding and object type versus reference type (sic).

Remember that any Java object that can pass more than one IS-A test can be considered polymorphic. Other than objects of type Object, all Java objects are polymorphic in that they pass the IS-A test for their own type and for class Object.

Remember, too, that the only way to access an object is through a reference variable. There are a few key things you should know about references:

   A reference variable can be of only one type, and once declared, that type can never be changed (although the object it references can change).

   A reference is a variable, so it can be reassigned to other objects (unless the reference is declared final).

   A reference variable’s type determines the methods that can be invoked on the object the variable is referencing.

   A reference variable can refer to any object of the same type as the declared reference, or—this is the big one—it can refer to any subtype of the declared type!

   A reference variable can be declared as a class type or an interface type. If the variable is declared as an interface type, it can reference any object of any class that implements the interface.

Earlier we created a GameShape class that was extended by two other classes, PlayerPiece and TilePiece. Now imagine you want to animate some of the shapes on the gameboard. But not all shapes are able to be animated, so what do you do with class inheritance?

Could we create a class with an animate() method and have only some of the GameShape subclasses inherit from that class? If we can, then we could have PlayerPiece, for example, extend both the GameShape class and Animatable class, whereas the TilePiece would extend only GameShape. But no, this won’t work! Java supports only single class inheritance! That means a class can have only one immediate superclass. In other words, if PlayerPiece is a class, there is no way to say something like this:

class PlayerPiece extends GameShape, Animatable { // NO!
  // more code
}

A class cannot extend more than one class: that means one parent per class. A class can have multiple ancestors, however, since class B could extend class A, and class C could extend class B, and so on. So any given class might have multiple classes up its inheritance tree, but that’s not the same as saying a class directly extends two classes.

Some languages (such as C++) allow a class to extend more than one other class. This capability is known as “multiple inheritance.” The reason that Java’s creators chose not to allow multiple class inheritance is that it can become quite messy. In a nutshell, the problem is that if a class extended two other classes, and both superclasses had, say, a doStuff() method, which version of doStuff() would the subclass inherit? This issue can lead to a scenario known as the “Deadly Diamond of Death,” because of the shape of the class diagram that can be created in a multiple inheritance design. The diamond is formed when classes B and C both extend A, and both B and C inherit a method from A. If class D extends both B and C, and both B and C have overridden the method in A, class D has, in theory, inherited two different implementations of the same method. Drawn as a class diagram, the shape of the four classes looks like a diamond.

So if that doesn’t work, what else could you do? You could simply put the animate() code in GameShape, and then disable the method in classes that can’t be animated. But that’s a bad design choice for many reasons—it’s more error prone; it makes the GameShape class less cohesive; and it means the GameShape API “advertises” that all shapes can be animated when, in fact, that’s not true since only some of the GameShape subclasses will be able to run the animate() method successfully.

So what else could you do? You already know the answer—create an Animatable interface, and have only the GameShape subclasses that can be animated implement that interface. Here’s the interface:

And here’s the modified PlayerPiece class that implements the interface:

So now we have a PlayerPiece that passes the IS-A test for both the GameShape class and the Animatable interface. That means a PlayerPiece can be treated polymorphically as one of four things at any given time, depending on the declared type of the reference variable:

   An Object (since any object inherits from Object)

   A GameShape (since PlayerPiece extends GameShape)

   A PlayerPiece (since that’s what it really is)

   An Animatable (since PlayerPiece implements Animatable)

The following are all legal declarations. Look closely:

There’s only one object here—an instance of type PlayerPiece—but there are four different types of reference variables, all referring to that one object on the heap. Pop quiz: Which of the preceding reference variables can invoke the displayShape() method? Hint: Only two of the four declarations can be used to invoke the displayShape() method.

Remember that method invocations allowed by the compiler are based solely on the declared type of the reference, regardless of the object type. So looking at the four reference types again—Object, GameShape, PlayerPiece, and Animatable—which of these four types know about the displayShape() method?

You guessed it—both the GameShape class and the PlayerPiece class are known (by the compiler) to have a displayShape() method, so either of those reference types can be used to invoke displayShape(). Remember that to the compiler, a PlayerPiece IS-A GameShape, so the compiler says, “I see that the declared type is PlayerPiece, and since PlayerPiece extends GameShape, that means PlayerPiece inherited the displayShape() method. Therefore, PlayerPiece can be used to invoke the displayShape() method.”

Which methods can be invoked when the PlayerPiece object is being referred to using a reference declared as type Animatable? Only the animate() method. Of course, the cool thing here is that any class from any inheritance tree can also implement Animatable, so that means if you have a method with an argument declared as type Animatable, you can pass in PlayerPiece objects, SpinningLogo objects, and anything else that’s an instance of a class that implements Animatable. And you can use that parameter (of type Animatable) to invoke the animate() method, but not the displayShape() method (which it might not even have), or anything other than what is known to the compiler based on the reference type. The compiler always knows, though, that you can invoke the methods of class Object on any object, so those are safe to call regardless of the reference—class or interface—used to refer to the object.

We’ve left out one big part of all this, which is that even though the compiler only knows about the declared reference type, the Java Virtual Machine (JVM) at runtime knows what the object really is. And that means that even if the PlayerPiece object’s displayShape() method is called using a GameShape reference variable, if the PlayerPiece overrides the displayShape() method, the JVM will invoke the PlayerPiece version! The JVM looks at the real object at the other end of the reference, “sees” that it has overridden the method of the declared reference variable type, and invokes the method of the object’s actual class. But there is one other thing to keep in mind:

Polymorphic method invocations apply only to instance methods. You can always refer to an object with a more general reference variable type (a superclass or interface), but at runtime, the ONLY things that are dynamically selected based on the actual object (rather than the reference type) are instance methods. Not static methods. Not variables. Only overridden instance methods are dynamically invoked based on the real object’s type.

Because this definition depends on a clear understanding of overriding and the distinction between static methods and instance methods, we’ll cover those later in the chapter.

CERTIFICATION OBJECTIVE

Overriding/Overloading (OCA Objectives 6.1 and 7.2)

6.1 Create methods with arguments and return values; including overloaded methods.

7.2 Develop code that demonstrates the use of polymorphism; including overriding and object type versus reference type (sic).

The exam will use overridden and overloaded methods on many, many questions. These two concepts are often confused (perhaps because they have similar names?), but each has its own unique and complex set of rules. It’s important to get really clear about which “over” uses which rules!

Overridden Methods

Any time a type inherits a method from a supertype, you have the opportunity to override the method (unless, as you learned earlier, the method is marked final). The key benefit of overriding is the ability to define behavior that’s specific to a particular subtype. The following example demonstrates a Horse subclass of Animal overriding the Animal version of the eat() method:

For abstract methods you inherit from a supertype, you have no choice: You must implement the method in the subtype unless the subtype is also abstract. Abstract methods must be implemented by the first concrete subclass, but this is a lot like saying the concrete subclass overrides the abstract methods of the supertype(s). So you could think of abstract methods as methods you’re forced to override—eventually.

The Animal class creator might have decided that for the purposes of polymorphism, all Animal subtypes should have an eat() method defined in a unique way. Polymorphically, when an Animal reference refers not to an Animal instance, but to an Animal subclass instance, the caller should be able to invoke eat() on the Animal reference, but the actual runtime object (say, a Horse instance) will run its own specific eat() method. Marking the eat() method abstract is the Animal programmer’s way of saying to all subclass developers, “It doesn’t make any sense for your new subtype to use a generic eat() method, so you have to come up with your own eat() method implementation!” A (nonabstract) example of using polymorphism looks like this:

In the preceding code, the test class uses an Animal reference to invoke a method on a Horse object. Remember, the compiler will allow only methods in class Animal to be invoked when using a reference to an Animal. The following would not be legal given the preceding code:

To reiterate, the compiler looks only at the reference type, not the instance type. Polymorphism lets you use a more abstract supertype (including an interface) reference to one of its subtypes (including interface implementers).

The overriding method cannot have a more restrictive access modifier than the method being overridden (for example, you can’t override a method marked public and make it protected). Think about it: If the Animal class advertises a public eat() method and someone has an Animal reference (in other words, a reference declared as type Animal), that someone will assume it’s safe to call eat() on the Animal reference regardless of the actual instance that the Animal reference is referring to. If a subtype were allowed to sneak in and change the access modifier on the overriding method, then suddenly at runtime—when the JVM invokes the true object’s (Horse) version of the method rather than the reference type’s (Animal) version—the program would die a horrible death. (Not to mention the emotional distress for the one who was betrayed by the rogue subtype.)

Let’s modify the polymorphic example we saw earlier in this section:

If this code compiled (which it doesn’t), the following would fail at runtime:

The variable b is of type Animal, which has a public eat() method. But remember that at runtime, Java uses virtual method invocation to dynamically select the actual version of the method that will run, based on the actual instance. An Animal reference can always refer to a Horse instance, because Horse IS-A(n) Animal. What makes that supertype reference to a subtype instance possible is that the subtype is guaranteed to be able to do everything the supertype can do. Whether the Horse instance overrides the inherited methods of Animal or simply inherits them, anyone with an Animal reference to a Horse instance is free to call all accessible Animal methods. For that reason, an overriding method must fulfill the contract of the superclass.

Note: In Chapter 5 we will explore exception handling in detail. Once you’ve studied Chapter 5, you’ll appreciate this single handy list of overriding rules. The rules for overriding a method are as follows:

   The argument list must exactly match that of the overridden method. If they don’t match, you can end up with an overloaded method you didn’t intend.

   The return type must be the same as, or a subtype of, the return type declared in the original overridden method in the superclass. (More on this in a few pages when we discuss covariant returns.)

   The access level can’t be more restrictive than that of the overridden method.

   The access level CAN be less restrictive than that of the overridden method.

   Instance methods can be overridden only if they are inherited by the subtype. A subtype within the same package as the instance’s supertype can override any supertype method that is not marked private or final. A subtype in a different package can override only those nonfinal methods marked public or protected (since protected methods are inherited by the subtype).

   The overriding method CAN throw any unchecked (runtime) exception, regardless of whether the overridden method declares the exception. (More in Chapter 5.)

   The overriding method must NOT throw checked exceptions that are new or broader than those declared by the overridden method. For example, a method that declares a FileNotFoundException cannot be overridden by a method that declares a SQLException, Exception, or any other nonruntime exception unless it’s a subclass of FileNotFoundException.

   The overriding method can throw narrower or fewer exceptions. Just because an overridden method “takes risks” doesn’t mean that the overriding subtype’s exception takes the same risks. Bottom line: an overriding method doesn’t have to declare any exceptions that it will never throw, regardless of what the overridden method declares.

   You cannot override a method marked final.

   You cannot override a method marked static. We’ll look at an example in a few pages when we discuss static methods in more detail.

   If a method can’t be inherited, you cannot override it. Remember that overriding implies that you’re reimplementing a method you inherited! For example, the following code is not legal, and even if you added an eat() method to Horse, it wouldn’t be an override of Animal’s eat() method:

Invoking a Supertype Version of an Overridden Method

Often, you’ll want to take advantage of some of the code in the supertype version of a method, yet still override it to provide some additional specific behavior. It’s like saying, “Run the supertype version of the method, and then come back down here and finish with my subtype additional method code.” (Note that there’s no requirement that the supertype version run before the subtype code.) It’s easy to do in code using the keyword super as follows:

In a similar way, you can access an interface’s overridden method with the syntax:

InterfaceX.super.doStuff();

Note: Using super to invoke an overridden method applies only to instance methods. (Remember that static methods can’t be overridden.) And you can use super only to access a method in a type’s supertype, not the supertype of the supertype—that is, you cannot say super.super.doStuff() and you cannot say: InterfaceX.super.super.doStuff().

Examples of Illegal Method Overrides

Let’s take a look at overriding the eat() method of Animal:

public class Animal {
  public void eat() { }
}

Table 2-2 lists examples of illegal overrides of the Animal eat() method, given the preceding version of the Animal class.

TABLE 2-2   Examples of Illegal Overrides

Overloaded Methods

Overloaded methods let you reuse the same method name in a class, but with different arguments (and, optionally, a different return type). Overloading a method often means you’re being a little nicer to those who call your methods because your code takes on the burden of coping with different argument types rather than forcing the caller to do conversions prior to invoking your method. The rules aren’t too complex:

   Overloaded methods MUST change the argument list.

   Overloaded methods CAN change the return type.

   Overloaded methods CAN change the access modifier.

   Overloaded methods CAN declare new or broader checked exceptions.

   A method can be overloaded in the same type or in a subtype. In other words, if class A defines a doStuff(int i) method, the subclass B could define a doStuff(String s) method without overriding the superclass version that takes an int. So two methods with the same name but in different types can still be considered overloaded if the subtype inherits one version of the method and then declares another overloaded version in its type definition.

Legal Overloads

Let’s look at a method we want to overload:

Invoking Overloaded Methods

In Chapter 6 we will look at how boxing and var-args impact overloading. (You still have to pay attention to what’s covered here, however.)

When a method is invoked, more than one method of the same name might exist for the object type you’re invoking a method on. For example, the Horse class might have three methods with the same name but with different argument lists, which means the method is overloaded.

Deciding which of the matching methods to invoke is based on the arguments. If you invoke the method with a String argument, the overloaded version that takes a String is called. If you invoke a method of the same name but pass it a float, the overloaded version that takes a float will run. If you invoke the method of the same name but pass it a Foo object, and there isn’t an overloaded version that takes a Foo, then the compiler will complain that it can’t find a match. The following are examples of invoking overloaded methods:

In this TestAdder code, the first call to a.addThem(b,c) passes two ints to the method, so the first version of addThem()—the overloaded version that takes two int arguments—is called. The second call to a.addThem(22.5, 9.3) passes two doubles to the method, so the second version of addThem()—the overloaded version that takes two double arguments—is called.

Invoking overloaded methods that take object references rather than primitives is a little more interesting. Say you have an overloaded method such that one version takes an Animal and one takes a Horse (subclass of Animal). If you pass a Horse object in the method invocation, you’ll invoke the overloaded version that takes a Horse. Or so it looks at first glance:

The output is what you expect:

In the Animal version
In the Horse version

But what if you use an Animal reference to a Horse object?

Animal animalRefToHorse = new Horse();
  ua.doStuff(animalRefToHorse);

Which of the overloaded versions is invoked? You might want to answer, “The one that takes a Horse since it’s a Horse object at runtime that’s being passed to the method.” But that’s not how it works. The preceding code would actually print this:

in the Animal version

Even though the actual object at runtime is a Horse and not an Animal, the choice of which overloaded method to call (in other words, the signature of the method) is NOT dynamically decided at runtime.

Just remember that the reference type (not the object type) determines which overloaded method is invoked!

To summarize, which overridden version of the method to call (in other words, from which class in the inheritance tree) is decided at runtime based on object type, but which overloaded version of the method to call is based on the reference type of the argument passed at compile time.

If you invoke a method passing it an Animal reference to a Horse object, the compiler knows only about the Animal, so it chooses the overloaded version of the method that takes an Animal. It does not matter that, at runtime, a Horse is actually being passed.

Polymorphism in Overloaded and Overridden Methods How does polymorphism work with overloaded methods? From what we just looked at, it doesn’t appear that polymorphism matters when a method is overloaded. If you pass an Animal reference, the overloaded method that takes an Animal will be invoked, even if the actual object passed is a Horse. Once the Horse masquerading as Animal gets in to the method, however, the Horse object is still a Horse despite being passed into a method expecting an Animal. So it’s true that polymorphism doesn’t determine which overloaded version is called; however, polymorphism does come into play when the decision is about which overridden version of a method is called. But sometimes a method is both overloaded and overridden. Imagine that the Animal and Horse classes look like this:

Notice that the Horse class has both overloaded and overridden the eat() method. Table 2-3 shows which version of the three eat() methods will run depending on how they are invoked.

TABLE 2-3   Examples of Legal and Illegal Overrides

Table 2-4 summarizes the difference between overloaded and overridden methods.

TABLE 2-4   Differences Between Overloaded and Overridden Methods

We’ll cover constructor overloading later in the chapter, where we’ll also cover the other constructor-related topics that are on the exam. Figure 2-4 illustrates the way overloaded and overridden methods appear in class relationships.

FIGURE 2-4   Overloaded and overridden methods in class relationships

CERTIFICATION OBJECTIVE

Casting (OCA Objectives 2.2 and 7.3)

2.2 Differentiate between object reference variables and primitive variables.

7.3 Determine when casting is necessary.

You’ve seen how it’s both possible and common to use general reference variable types to refer to more specific object types. It’s at the heart of polymorphism. For example, this line of code should be second nature by now:

Animal animal = new Dog();

But what happens when you want to use that animal reference variable to invoke a method that only class Dog has? You know it’s referring to a Dog, and you want to do a Dog-specific thing? In the following code, we’ve got an array of Animals, and whenever we find a Dog in the array, we want to do a special Dog thing. Let’s agree for now that all this code is okay, except that we’re not sure about the line of code that invokes the playDead method.

When we try to compile this code, the compiler says something like this:

cannot find symbol

The compiler is saying, “Hey, class Animal doesn’t have a playDead() method.” Let’s modify the if code block:

The new and improved code block contains a cast, which in this case is sometimes called a downcast, because we’re casting down the inheritance tree to a more specific class. Now the compiler is happy. Before we try to invoke playDead, we cast the animal variable to type Dog. What we’re saying to the compiler is, “We know it’s really referring to a Dog object, so it’s okay to make a new Dog reference variable to refer to that object.” In this case we’re safe, because before we ever try the cast, we do an instanceof test to make sure.

It’s important to know that the compiler is forced to trust us when we do a downcast, even when we screw up:

It can be maddening! This code compiles! But when we try to run it, we’ll get an exception, something like this:

java.lang.ClassCastException

Why can’t we trust the compiler to help us out here? Can’t it see that animal is of type Animal? All the compiler can do is verify that the two types are in the same inheritance tree, so that depending on whatever code might have come before the downcast, it’s possible that animal is of type Dog. The compiler must allow things that might possibly work at runtime. However, if the compiler knows with certainty that the cast could not possibly work, compilation will fail. The following replacement code block will NOT compile:

Animal animal = new Animal();
Dog d = (Dog) animal;
String s = (String) animal; // animal can’t EVER be a String

In this case, you’ll get an error like this:

inconvertible types

Unlike downcasting, upcasting (casting up the inheritance tree to a more general type) works implicitly (that is, you don’t have to type in the cast) because when you upcast you’re implicitly restricting the number of methods you can invoke, as opposed to downcasting, which implies that later on, you might want to invoke a more specific method. Here’s an example:

Both of the previous upcasts will compile and run without exception because a Dog IS-A(n) Animal, which means that anything an Animal can do, a Dog can do. A Dog can do more, of course, but the point is that anyone with an Animal reference can safely call Animal methods on a Dog instance. The Animal methods may have been overridden in the Dog class, but all we care about now is that a Dog can always do at least everything an Animal can do. The compiler and JVM know it, too, so the implicit upcast is always legal for assigning an object of a subtype to a reference of one of its supertype classes (or interfaces). If Dog implements Pet and Pet defines beFriendly(), then a Dog can be implicitly cast to a Pet, but the only Dog method you can invoke then is beFriendly(), which Dog was forced to implement because Dog implements the Pet interface.

One more thing…if Dog implements Pet, then, if Beagle extends Dog but Beagle does not declare that it implements Pet, Beagle is still a Pet! Beagle is a Pet simply because it extends Dog, and Dog’s already taken care of the Pet parts for itself and for all its children. The Beagle class can always override any method it inherits from Dog, including methods that Dog implemented to fulfill its interface contract.

And just one more thing…if Beagle does declare that it implements Pet, just so that others looking at the Beagle class API can easily see that Beagle IS-A Pet without having to look at Beagle’s superclasses, Beagle still doesn’t need to implement the beFriendly() method if the Dog class (Beagle’s superclass) has already taken care of that. In other words, if Beagle IS-A Dog, and Dog IS-A Pet, then Beagle IS-A Pet and has already met its Pet obligations for implementing the beFriendly() method since it inherits the beFriendly() method. The compiler is smart enough to say, “I know Beagle already IS a Dog, but it’s okay to make it more obvious by adding a cast.”

So don’t be fooled by code that shows a concrete class that declares it implements an interface but doesn’t implement the methods of the interface. Before you can tell whether the code is legal, you must know what the supertypes of this implementing class have declared. If any supertype in its inheritance tree has already provided concrete (that is, nonabstract) method implementations, then regardless of whether the supertype declares that it implements the interface, the subclass is under no obligation to reimplement (override) those methods.

CERTIFICATION OBJECTIVE

Implementing an Interface (OCA Objective 7.5)

7.5 Use abstract classes and interfaces.

When you implement an interface, you’re agreeing to adhere to the contract defined in the interface. That means you’re agreeing to provide legal implementations for every abstract method defined in the interface, and that anyone who knows what the interface methods look like (not how they’re implemented, but how they can be called and what they return) can rest assured that they can invoke those methods on an instance of your implementing class.

For example, if you create a class that implements the Runnable interface (so that your code can be executed by a specific thread), you must provide the public void run() method. Otherwise, the poor thread could be told to go execute your Runnable object’s code and—surprise, surprise—the thread then discovers the object has no run() method! (At which point, the thread would blow up and the JVM would crash in a spectacular yet horrible explosion.) Thankfully, Java prevents this meltdown from occurring by running a compiler check on any class that claims to implement an interface. If the class says it’s implementing an interface, it darn well better have an implementation for each abstract method in the interface (with a few exceptions that we’ll look at in a moment).

Assuming an interface Bounceable with two methods, bounce() and setBounceFactor(), the following class will compile:

Okay, we know what you’re thinking: “This has got to be the worst implementation class in the history of implementation classes.” It compiles, though. And it runs. The interface contract guarantees that a class will have the method (in other words, others can call the method subject to access control), but it never guaranteed a good implementation—or even any actual implementation code in the body of the method. (Keep in mind, though, that if the interface declares that a method is NOT void, your class’s implementation code has to include a return statement.) The compiler will never say, “Um, excuse me, but did you really mean to put nothing between those curly braces? HELLO. This is a method after all, so shouldn’t it do something?”

Implementation classes must adhere to the same rules for method implementation as a class extending an abstract class. To be a legal implementation class, a nonabstract implementation class must do the following:

   Provide concrete (nonabstract) implementations for all abstract methods from the declared interface.

   Follow all the rules for legal overrides, such as the following:

   Declare no checked exceptions on implementation methods other than those declared by the interface method or subclasses of those declared by the interface method.

   Maintain the signature of the interface method, and maintain the same return type (or a subtype). (But it does not have to declare the exceptions declared in the interface method declaration.)

But wait, there’s more! An implementation class can itself be abstract! For example, the following is legal for a class Ball implementing Bounceable:

abstract class Ball implements Bounceable { }

Notice anything missing? We never provided the implementation methods. And that’s okay. If the implementation class is abstract, it can simply pass the buck to its first concrete subclass. For example, if class BeachBall extends Ball, and BeachBall is not abstract, then BeachBall has to provide an implementation for all the abstract methods from Bounceable:

Look for classes that claim to implement an interface but don’t provide the correct method implementations. Unless the implementing class is abstract, the implementing class must provide implementations for all abstract methods defined in the interface.

You need to know two more rules, and then we can put this topic to sleep (or put you to sleep; we always get those two confused):

1.   A class can implement more than one interface. It’s perfectly legal to say, for example, the following:

public class Ball implements Bounceable, Serializable, Runnable { ... }

You can extend only one class, but you can implement many interfaces (which, as of Java 8, means a form of multiple inheritance, which we’ll discuss shortly). In other words, subclassing defines who and what you are, whereas implementing defines a role you can play or a hat you can wear, despite how different you might be from some other class implementing the same interface (but from a different inheritance tree). For example, a Person extends HumanBeing (although for some, that’s debatable). But a Person may also implement Programmer, Snowboarder, Employee, Parent, or PersonCrazyEnoughToTakeThisExam.

2.   An interface can itself extend another interface. The following code is perfectly legal:

public interface Bounceable extends Moveable { } // ok!

What does that mean? The first concrete (nonabstract) implementation class of Bounceable must implement all the abstract methods of Bounceable, plus all the abstract methods of Moveable! The subinterface, as we call it, simply adds more requirements to the contract of the superinterface. You’ll see this concept applied in many areas of Java, especially Java EE, where you’ll often have to build your own interface that extends one of the Java EE interfaces.

Hold on, though, because here’s where it gets strange. An interface can extend more than one interface! Think about that for a moment. You know that when we’re talking about classes, the following is illegal:

public class Programmer extends Employee, Geek { } // Illegal!

As we mentioned earlier, a class is not allowed to extend multiple classes in Java. An interface, however, is free to extend multiple interfaces:

In the next example, Ball is required to implement Bounceable, plus all abstract methods from the interfaces that Bounceable extends (including any interfaces those interfaces extend, and so on, until you reach the top of the stack—or is it the bottom of the stack?). So Ball would need to look like the following:

If class Ball fails to implement any of the abstract methods from Bounceable, Moveable, or Spherical, the compiler will jump up and down wildly, red in the face, until it does. Unless, that is, class Ball is marked abstract. In that case, Ball could choose to implement any, all, or none of the abstract methods from any of the interfaces, thus leaving the rest of the implementations to a concrete subclass of Ball, as follows:

Figure 2-5 compares concrete and abstract examples of extends and implements for both classes and interfaces.

FIGURE 2-5   Comparing concrete and abstract examples of extends and implements

Java 8—Now with Multiple Inheritance!

It might have already occurred to you that since interfaces can now have concrete methods and classes can implement multiple interfaces, the spectre of multiple inheritance and the Deadly Diamond of Death can rear its ugly head! Well, you’re partly correct. A class CAN implement interfaces with duplicate, concrete method signatures! But the good news is that the compiler’s got your back, and if you DO want to implement both interfaces, you’ll have to provide an overriding method in your class. Let’s look at the following code:

As the code stands, it WILL NOT COMPILE because it’s not clear which version of doStuff() should be used. In order to make the code compile, you need to override doStuff() in the class. Uncommenting the class’s doStuff() method would allow the code to compile and when run produce the following output:

3

CERTIFICATION OBJECTIVE

Legal Return Types (OCA Objectives 2.2 and 6.1)

2.2 Differentiate between object reference variables and primitive variables.

6.1 Create methods with arguments and return values; including overloaded methods.

This section covers two aspects of return types: what you can declare as a return type, and what you can actually return as a value. What you can and cannot declare is pretty straightforward, but it all depends on whether you’re overriding an inherited method or simply declaring a new method (which includes overloaded methods). We’ll take just a quick look at the difference between return type rules for overloaded and overriding methods, because we’ve already covered that in this chapter. We’ll cover a small bit of new ground, though, when we look at polymorphic return types and the rules for what is and is not legal to actually return.

Return Type Declarations

This section looks at what you’re allowed to declare as a return type, which depends primarily on whether you are overriding, overloading, or declaring a new method.

Return Types on Overloaded Methods

Remember that method overloading is not much more than name reuse. The overloaded method is a completely different method from any other method of the same name. So if you inherit a method but overload it in a subtype, you’re not subject to the restrictions of overriding, which means you can declare any return type you like. What you can’t do is change only the return type. To overload a method, remember, you must change the argument list. The following code shows an overloaded method:

Notice that the Bar version of the method uses a different return type. That’s perfectly fine. As long as you’ve changed the argument list, you’re overloading the method, so the return type doesn’t have to match that of the supertype version. What you’re NOT allowed to do is this:

Overriding and Return Types and Covariant Returns

When a subtype wants to change the method implementation of an inherited method (an override), the subtype must define a method that matches the inherited version exactly. Or, since Java 5, you’re allowed to change the return type in the overriding method as long as the new return type is a subtype of the declared return type of the overridden (superclass) method.

Let’s look at a covariant return in action:

Since Java 5, this code compiles. If you were to attempt to compile this code with a 1.4 compiler or with the source flag as follows,

javac -source 1.4 Beta.java

you would get a compiler error like this:

attempting to use incompatible return type

Other rules apply to overriding, including those for access modifiers and declared exceptions, but those rules aren’t relevant to the return type discussion.

Returning a Value

You have to remember only six rules for returning a value:

1.   You can return null in a method with an object reference return type.

2.   An array is a perfectly legal return type.

3.   In a method with a primitive return type, you can return any value or variable that can be implicitly converted to the declared return type.

4.   In a method with a primitive return type, you can return any value or variable that can be explicitly cast to the declared return type.

5.   You must not return anything from a method with a void return type.

6.   In a method with an object reference return type, you can return any object type that can be implicitly cast to the declared return type.

CERTIFICATION OBJECTIVE

Constructors and Instantiation
(OCA Objectives 6.3 and 7.4)

6.3 Create and overload constructors; including impact on default constructors (sic)

7.4 Use super and this to access objects and constructors.

Objects are constructed. You CANNOT make a new object without invoking a constructor. In fact, you can’t make a new object without invoking not just the constructor of the object’s actual class type, but also the constructor of each of its superclasses! Constructors are the code that runs whenever you use the keyword new. (Okay, to be a bit more accurate, there can also be initialization blocks that run when you say new, and we’re going to cover init blocks and their static initialization counterparts after we discuss constructors.) We’ve got plenty to talk about here—we’ll look at how constructors are coded, who codes them, and how they work at runtime. So grab your hardhat and a hammer, and let’s do some object building.

Constructor Basics

Every class, including abstract classes, MUST have a constructor. Burn that into your brain. But just because a class must have a constructor doesn’t mean the programmer has to type it. A constructor looks like this:

class Foo {
  Foo() { } // The constructor for the Foo class
}

Notice what’s missing? There’s no return type! Two key points to remember about constructors are that they have no return type, and their names must exactly match the class name. Typically, constructors are used to initialize the instance variable state, as follows:

In the preceding code example, the Foo class does not have a no-arg constructor. That means the following will fail to compile:

Foo f = new Foo(); // Won’t compile, no matching constructor

but the following will compile:

So it’s very common (and desirable) for a class to have a no-arg constructor, regardless of how many other overloaded constructors are in the class (yes, constructors can be overloaded). You can’t always make that work for your classes; occasionally you have a class where it makes no sense to create an instance without supplying information to the constructor. A java.awt.Color object, for example, can’t be created by calling a no-arg constructor, because that would be like saying to the JVM, “Make me a new Color object, and I really don’t care what color it is…you pick.” Do you seriously want the JVM making your style decisions?

Constructor Chaining

We know that constructors are invoked at runtime when you say new on some class type as follows:

Horse h = new Horse();

But what really happens when you say new Horse()? (Assume Horse extends Animal and Animal extends Object.)

1.   The Horse constructor is invoked. Every constructor invokes the constructor of its superclass with an (implicit) call to super(), unless the constructor invokes an overloaded constructor of the same class (more on that in a minute).

2.   The Animal constructor is invoked (Animal is the superclass of Horse).

3.   The Object constructor is invoked (Object is the ultimate superclass of all classes, so class Animal extends Object even though you don’t actually type “extends Object” into the Animal class declaration; it’s implicit.) At this point we’re on the top of the stack.

4.   If class Object had any instance variables, then they would be given their explicit values. By explicit values, we mean values that are assigned at the time the variables are declared, such as int x = 27, where 27 is the explicit value (as opposed to the default value) of the instance variable.

5.   The Object constructor completes.

6.   The Animal instance variables are given their explicit values (if any).

7.   The Animal constructor completes.

8.   The Horse instance variables are given their explicit values (if any).

9.   The Horse constructor completes.

Figure 2-6 shows how constructors work on the call stack.

FIGURE 2-6   Constructors on the call stack

Rules for Constructors

The following list summarizes the rules you’ll need to know for the exam (and to understand the rest of this section). You MUST remember these, so be sure to study them more than once.

   Constructors can use any access modifier, including private. (A private constructor means only code within the class itself can instantiate an object of that type, so if the private constructor class wants to allow an instance of the class to be used, the class must provide a static method or variable that allows access to an instance created from within the class.)

   The constructor name must match the name of the class.

   Constructors must not have a return type.

   It’s legal (but stupid) to have a method with the same name as the class, but that doesn’t make it a constructor. If you see a return type, it’s a method rather than a constructor. In fact, you could have both a method and a constructor with the same name—the name of the class—in the same class, and that’s not a problem for Java. Be careful not to mistake a method for a constructor—be sure to look for a return type.

   If you don’t type a constructor into your class code, a default constructor will be automatically generated by the compiler.

   The default constructor is ALWAYS a no-arg constructor.

   If you want a no-arg constructor and you’ve typed any other constructor(s) into your class code, the compiler won’t provide the no-arg constructor (or any other constructor) for you. In other words, if you’ve typed in a constructor with arguments, you won’t have a no-arg constructor unless you typed it in yourself!

   Every constructor has, as its first statement, either a call to an overloaded constructor (this()) or a call to the superclass constructor (super()), although remember that this call can be inserted by the compiler.

   If you do type in a constructor (as opposed to relying on the compiler-generated default constructor), and you do not type in the call to super() or a call to this(), the compiler will insert a no-arg call to super() for you as the very first statement in the constructor.

   A call to super() can either be a no-arg call or can include arguments passed to the super constructor.

   A no-arg constructor is not necessarily the default (that is, compiler-supplied) constructor, although the default constructor is always a no-arg constructor. The default constructor is the one the compiler provides! Although the default constructor is always a no-arg constructor, you’re free to put in your own no-arg constructor.

   You cannot make a call to an instance method or access an instance variable until after the super constructor runs.

   Only static variables and methods can be accessed as part of the call to super() or this(). (Example: super(Animal.NAME) is OK, because NAME is declared as a static variable.)

   Abstract classes have constructors, and those constructors are always called when a concrete subclass is instantiated.

   Interfaces do not have constructors. Interfaces are not part of an object’s inheritance tree.

   The only way a constructor can be invoked is from within another constructor. In other words, you can’t write code that actually calls a constructor as follows:

Determine Whether a Default Constructor Will Be Created

The following example shows a Horse class with two constructors:

Will the compiler put in a default constructor for this class? No!

How about for the following variation of the class?

class Horse {
   Horse(String name) { }
}

Now will the compiler insert a default constructor? No!

What about this class?

class Horse { }

Now we’re talking. The compiler will generate a default constructor for this class because the class doesn’t have any constructors defined.

Okay, what about this class?

class Horse {
   void Horse() { }
}

It might look like the compiler won’t create a constructor, since one is already in the Horse class. Or is it? Take another look at the preceding Horse class.

What’s wrong with the Horse() constructor? It isn’t a constructor at all! It’s simply a method that happens to have the same name as the class. Remember, the return type is a dead giveaway that we’re looking at a method, not a constructor.

How do you know for sure whether a default constructor will be created? Because you didn’t write any constructors in your class.

How do you know what the default constructor will look like? Because...

   The default constructor has the same access modifier as the class.

   The default constructor has no arguments.

   The default constructor includes a no-arg call to the super constructor (super()).

Table 2-5 shows what the compiler will (or won’t) generate for your class.

TABLE 2-5   Compiler-Generated Constructor Code

What happens if the super constructor has arguments? Constructors can have arguments just as methods can, and if you try to invoke a method that takes, say, an int, but you don’t pass anything to the method, the compiler will complain as follows:

The compiler will complain that you can’t invoke takeInt() without passing an int. Of course, the compiler enjoys the occasional riddle, so the message it spits out on some versions of the JVM (your mileage may vary) is less than obvious:

UseBar.java:7: takeInt(int) in Bar cannot be applied to ()
   b.takeInt();
   ^

But you get the idea. The bottom line is that there must be a match for the method. And by match, we mean the argument types must be able to accept the values or variables you’re passing and in the order you’re passing them. Which brings us back to constructors (and here you were thinking we’d never get there), which work exactly the same way.

So if your super constructor (that is, the constructor of your immediate superclass/parent) has arguments, you must type in the call to super(), supplying the appropriate arguments. Crucial point: if your superclass does not have a no-arg constructor, you must type a constructor in your class (the subclass) because you need a place to put in the call to super() with the appropriate arguments.

The following is an example of the problem:

And once again the compiler treats us with stunning lucidity:

If you’re lucky (and it’s a full moon), your compiler might be a little more explicit. But again, the problem is that there just isn’t a match for what we’re trying to invoke with super()—an Animal constructor with no arguments.

Another way to put this is that if your superclass does not have a no-arg constructor, then in your subclass you will not be able to use the default constructor supplied by the compiler. It’s that simple. Because the compiler can only put in a call to a no-arg super(), you won’t even be able to compile something like this:

Trying to compile this code gives us exactly the same error we got when we put a constructor in the subclass with a call to the no-arg version of super():

In fact, the preceding Clothing and TShirt code is implicitly the same as the following code, where we’ve supplied a constructor for TShirt that’s identical to the default constructor supplied by the compiler:

One last point on the whole default constructor thing (and it’s probably very obvious, but we have to say it or we’ll feel guilty for years), constructors are never inherited. They aren’t methods. They can’t be overridden (because they aren’t methods, and only instance methods can be overridden). So the type of constructor(s) your superclass has in no way determines the type of default constructor you’ll get. Some folks mistakenly believe that the default constructor somehow matches the super constructor, either by the arguments the default constructor will have (remember, the default constructor is always a no-arg) or by the arguments used in the compiler-supplied call to super().

So although constructors can’t be overridden, you’ve already seen that they can be overloaded, and typically are.

Overloaded Constructors

Overloading a constructor means typing in multiple versions of the constructor, each having a different argument list, like the following examples:

The preceding Foo class has two overloaded constructors: one that takes a string, and one with no arguments. Because there’s no code in the no-arg version, it’s actually identical to the default constructor the compiler supplies—but remember, since there’s already a constructor in this class (the one that takes a string), the compiler won’t supply a default constructor. If you want a no-arg constructor to overload the with-args version you already have, you’re going to have to type it yourself, just as in the Foo example.

Overloading a constructor is typically used to provide alternate ways for clients to instantiate objects of your class. For example, if a client knows the animal name, they can pass that to an Animal constructor that takes a string. But if they don’t know the name, the client can call the no-arg constructor, and that constructor can supply a default name. Here’s what it looks like:

Running the code four times produces this output:

There’s a lot going on in the preceding code. Figure 2-7 shows the call stack for constructor invocations when a constructor is overloaded.

FIGURE 2-7   Overloaded constructors on the call stack

Take a look at the call stack, and then let’s walk through the code straight from the top.

   Line 2 Declare a String instance variable name.

   Lines 3–5 Constructor that takes a String and assigns it to instance variable name.

   Line 7 Here’s where it gets fun. Assume every animal needs a name, but the client (calling code) might not always know what the name should be, so the Animal class will assign a random name. The no-arg constructor generates a name by invoking the makeRandomName() method.

   Line 8 The no-arg constructor invokes its own overloaded constructor that takes a String, in effect calling it the same way it would be called if client code were doing a new to instantiate an object, passing it a String for the name. The overloaded invocation uses the keyword this, but uses it as though it were a method named this(). So line 8 is simply calling the constructor on line 3, passing it a randomly selected String rather than a client-code chosen name.

   Line 11 Notice that the makeRandomName() method is marked static! That’s because you cannot invoke an instance (in other words, nonstatic) method (or access an instance variable) until after the super constructor has run. And since the super constructor will be invoked from the constructor on line 3, rather than from the one on line 7, line 8 can use only a static method to generate the name. If we wanted all animals not specifically named by the caller to have the same default name, say, “Fred,” then line 8 could have read this(“Fred”); rather than calling a method that returns a string with the randomly chosen name.

   Line 12 This doesn’t have anything to do with constructors, but since we’re all here to learn, it generates a random integer between 0 and 4.

   Line 13 Weird syntax, we know. We’re creating a new String object (just a single String instance), but we want the string to be selected randomly from a list. Except we don’t have the list, so we need to make it. So in that one line of code we

1. Declare a String variable name.

2. Create a String array (anonymously—we don’t assign the array itself to a variable).

3. Retrieve the string at index [x] (x being the random number generated on line 12) of the newly created String array.

4. Assign the string retrieved from the array to the declared instance variable name. We could have made it much easier to read if we’d just written

But where’s the fun in that? Throwing in unusual syntax (especially for code wholly unrelated to the real question) is in the spirit of the exam. Don’t be startled! (Okay, be startled, but then just say to yourself, “Whoa!” and get on with it.)

   Line 18 We’re invoking the no-arg version of the constructor (causing a random name from the list to be passed to the other constructor).

   Line 20 We’re invoking the overloaded constructor that takes a string representing the name.

The key point to get from this code example is in line 8. Rather than calling super(), we’re calling this(), and this() always means a call to another constructor in the same class. Okay, fine, but what happens after the call to this()? Sooner or later the super() constructor gets called, right? Yes, indeed. A call to this() just means you’re delaying the inevitable. Some constructor, somewhere, must make the call to super().

Key Rule: The first line in a constructor must be a call to super()or a call to this().

No exceptions. If you have neither of those calls in your constructor, the compiler will insert the no-arg call to super(). In other words, if constructor A() has a call to this(), the compiler knows that constructor A() will not be the one to invoke super().

The preceding rule means a constructor can never have both a call to super() and a call to this(). Because each of those calls must be the first statement in a constructor, you can’t legally use both in the same constructor. That also means the compiler will not put a call to super() in any constructor that has a call to this().

Thought question: What do you think will happen if you try to compile the following code?

Your compiler may not actually catch the problem (it varies depending on your compiler, but most won’t catch the problem). It assumes you know what you’re doing. Can you spot the flaw? Given that a super constructor must always be called, where would the call to super() go? Remember, the compiler won’t put in a default constructor if you’ve already got one or more constructors in your class. And when the compiler doesn’t put in a default constructor, it still inserts a call to super() in any constructor that doesn’t explicitly have a call to the super constructor—unless, that is, the constructor already has a call to this(). So in the preceding code, where can super() go? The only two constructors in the class both have calls to this(), and, in fact, you’ll get exactly what you’d get if you typed the following method code:

public void go() {
   doStuff();
}

public void doStuff() {
   go();
}

Now can you see the problem? Of course you can. The stack explodes! It gets higher and higher and higher until it just bursts open and method code goes spilling out, oozing out of the JVM right onto the floor. Two overloaded constructors both calling this() are two constructors calling each other—over and over and over, resulting in this:

% java A
Exception in thread “main” java.lang.StackOverflowError

The benefit of having overloaded constructors is that you offer flexible ways to instantiate objects from your class. The benefit of having one constructor invoke another overloaded constructor is to avoid code duplication. In the Animal example, there wasn’t any code other than setting the name, but imagine if after line 4 there was still more work to be done in the constructor. By putting all the other constructor work in just one constructor, and then having the other constructors invoke it, you don’t have to write and maintain multiple versions of that other important constructor code. Basically, each of the other not-the-real-one overloaded constructors will call another overloaded constructor, passing it whatever data it needs (data the client code didn’t supply).

Constructors and instantiation become even more exciting (just when you thought it was safe) when you get to inner classes, but we know you can stand to have only so much fun in one chapter, and besides, you don’t have to deal with inner classes until you tackle the OCP exam.

CERTIFICATION OBJECTIVE

Initialization Blocks (OCA Objectives 1.2 and 6.3-ish)

1.2 Define the structure of a Java class

6.3 Create and overload constructors; including impact on default constructors

We’ve talked about two places in a class where you can put code that performs operations: methods and constructors. Initialization blocks are the third place in a Java program where operations can be performed. Static initialization blocks run when the class is first loaded, and instance initialization blocks run whenever an instance is created (a bit similar to a constructor). Let’s look at an example:

As you can see, the syntax for initialization blocks is pretty terse. They don’t have names, they can’t take arguments, and they don’t return anything. A static initialization block runs once when the class is first loaded. An instance initialization block runs once every time a new instance is created. Remember when we talked about the order in which constructor code executed? Instance init block code runs right after the call to super() in a constructor—in other words, after all super constructors have run.

You can have many initialization blocks in a class. It is important to note that unlike methods or constructors, the order in which initialization blocks appear in a class matters. When it’s time for initialization blocks to run, if a class has more than one, they will run in the order in which they appear in the class file—in other words, from the top down. Based on the rules we just discussed, can you determine the output of the following program?

To figure this out, remember these rules:

   init blocks execute in the order in which they appear.

   Static init blocks run once, when the class is first loaded.

   Instance init blocks run every time a class instance is created.

   Instance init blocks run after the constructor’s call to super().

With those rules in mind, the following output should make sense:

As you can see, the instance init blocks each ran twice. Instance init blocks are often used as a place to put code that all the constructors in a class should share. That way, the code doesn’t have to be duplicated across constructors.

Finally, if you make a mistake in your static init block, the JVM can throw an ExceptionInInitializerError. Let’s look at an example:

It produces something like this:

CERTIFICATION OBJECTIVE

Statics (OCA Objective 6.2)

6.2 Apply the static keyword to methods and fields.

Static Variables and Methods

The static modifier has such a profound impact on the behavior of a method or variable that we’re treating it as a concept entirely separate from the other modifiers. To understand the way a static member works, we’ll look first at a reason for using one. Imagine you’ve got a utility class or interface with a method that always runs the same way; its sole function is to return, say, a random number. It wouldn’t matter which instance of the class performed the method—it would always behave exactly the same way. In other words, the method’s behavior has no dependency on the state (instance variable values) of an object. So why, then, do you need an object when the method will never be instance-specific? Why not just ask the type itself to run the method?

Let’s imagine another scenario: Suppose you want to keep a running count of all instances instantiated from a particular class. Where do you actually keep that variable? It won’t work to keep it as an instance variable within the class whose instances you’re tracking, because the count will just be initialized back to a default value with each new instance. The answer to both the utility-method-always-runs-the-same scenario and the keep-a-running-total-of-instances scenario is to use the static modifier. Variables and methods marked static belong to the type, rather than to any particular instance. In fact, for classes, you can use a static method or variable without having any instances of that class at all. You need only have the type available to be able to invoke a static method or access a static variable. static variables, too, can be accessed without having an instance of a class. But if there are instances, a static variable of a class will be shared by all instances of that class; there is only one copy.

The following code declares and uses a static counter variable:

In the preceding code, the static frogCount variable is set to zero when the Frog class is first loaded by the JVM, before any Frog instances are created! (By the way, you don’t actually need to initialize a static variable to zero; static variables get the same default values instance variables get.) Whenever a Frog instance is created, the Frog constructor runs and increments the static frogCount variable. When this code executes, three Frog instances are created in main(), and the result is

Frog count is now 3

Now imagine what would happen if frogCount were an instance variable (in other words, nonstatic):

When this code executes, it should still create three Frog instances in main(), but the result is…a compiler error! We can’t get this code to compile, let alone run.

The JVM doesn’t know which Frog object’s frogCount you’re trying to access. The problem is that main() is itself a static method and thus isn’t running against any particular instance of the class; instead it’s running on the class itself. A static method can’t access a nonstatic (instance) variable because there is no instance! That’s not to say there aren’t instances of the class alive on the heap, but rather that even if there are, the static method doesn’t know anything about them. The same applies to instance methods; a static method can’t directly invoke a nonstatic method. Think static = class, nonstatic = instance. Making the method called by the JVM (main()) a static method means the JVM doesn’t have to create an instance of your class just to start running code.

Accessing Static Methods and Variables

Since you don’t need to have an instance in order to invoke a static method or access a static variable, how do you invoke or use a static member? What’s the syntax? We know that with a regular old instance method, you use the dot operator on a reference to an instance:

In the preceding code, we instantiate a Frog, assign it to the reference variable f, and then use that f reference to invoke a method on the Frog instance we just created. In other words, the getFrogSize() method is being invoked on a specific Frog object on the heap.

But this approach (using a reference to an object) isn’t appropriate for accessing a static method, because there might not be any instances of the class at all! So the way we access a static method (or static variable) is to use the dot operator on the type name, as opposed to using it on a reference to an instance, as follows:

which produces the output:

from static 3
from instance 4
use ref var 5

But just to make it really confusing, the Java language also allows you to use an object reference variable to access a static member. Did you catch the last line of main()? It included this invocation:

f.getCount(); // Access a static using an instance variable

In the preceding code, we instantiate a Frog, assign the new Frog object to the reference variable f, and then use the f reference to invoke a static method! But even though we are using a specific Frog instance to access the static method, the rules haven’t changed. This is merely a syntax trick to let you use an object reference variable (but not the object it refers to) to get to a static method or variable, but the static member is still unaware of the particular instance used to invoke the static member. In the Frog example, the compiler knows that the reference variable f is of type Frog, and so the Frog class static method is run with no awareness or concern for the Frog instance at the other end of the f reference. In other words, the compiler cares only that reference variable f is declared as type Frog.

Invoking static methods from interfaces is almost the same as invoking static methods from classes, except the “instance variable syntax trick” just discussed works only for static methods in classes. The following code demonstrates how interface static methods can and cannot be invoked:

Let’s review the code:

   Line 1 is a legal invocation of an interface’s default method.

   Line 2 is an illegal attempt to invoke an interface’s static method.

   Line 3 is THE legal way to invoke an interface’s static method.

   Line 4 is another illegal attempt to invoke an interface’s static method.

Figure 2-8 illustrates the effects of the static modifier on methods and variables.

FIGURE 2-8   The effects of static on methods and variables

Finally, remember that static methods can’t be overridden! This doesn’t mean they can’t be redefined in a subclass, but redefining and overriding aren’t the same thing. Let’s look at an example of a redefined (remember, not overridden) static method:

Running this code produces this output:

a a a d

Remember, the syntax a [x].doStuff() is just a shortcut (the syntax trick)—the compiler is going to substitute something like Animal.doStuff() instead. Notice also that you can invoke a static method by using the class name.

Notice that we didn’t use the enhanced for loop here (covered in Chapter 5), even though we could have. Expect to see a mix of both Java 1.4 and Java 5–8 coding styles and practices on the exam.

CERTIFICATION SUMMARY

We started the chapter by discussing the importance of encapsulation in good OO design, and then we talked about how good encapsulation is implemented: with private instance variables and public getters and setters.

Next, we covered the importance of inheritance, so that you can grasp overriding, overloading, polymorphism, reference casting, return types, and constructors.

We covered IS-A and HAS-A. IS-A is implemented using inheritance, and HAS-A is implemented by using instance variables that refer to other objects.

Polymorphism was next. Although a reference variable’s type can’t be changed, it can be used to refer to an object whose type is a subtype of its own. We learned how to determine what methods are invocable for a given reference variable.

We looked at the difference between overridden and overloaded methods, learning that an overridden method occurs when a subtype inherits a method from a supertype and then reimplements the method to add more specialized behavior. We learned that, at runtime, the JVM will invoke the subtype version on an instance of a subtype and the supertype version on an instance of the supertype. Abstract methods must be “overridden” (technically, abstract methods must be implemented, as opposed to overridden, since there really isn’t anything to override).

We saw that overriding methods must declare the same argument list and return type or they can return a subtype of the declared return type of the supertype’s overridden method), and that the access modifier can’t be more restrictive. The overriding method also can’t throw any new or broader checked exceptions that weren’t declared in the overridden method. You also learned that the overridden method can be invoked using the syntax super.doSomething();.

Overloaded methods let you reuse the same method name in a class, but with different arguments (and, optionally, a different return type). Whereas overriding methods must not change the argument list, overloaded methods must. But unlike overriding methods, overloaded methods are free to vary the return type, access modifier, and declared exceptions any way they like.

We learned the mechanics of casting (mostly downcasting) reference variables and when it’s necessary to do so.

Implementing interfaces came next. An interface describes a contract that the implementing class must follow. The rules for implementing an interface are similar to those for extending an abstract class. As of Java 8, interfaces can have concrete methods, which are labeled default. Also, remember that a class can implement more than one interface and that interfaces can extend another interface.

We also looked at method return types and saw that you can declare any return type you like (assuming you have access to a class for an object reference return type), unless you’re overriding a method. Barring a covariant return, an overriding method must have the same return type as the overridden method of the superclass. We saw that, although overriding methods must not change the return type, overloaded methods can (as long as they also change the argument list).

Finally, you learned that it is legal to return any value or variable that can be implicitly converted to the declared return type. So, for example, a short can be returned when the return type is declared as an int. And (assuming Horse extends Animal), a Horse reference can be returned when the return type is declared an Animal.

We covered constructors in detail, learning that if you don’t provide a constructor for your class, the compiler will insert one. The compiler-generated constructor is called the default constructor, and it is always a no-arg constructor with a no-arg call to super(). The default constructor will never be generated if even a single constructor exists in your class (regardless of the arguments of that constructor); so if you need more than one constructor in your class and you want a no-arg constructor, you’ll have to write it yourself. We also saw that constructors are not inherited and that you can be confused by a method that has the same name as the class (which is legal). The return type is the giveaway that a method is not a constructor because constructors do not have return types.

We saw how all the constructors in an object’s inheritance tree will always be invoked when the object is instantiated using new. We also saw that constructors can be overloaded, which means defining constructors with different argument lists. A constructor can invoke another constructor of the same class using the keyword this(), as though the constructor were a method named this(). We saw that every constructor must have either this() or super() as the first statement (although the compiler can insert it for you).

After constructors, we discussed the two kinds of initialization blocks and how and when their code runs.

We looked at static methods and variables. static members are tied to the class or interface, not an instance, so there is only one copy of any static member. A common mistake is to attempt to reference an instance variable from a static method. Use the respective class or interface name with the dot operator to access static members.

And, once again, you learned that the exam includes tricky questions designed largely to test your ability to recognize just how tricky the questions can be.

TWO-MINUTE DRILL

Here are some of the key points from each certification objective in this chapter.

Encapsulation, IS-A, HAS-A* (OCA Objective 6.5)

   Encapsulation helps hide implementation behind an interface (or API).

   Encapsulated code has two features:

   Instance variables are kept protected (usually with the private modifier).

   Getter and setter methods provide access to instance variables.

   IS-A refers to inheritance or implementation.

   IS-A is expressed with the keyword extends or implements.

   IS-A, “inherits from,” and “is a subtype of” are all equivalent expressions.

   HAS-A means an instance of one class “has a” reference to an instance of another class or another instance of the same class. *HAS-A is NOT on the exam, but it’s good to know.

Inheritance (OCA Objective 7.1)

   Inheritance allows a type to be a subtype of a supertype and thereby inherit public and protected variables and methods of the supertype.

   Inheritance is a key concept that underlies IS-A, polymorphism, overriding, overloading, and casting.

   All classes (except class Object) are subclasses of type Object, and therefore they inherit Object’s methods.

Polymorphism (OCA Objective 7.2)

   Polymorphism means “many forms.”

   A reference variable is always of a single, unchangeable type, but it can refer to a subtype object.

   A single object can be referred to by reference variables of many different types—as long as they are the same type or a supertype of the object.

   The reference variable’s type (not the object’s type) determines which methods can be called!

   Polymorphic method invocations apply only to overridden instance methods.

Overriding and Overloading (OCA Objectives 6.1 and 7.2)

   Methods can be overridden or overloaded; constructors can be overloaded but not overridden.

   With respect to the method it overrides, the overriding method

   Must have the same argument list

   Must have the same return type or a subclass (known as a covariant return)

   Must not have a more restrictive access modifier

   May have a less restrictive access modifier

   Must not throw new or broader checked exceptions

   May throw fewer or narrower checked exceptions, or any unchecked exception

   final methods cannot be overridden.

   Only inherited methods may be overridden, and remember that private methods are not inherited.

   A subclass uses super.overriddenMethodName() to call the superclass version of an overridden method.

   A subclass uses MyInterface.super.overriddenMethodName() to call the super interface version on an overridden method.

   Overloading means reusing a method name but with different arguments.

   Overloaded methods

   Must have different argument lists

   May have different return types, if argument lists are also different

   May have different access modifiers

   May throw different exceptions

   Methods from a supertype can be overloaded in a subtype.

   Polymorphism applies to overriding, not to overloading.

   Object type (not the reference variable’s type) determines which overridden method is used at runtime.

   Reference type determines which overloaded method will be used at compile time.

Reference Variable Casting (OCA Objective 7.3)

   There are two types of reference variable casting: downcasting and upcasting.

   Downcasting If you have a reference variable that refers to a subtype object, you can assign it to a reference variable of the subtype. You must make an explicit cast to do this, and the result is that you can access the subtype’s members with this new reference variable.

   Upcasting You can assign a reference variable to a supertype reference variable explicitly or implicitly. This is an inherently safe operation because the assignment restricts the access capabilities of the new variable.

Implementing an Interface (OCA Objective 7.5)

   When you implement an interface, you are fulfilling its contract.

   You implement an interface by properly and concretely implementing all the abstract methods defined by the interface.

   A single class can implement many interfaces.

Return Types (OCA Objectives 7.2 and 7.5)

   Overloaded methods can change return types; overridden methods cannot, except in the case of covariant returns.

   Object reference return types can accept null as a return value.

   An array is a legal return type, both to declare and return as a value.

   For methods with primitive return types, any value that can be implicitly converted to the return type can be returned.

   Nothing can be returned from a void, but you can return nothing. You’re allowed to simply say return in any method with a void return type to bust out of a method early. But you can’t return nothing from a method with a non-void return type.

   Methods with an object reference return type can return a subtype.

   Methods with an interface return type can return any implementer.

Constructors and Instantiation (OCA Objectives 6.3 and 7.4)

   A constructor is always invoked when a new object is created.

   Each superclass in an object’s inheritance tree will have a constructor called.

   Every class, even an abstract class, has at least one constructor.

   Constructors must have the same name as the class.

   Constructors don’t have a return type. If you see code with a return type, it’s a method with the same name as the class; it’s not a constructor.

   Typical constructor execution occurs as follows:

   The constructor calls its superclass constructor, which calls its superclass constructor, and so on all the way up to the Object constructor.

   The Object constructor executes and then returns to the calling constructor, which runs to completion and then returns to its calling constructor, and so on back down to the completion of the constructor of the actual instance being created.

   Constructors can use any access modifier (even private!).

   The compiler will create a default constructor if you don’t create any constructors in your class.

   The default constructor is a no-arg constructor with a no-arg call to super().

   The first statement of every constructor must be a call either to this() (an overloaded constructor) or to super().

   The compiler will add a call to super() unless you have already put in a call to this() or super().

   Instance members are accessible only after the super constructor runs.

   Abstract classes have constructors that are called when a concrete subclass is instantiated.

   Interfaces do not have constructors.

   If your superclass does not have a no-arg constructor, you must create a constructor and insert a call to super() with arguments matching those of the superclass constructor.

   Constructors are never inherited; thus they cannot be overridden.

   A constructor can be directly invoked only by another constructor (using a call to super() or this()).

   Regarding issues with calls to this():

   They may appear only as the first statement in a constructor.

   The argument list determines which overloaded constructor is called.

   Constructors can call constructors, and so on, but sooner or later one of them better call super() or the stack will explode.

   Calls to this() and super() cannot be in the same constructor. You can have one or the other, but never both.

Initialization Blocks (OCA Objective 1.2 and 6.3-ish)

   Use static init blocks—static { /* code here */ }—for code you want to have run once, when the class is first loaded. Multiple blocks run from the top down.

   Use normal init blocks—{ /* code here }—for code you want to have run for every new instance, right after all the super constructors have run. Again, multiple blocks run from the top of the class down.

Statics (OCA Objective 6.2)

   Use static methods to implement behaviors that are not affected by the state of any instances.

   Use static variables to hold data that is class specific as opposed to instance specific—there will be only one copy of a static variable.

   All static members belong to the class, not to any instance.

   A static method can’t access an instance variable directly.

   Use the dot operator to access static members, but remember that using a reference variable with the dot operator is really a syntax trick, and the compiler will substitute the class name for the reference variable; for instance:

d.doStuff();

becomes

Dog.doStuff();

   To invoke an interface’s static method use MyInterface.doStuff() syntax.

   static methods can’t be overridden, but they can be redefined.

SELF TEST

1. Given:

public abstract interface Frobnicate { public void twiddle(String s); }

Which is a correct class? (Choose all that apply.)

2.   Given:

What is the result?

A.   BD

B.   DB

C.   BDC

D.   DBC

E.   Compilation fails

3.   Given:

What is the result?

A.   Clidlet

B.   Clidder

C.   Clidder
    Clidlet

D.   Clidlet
    Clidder

E.   Compilation fails

Special Note: The next question crudely simulates a style of question known as “drag-and-drop.” Up through the SCJP 6 exam, drag-and-drop questions were included on the exam. As of spring 2014, Oracle DOES NOT include any drag-and-drop questions on its Java exams, but just in case Oracle’s policy changes, we left a few in the book.

4.   Using the fragments below, complete the following code so it compiles. Note that you may not have to fill in all of the slots.

Code:

Fragments: Use the following fragments zero or more times:

5.   Given:

What is the result?

A.   pre b1 b2 r3 r2 hawk

B.   pre b2 b1 r2 r3 hawk

C.   pre b2 b1 r2 r3 hawk r1 r4

D.   r1 r4 pre b1 b2 r3 r2 hawk

E.   r1 r4 pre b2 b1 r2 r3 hawk

F.   pre r1 r4 b1 b2 r3 r2 hawk

G.   pre r1 r4 b2 b1 r2 r3 hawk

H.   The order of output cannot be predicted

  I.   Compilation fails

Note: You’ll probably never see this many choices on the real exam!

6.   Given the following:

Which of the following, inserted at line 9, will compile? (Choose all that apply.)

A.   x2.do2();

B.   (Y)x2.do2();

C.   ((Y)x2).do2();

D.   None of the above statements will compile

7.   Given:

What is the result? (Choose all that apply.)

A.   2 will be included in the output

B.   3 will be included in the output

C.   hi will be included in the output

D.   Compilation fails

E.   An exception is thrown at runtime

8.   Given:

What is the result? (Choose all that apply.)

A.   howl howl sniff

B.   howl woof sniff

C.   howl howl followed by an exception

D.   howl woof followed by an exception

E.   Compilation fails with an error at line 14

F.   Compilation fails with an error at line 15

9.   Given:

What is the result? (Choose all that apply.)

A.   An exception is thrown at runtime

B.   The code compiles and runs with no output

C.   Compilation fails with an error at line 8

D.   Compilation fails with an error at line 9

E.   Compilation fails with an error at line 12

F.   Compilation fails with an error at line 13

10.   Given:

What is the result?

A.   fa fa

B.   fa la

C.   la la

D.   Compilation fails

E.   An exception is thrown at runtime

11.   Given:

What is the result?

A.   subsub

B.   sub subsub

C.   alpha subsub

D.   alpha sub subsub

E.   Compilation fails

F.   An exception is thrown at runtime

12.   Given:

What is the result?

A.   h hn x

B.   hn x h

C.   b h hn x

D.   b hn x h

E.   bn x h hn x

F.   b bn x h hn x

G.   bn x b h hn x

H.   Compilation fails

13.   Given:

What is the result?

A.   furry bray

B.   stripes bray

C.   furry generic noise

D.   stripes generic noise

E.   Compilation fails

F.   An exception is thrown at runtime

14.   Given:

What is the result? (Choose all that apply.)

A.   hopping 212

B.   Compilation fails due to an error on line 2

C.   Compilation fails due to an error on line 5

D.   Compilation fails due to an error on line 12

E.   Compilation fails due to an error on line 13

F.   Compilation fails due to an error on line 14

G.   Compilation fails due to an error on line 16

15.   Given:

What is the result?

A.   1

B.   2

C.   3

D.   The output is unpredictable

E.   Compilation fails

F.   An exception is thrown at runtime

16.   Given:

Which line(s) of code, inserted independently at // INSERT CODE HERE, will allow the code to compile? (Choose all that apply.)

A.   System.out.println(“class: ” + doStuff());

B.   System.out.println(“iface: ” + super.doStuff());

C.   System.out.println(“iface: ” + MyInterface.super.doStuff());

D.   System.out.println(“iface: ” + MyInterface.doStuff());

E.   System.out.println(“iface: ” + super.MyInterface.doStuff());

F.   None of the lines, A–E will allow the code to compile

SELF TEST ANSWERS

1.    B and E are correct. B is correct because an abstract class need not implement any or all of an interface’s methods. E is correct because the class implements the interface method and additionally overloads the twiddle() method.
A, C, and D are incorrect. A is incorrect because abstract methods have no body. C is incorrect because classes implement interfaces; they don’t extend them. D is incorrect because overloading a method is not implementing it. (OCA Objectives 7.1 and 7.5)

2.    E is correct. The implied super() call in Bottom2’s constructor cannot be satisfied because there is no no-arg constructor in Top. A default, no-arg constructor is generated by the compiler only if the class has no constructor defined explicitly.
A, B, C, and D are incorrect based on the above. (OCA Objective 6.3)

3.    A is correct. Although a final method cannot be overridden, in this case, the method is private and, therefore, hidden. The effect is that a new, accessible, method flipper is created. Therefore, no polymorphism occurs in this example, the method invoked is simply that of the child class, and no error occurs.
B, C, D, and E are incorrect based on the preceding. (OCA Objective 7.2)

Special Note: This next question crudely simulates a style of question known as “drag-and-drop.” Up through the SCJP 6 exam, drag-and-drop questions were included on the exam. As of spring 2014, Oracle DOES NOT include any drag-and-drop questions on its Java exams, but just in case Oracle’s policy changes, we left a few in the book.

4. Here is the answer:

As there is no droppable tile for the variable x and the parentheses (in the Kinder constructor) are already in place and empty, there is no way to construct a call to the superclass constructor that takes an argument. Therefore, the only remaining possibility is to create a call to the no-arg superclass constructor. This is done as super();. The line cannot be left blank, as the parentheses are already in place. Further, since the superclass constructor called is the no-arg version, this constructor must be created. It will not be created by the compiler because another constructor is already present. (OCA Objectives 6.3 and 7.4) Note: As you can see, many questions test for OCA Objective 7.1, we’re going to stop mentioning objective 7.1.

5.    D is correct. Static init blocks are executed at class loading time; instance init blocks run right after the call to super() in a constructor. When multiple init blocks of a single type occur in a class, they run in order, from the top down.
A, B, C, E, F, G, H, and I are incorrect based on the above. Note: You’ll probably never see this many choices on the real exam! (OCA Objective 6.3)

6.    C is correct. Before you can invoke Y’s do2 method, you have to cast x2 to be of type Y.
A, B, and D are incorrect based on the preceding. B looks like a proper cast, but without the second set of parentheses, the compiler thinks it’s an incomplete statement. (OCA Objective 7.3)

7.    A is correct. It’s legal to overload main(). Since no instances of Locomotive are created, the constructor does not run and the overloaded version of main() does not run.
B, C, D, and E are incorrect based on the preceding. (OCA Objectives 1.3 and 6.3)

8.    F is correct. Class Dog doesn’t have a sniff method.
A, B, C, D, and E are incorrect based on the above information. (OCA Objectives 7.2 and 7.3)

9.    A is correct. A ClassCastException will be thrown when the code attempts to downcast a Tree to a Redwood.
B, C, D, E, and F are incorrect based on the above information. (OCA Objective 7.3)

10.    B is correct. The code is correct, but polymorphism doesn’t apply to static methods.
A, C, D, and E are incorrect based on the above information. (OCA Objectives 6.2 and 7.2)

11.    C is correct. Watch out, because SubSubAlpha extends Alpha! Because the code doesn’t attempt to make a SubAlpha, the private constructor in SubAlpha is okay.
A, B, D, E, and F are incorrect based on the above information. (OCA Objectives 6.3 and 7.2)

12.    C is correct. Remember that constructors call their superclass constructors, which execute first, and that constructors can be overloaded.
A, B, D, E, F, G, and H are incorrect based on the above information. (OCA Objectives 6.3 and 7.4)

13.    A is correct. Polymorphism is only for instance methods, not instance variables.
B, C, D, E, and F are incorrect based on the above information. (OCA Objective 6.3)

14.    E and G are correct. Neither of these lines of code uses the correct syntax to invoke an interface’s static method.
A, B, C, D, and F are incorrect based on the above information. (OCP Objectives 6.2 and 7.5)

15.    E is correct. This is kind of a trick question; the implementing method must be marked public. If it was, all the other code is legal, and the output would be 3. If you understood all the multiple inheritance rules and just missed the access modifier, give yourself half credit.
A, B, C, D, and F are incorrect based on the above information. (OCP Objective 7.5)

16.    A and C are correct. A uses correct syntax to invoke the class’s method, and C uses the correct syntax to invoke the interface’s overloaded default method.
B, D, E, and F are incorrect. (OCP Objective 7.5)