Classes, Superclasses, and Subclasses

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Let's return to the Employee class that we discussed in the previous chapter. Suppose (alas) you work for a company at which managers are treated differently from other employees. Managers are, of course, just like employees in many respects. Both employees and managers are paid a salary. However, while employees are expected to complete their assigned tasks in return for receiving their salary, managers get bonuses if they actually achieve what they are supposed to do. This is the kind of situation that cries out for inheritance. Why? Well, you need to define a new class, Manager, and add functionality. But you can retain some of what you have already programmed in the Employee class, and all the fields of the original class can be preserved. More abstractly, there is an obvious "is a" relationship between Manager and Employee. Every manager is an employee: this "is a" relationship is the hallmark of inheritance.

Here is how you define a Manager class that inherits from the Employee class. You use the Java keyword extends to denote inheritance.

class Manager extends Employee
{
   added methods and fields
}

C++ NOTE

Inheritance is similar in Java and C++. Java uses the extends keyword instead of the : token. All inheritance in Java is public inheritance; there is no analog to the C++ features of private and protected inheritance.

The keyword extends indicates that you are making a new class that derives from an existing class. The existing class is called the superclass, base class, or parent class. The new class is called the subclass, derived class, or child class. The terms superclass and subclass are those most commonly used by Java programmers, although some programmers prefer the parent/child analogy, which also ties in nicely with the "inheritance" theme.

The Employee class is a superclass, but not because it is superior to its subclass or contains more functionality. In fact, the opposite is true: subclasses have more functionality than their superclasses. For example, as you will see when we go over the rest of the Manager class code, the Manager class encapsulates more data and has more functionality than its superclass Employee.

NOTE

The prefixes super and sub come from the language of sets used in theoretical computer science and mathematics. The set of all employees contains the set of all managers, and this is described by saying it is a superset of the set of managers. Or, put it another way, the set of all managers is a subset of the set of all employees.

Our Manager class has a new field to store the bonus, and a new method to set it:

 class Manager extends Employee {    . . .    public void setBonus(double b)    {       bonus = b;    }    private double bonus; } 

There is nothing special about these methods and fields. If you have a Manager object, you can simply apply the setBonus method.

 Manager boss = . . .; boss.setBonus(5000); 

Of course, if you have an Employee object, you cannot apply the setBonus method it is not among the methods that are defined in the Employee class.

However, you can use methods such as getName and getHireDay with Manager objects. Even though these methods are not explicitly defined in the Manager class, they are automatically inherited from the Employee superclass.

Similarly, the fields name, salary, and hireDay are inherited from the superclass. Every Manager object has four fields: name, salary, hireDay, and bonus.

When defining a subclass by extending its superclass, you only need to indicate the differences between the subclass and the superclass. When designing classes, you place the most general methods into the superclass and more specialized methods in the subclass. Factoring out common functionality by moving it to a superclass is common in object-oriented programming.

However, some of the superclass methods are not appropriate for the Manager subclass. In particular, the getSalary method should return the sum of the base salary and the bonus. You need to supply a new method to override the superclass method:

 class Manager extends Employee {    . . .    public double getSalary()    {       . . .    }    . . . } 

How can you implement this method? At first glance, it appears to be simple: just return the sum of the salary and bonus fields:

 public double getSalary() {    return salary + bonus; // won't work } 

However, that won't work. The getSalary method of the Manager class has no direct access to the private fields of the superclass. This means that the getSalary method of the Manager class cannot directly access the salary field, even though every Manager object has a field called salary. Only the methods of the Employee class have access to the private fields. If the Manager methods want to access those private fields, they have to do what every other method does use the public interface, in this case, the public getSalary method of the Employee class.

So, let's try this again. You need to call getSalary instead of simply accessing the salary field.

 public double getSalary() {    double baseSalary = getSalary(); // still won't work    return baseSalary + bonus; } 

The problem is that the call to getSalary simply calls itself, because the Manager class has a getSalary method (namely, the method we are trying to implement). The consequence is an infinite set of calls to the same method, leading to a program crash.

We need to indicate that we want to call the getSalary method of the Employee superclass, not the current class. You use the special keyword super for this purpose. The call

 super.getSalary() 

calls the getSalary method of the Employee class. Here is the correct version of the getSalary method for the Manager class:

 public double getSalary() {    double baseSalary = super.getSalary();    return baseSalary + bonus; } 

NOTE

Some people think of super as being analogous to the this reference. However, that analogy is not quite accurate super is not a reference to an object. For example, you cannot assign the value super to another object variable. Instead, super is a special keyword that directs the compiler to invoke the superclass method.

As you saw, a subclass can add fields, and it can add or override methods of the superclass. However, inheritance can never take away any fields or methods.

C++ NOTE

Java uses the keyword super to call a superclass method. In C++, you would use the name of the superclass with the :: operator instead. For example, the getSalary method of the Manager class would call Employee::getSalary instead of super.getSalary.

Finally, let us supply a constructor.

 public Manager(String n, double s, int year, int month, int day) {    super(n, s, year, month, day);    bonus = 0; } 

Here, the keyword super has a different meaning. The instruction

 super(n, s, year, month, day); 

is shorthand for "call the constructor of the Employee superclass with n, s, year, month, and day as parameters."

Because the Manager constructor cannot access the private fields of the Employee class, it must initialize them through a constructor. The constructor is invoked with the special super syntax. The call using super must be the first statement in the constructor for the subclass.

If the subclass constructor does not call a superclass constructor explicitly, then the default (no-parameter) constructor of the superclass is invoked. If the superclass has no default constructor and the subclass constructor does not call another superclass constructor explicitly, then the Java compiler reports an error.

NOTE

Recall that the this keyword has two meanings: to denote a reference to the implicit parameter and to call another constructor of the same class. Likewise, the super keyword has two meanings: to invoke a superclass method and to invoke a superclass constructor. When used to invoke constructors, the this and super keywords are closely related. The constructor calls can only occur as the first statement in another constructor. The construction parameters are either passed to another constructor of the same class (this) or a constructor of the superclass (super).

C++ NOTE

In a C++ constructor, you do not call super, but you use the initializer list syntax to construct the superclass. The Manager constructor looks like this in C++:

 Manager::Manager(String n, double s, int year, int month, int day) // C++ : Employee(n, s, year, month, day) {    bonus = 0; } 

Having redefined the getSalary method for Manager objects, managers will automatically have the bonus added to their salaries.

Here's an example of this at work: we make a new manager and set the manager's bonus:

 Manager boss = new Manager("Carl Cracker", 80000, 1987, 12, 15); boss.setBonus(5000); 

We make an array of three employees:

 Employee[] staff = new Employee[3]; 

We populate the array with a mix of managers and employees:

 staff[0] = boss; staff[1] = new Employee("Harry Hacker", 50000, 1989, 10, 1); staff[2] = new Employee("Tony Tester", 40000, 1990, 3, 15); 

We print out everyone's salary:

 for (Employee e : staff)    System.out.println(e.getName() + " " + e.getSalary()); 

This loop prints the following data:

 Carl Cracker 85000.0 Harry Hacker 50000.0 Tommy Tester 40000.0 

Now staff[1] and staff[2] each print their base salary because they are Employee objects. However, staff[0] is a Manager object and its getSalary method adds the bonus to the base salary.

What is remarkable is that the call

 e.getSalary() 

picks out the correct getSalary method. Note that the declared type of e is Employee, but the actual type of the object to which e refers can be either Employee (that is, when i is 1 or 2) or Manager (when i is 0).

When e refers to an Employee object, then the call e.getSalary() calls the getSalary method of the Employee class. However, when e refers to a Manager object, then the getSalary method of the Manager class is called instead. The virtual machine knows about the actual type of the object to which e refers, and therefore can invoke the correct method.

The fact that an object variable (such as the variable e) can refer to multiple actual types is called polymorphism. Automatically selecting the appropriate method at run time is called dynamic binding. We discuss both topics in more detail in this chapter.

C++ NOTE

In Java, you do not need to declare a method as virtual. Dynamic binding is the default behavior. If you do not want a method to be virtual, you tag it as final. (We discuss the final keyword later in this chapter.)

Example 5-1 contains a program that shows how the salary computation differs for Employee and Manager objects.

Example 5-1. ManagerTest.java

  1. import java.util.*;  2.  3. public class ManagerTest  4. {  5.    public static void main(String[] args)  6.    {  7.       // construct a Manager object  8.       Manager boss = new Manager("Carl Cracker", 80000, 1987, 12, 15);  9.       boss.setBonus(5000); 10. 11.       Employee[] staff = new Employee[3]; 12. 13.       // fill the staff array with Manager and Employee objects 14. 15.       staff[0] = boss; 16.       staff[1] = new Employee("Harry Hacker", 50000, 1989, 10, 1); 17.       staff[2] = new Employee("Tommy Tester", 40000, 1990, 3, 15); 18. 19.       // print out information about all Employee objects 20.       for (Employee e : staff) 21.          System.out.println("name=" + e.getName() 22.             + ",salary=" + e.getSalary()); 23.    } 24. } 25. 26. class Employee 27. { 28.    public Employee(String n, double s, int year, int month, int day) 29.    { 30.       name = n; 31.       salary = s; 32.       GregorianCalendar calendar = new GregorianCalendar(year, month - 1, day); 33.       hireDay = calendar.getTime(); 34.    } 35. 36.    public String getName() 37.    { 38.       return name; 39.    } 40. 41.    public double getSalary() 42.    { 43.       return salary; 44.    } 45. 46.    public Date getHireDay() 47.    { 48.       return hireDay; 49.    } 50. 51.    public void raiseSalary(double byPercent) 52.    { 53.       double raise = salary * byPercent / 100; 54.       salary += raise; 55.    } 56. 57.    private String name; 58.    private double salary; 59.    private Date hireDay; 60. } 61. 62. class Manager extends Employee 63. { 64.    /** 65.       @param n the employee's name 66.       @param s the salary 67.       @param year the hire year 68.       @param month the hire month 69.       @param day the hire day 70.    */ 71.    public Manager(String n, double s, int year, int month, int day) 72.    { 73.       super(n, s, year, month, day); 74.       bonus = 0; 75.    } 76. 77.    public double getSalary() 78.    { 79.       double baseSalary = super.getSalary(); 80.       return baseSalary + bonus; 81.    } 82. 83.    public void setBonus(double b) 84.    { 85.       bonus = b; 86.    } 87. 88.    private double bonus; 89. } 

Inheritance Hierarchies

Inheritance need not stop at deriving one layer of classes. We could have an Executive class that extends Manager, for example. The collection of all classes extending from a common superclass is called an inheritance hierarchy, as shown in Figure 5-1. The path from a particular class to its ancestors in the inheritance hierarchy is its inheritance chain.

Figure 5-1. Employee inheritance hierarchy

There is usually more than one chain of descent from a distant ancestor class. You could form a subclass Programmer or Secretary that extends Employee, and they would have nothing to do with the Manager class (or with each other). This process can continue as long as is necessary.

C++ NOTE

Java does not support multiple inheritance. (For ways to recover much of the functionality of multiple inheritance, see the section on Interfaces in the next chapter.)

Polymorphism

A simple rule enables you to know whether or not inheritance is the right design for your data. The "is-a" rule states that every object of the subclass is an object of the superclass. For example, every manager is an employee. Thus, it makes sense for the Manager class to be a subclass of the Employee class. Naturally, the opposite is not true not every employee is a manager.

Another way of formulating the "is-a" rule is the substitution principle. That principle states that you can use a subclass object whenever the program expects a superclass object.

For example, you can assign a subclass object to a superclass variable.

 Employee e; e = new Employee(. . .);  // Employee object expected e = new Manager(. . .); // OK, Manager can be used as well 

In the Java programming language, object variables are polymorphic. A variable of type Employee can refer to an object of type Employee or to an object of any subclass of the Employee class (such as Manager, Executive, Secretary, and so on).

We took advantage of this principle in Example 5-1:

 Manager boss = new Manager(. . .); Employee[] staff = new Employee[3]; staff[0] = boss; 

In this case, the variables staff[0] and boss refer to the same object. However, staff[0] is considered to be only an Employee object by the compiler.

That means, you can call

 boss.setBonus(5000); // OK 

but you can't call

 staff[0].setBonus(5000); // ERROR 

The declared type of staff[0] is Employee, and the setBonus method is not a method of the Employee class.

However, you cannot assign a superclass reference to a subclass variable. For example, it is not legal to make the assignment

 Manager m = staff[i]; // ERROR 

The reason is clear: Not all employees are managers. If this assignment were to succeed and m were to refer to an Employee object that is not a manager, then it would later be possible to call m.setBonus(...) and a runtime error would occur.

CAUTION

In Java, arrays of subclass references can be converted to arrays of superclass references without a cast. For example, consider an array of managers

 Manager[] managers = new Manager[10]; 

It is legal to convert this array to an Employee[] array:

 Employee[] staff = managers; // OK 

Sure, why not, you may think. After all, if manager[i] is a Manager, it is also an Employee. But actually, something surprising is going on. Keep in mind that managers and staff are references to the same array. Now consider the statement

 staff[0] = new Employee("Harry Hacker", ...); 

The compiler will cheerfully allow this assignment. But staff[0] and manager[0] are the same reference, so it looks as if we managed to smuggle a mere employee into the management ranks. That would be very bad calling managers[0].setBonus(1000) would try to access a nonexistent instance field and would corrupt neighboring memory.

To make sure no such corruption can occur, all arrays remember the element type with which they were created, and they monitor that only compatible references are stored into them. For example, the array created as new Manager[10] remembers that it is an array of managers. Attempting to store an Employee reference causes an ArrayStoreException.

Dynamic Binding

It is important to understand what happens when a method call is applied to an object. Here are the details:

1. The compiler looks at the declared type of the object and the method name. Let's say we call x.f(param), and the implicit parameter x is declared to be an object of class C. Note that there may be multiple methods, all with the same name, f, but with different parameter types. For example, there may be a method f(int) and a method f(String). The compiler enumerates all methods called f in the class C and all public methods called f in the superclasses of C.

   Now the compiler knows all possible candidates for the method to be called.

2. Next, the compiler determines the types of the parameters that are supplied in the method call. If among all the methods called f there is a unique method whose parameter types are a best match for the supplied parameters, then that method is chosen to be called. This process is called overloading resolution. For example, in a call x.f("Hello"), the compiler picks f(String) and not f(int). The situation can get complex because of type conversions (int to double, Manager to Employee, and so on). If the compiler cannot find any method with matching parameter types or if multiple methods all match after applying conversions, then the compiler reports an error.

   Now the compiler knows the name and parameter types of the method that needs to be called.

   NOTE

   

   Recall that the name and parameter type list for a method is called the method's signature. For example, f(int) and f(String) are two methods with the same name but different signatures. If you define a method in a subclass that has the same signature as a superclass method, then you override that method.

   The return type is not part of the signature. However, when you override a method, you need to keep the return type compatible. Prior to JDK 5.0, the return types had to be identical. However, it is now legal for the subclass to change the return type of an overridden method to a subtype of the original type. For example, suppose that the Employee class has a

    public Employee getBuddy() { ... } 

   
   

   Then the Manager subclass can override this method as

    public Manager getBuddy() { ... } // OK in JDK 5.0 

   
   

   We say that the two getBuddy methods have covariant return types.

   
   

3. If the method is private, static, final, or a constructor, then the compiler knows exactly which method to call. (The final modifier is explained in the next section.) This is called static binding. Otherwise, the method to be called depends on the actual type of the implicit parameter, and dynamic binding must be used at run time. In our example, the compiler would generate an instruction to call f(String) with dynamic binding.

4. When the program runs and uses dynamic binding to call a method, then the virtual machine must call the version of the method that is appropriate for the actual type of the object to which x refers. Let's say the actual type is D, a subclass of C. If the class D defines a method f(String), that method is called. If not, D's superclass is searched for a method f(String), and so on.

   It would be time consuming to carry out this search every time a method is called. Therefore, the virtual machine precomputes for each class a method table that lists all method signatures and the actual methods to be called. When a method is actually called, the virtual machine simply makes a table lookup. In our example, the virtual machine consults the method table for the class D and looks up the method to call for f(String). That method may be D.f(String) or X.f(String), where X is some superclass of D.

   There is one twist to this scenario. If the call is super.f(param), then the compiler consults the method table of the superclass of the implicit parameter.

Let's look at this process in detail in the call e.getSalary() in Example 5-1. The declared type of e is Employee. The Employee class has a single method, called getSalary, with no method parameters. Therefore, in this case, we don't worry about overloading resolution.

Because the getSalary method is not private, static, or final, it is dynamically bound. The virtual machine produces method tables for the Employee and Manager classes. The Employee table shows that all methods are defined in the Employee class itself:

 Employee:    getName() -> Employee.getName()    getSalary() -> Employee.getSalary()    getHireDay() -> Employee.getHireDay()    raiseSalary(double) -> Employee.raiseSalary(double) 

Actually, that isn't the whole story as you will see later in this chapter, the Employee class has a superclass Object from which it inherits a number of methods. We ignore the Object methods for now.

The Manager method table is slightly different. Three methods are inherited, one method is redefined, and one method is added.

 Manager:    getName() -> Employee.getName()    getSalary() -> Manager.getSalary()    getHireDay() -> Employee.getHireDay()    raiseSalary(double) -> Employee.raiseSalary(double)    setBonus(double) -> Manager.setBonus(double) 

At run time, the call e.getSalary() is resolved as follows.

1.	First, the virtual machine fetches the method table for the actual type of e. That may be the table for Employee, Manager, or another subclass of Employee.
2.	Then, the virtual machine looks up the defining class for the getSalary() signature. Now it knows which method to call.
3.	Finally, the virtual machine calls the method.

Dynamic binding has a very important property: it makes programs extensible without the need for modifying existing code. Suppose a new class Executive is added and there is the possibility that the variable e refers to an object of that class. The code containing the call e.getSalary() need not be recompiled. The Executive.getSalary() method is called automatically if e happens to refer to an object of type Executive.

CAUTION

When you override a method, the subclass method must be at least as visible as the superclass method. In particular, if the superclass method is public, then the subclass method must also be declared as public. It is a common error to accidentally omit the public specifier for the subclass method. The compiler then complains that you try to supply a weaker access privilege.

Preventing Inheritance: Final Classes and Methods

Occasionally, you want to prevent someone from forming a subclass from one of your classes. Classes that cannot be extended are called final classes, and you use the final modifier in the definition of the class to indicate this. For example, let us suppose we want to prevent others from subclassing the Executive class. Then, we simply declare the class by using the final modifier as follows:

 final class Executive extends Manager {    . . . } 

You can also make a specific method in a class final. If you do this, then no subclass can override that method. (All methods in a final class are automatically final.) For example,

 class Employee {    . . .    public final String getName()    {       return name;    }    . . . } 

NOTE

Recall that fields can also be declared as final. A final field cannot be changed after the object has been constructed. However, if a class is declared as final, only the methods, not the fields, are automatically final.

There is only one good reason to make a method or class final: to make sure that the semantics cannot be changed in a subclass. For example, the getTime and setTime methods of the Calendar class are final. This indicates that the designers of the Calendar class have taken over responsibility for the conversion between the Date class and the calendar state. No subclass should be allowed to mess up this arrangement. Similarly, the String class is a final class. That means nobody can define a subclass of String. In other words, if you have a String reference, then you know it refers to a String and nothing but a String.

Some programmers believe that you should declare all methods as final unless you have a good reason that you want polymorphism. In fact, in C++ and C#, methods do not use polymorphism unless you specifically request it. That may be a bit extreme, but we agree that it is a good idea to think carefully about final methods and classes when you design a class hierarchy.

In the early days of Java, some programmers used the final keyword in the hope of avoiding the overhead of dynamic binding. If a method is not overridden, and it is short, then a compiler can optimize the method call away a process called inlining. For example, inlining the call e.getName() replaces it with the field access e.name. This is a worthwhile improvement CPUs hate branching because it interferes with their strategy of prefetching instructions while processing the current one. However, if getName can be overridden in another class, then the compiler cannot inline it because it has no way of knowing what the overriding code may do.

Fortunately, the just-in-time compiler in the virtual machine can do a better job than a traditional compiler. It knows exactly which classes extend a given class, and it can check whether any class actually overrides a given method. If a method is short, frequently called, and not actually overridden, the just-in-time compiler can inline the method. What happens if the virtual machine loads another subclass that overrides an inlined method? Then the optimizer must undo the inlining. That's slow, but it happens rarely.

C++ NOTE

In C++, a method is not dynamically bound by default, and you can tag it as inline to have method calls replaced with the method source code. However, there is no mechanism that would prevent a subclass from overriding a superclass method. In C++, you can write classes from which no other class can derive, but doing so requires an obscure trick, and there are few reasons to write such a class. (The obscure trick is left as an exercise to the reader. Hint: Use a virtual base class.)

Casting

Recall from Chapter 3 that the process of forcing a conversion from one type to another is called casting. The Java programming language has a special notation for casts. For example:

 double x = 3.405; int nx = (int) x; 

converts the value of the expression x into an integer, discarding the fractional part.

Just as you occasionally need to convert a floating-point number to an integer, you also need to convert an object reference from one class to another. To actually make a cast of an object reference, you use a syntax similar to what you use for casting a numeric expression. Surround the target class name with parentheses and place it before the object reference you want to cast. For example:

 Manager boss = (Manager) staff[0]; 

There is only one reason why you would want to make a cast to use an object in its full capacity after its actual type has been temporarily forgotten. For example, in the ManagerTest class, the staff array had to be an array of Employee objects because some of its entries were regular employees. We would need to cast the managerial elements of the array back to Manager to access any of its new variables. (Note that in the sample code for the first section, we made a special effort to avoid the cast. We initialized the boss variable with a Manager object before storing it in the array. We needed the correct type to set the bonus of the manager.)

As you know, in Java every object variable has a type. The type describes the kind of object the variable refers to and what it can do. For example, staff[i] refers to an Employee object (so it can also refer to a Manager object).

The compiler checks that you do not promise too much when you store a value in a variable. If you assign a subclass reference to a superclass variable, you are promising less, and the compiler will simply let you do it. If you assign a superclass reference to a subclass variable, you are promising more. Then you must use a cast so that your promise can be checked at run time.

What happens if you try to cast down an inheritance chain and you are "lying" about what an object contains?

 Manager boss = (Manager) staff[1]; // ERROR 

When the program runs, the Java runtime system notices the broken promise and generates a ClassCastException. If you do not catch the exception, your program terminates. Thus, it is good programming practice to find out whether a cast will succeed before attempting it. Simply use the instanceof operator. For example:

 if (staff[1] instanceof Manager) {    boss = (Manager) staff[1];    . . . } 

Finally, the compiler will not let you make a cast if there is no chance for the cast to succeed. For example, the cast

 Date c = (Date) staff[1]; 

is a compile-time error because Date is not a subclass of Employee.

To sum up:

• You can cast only within an inheritance hierarchy.

• Use instanceof to check before casting from a superclass to a subclass.

NOTE

The test

 x instanceof C 

does not generate an exception if x is null. It simply returns false. That makes sense. Because null refers to no object, it certainly doesn't refer to an object of type C.

Actually, converting the type of an object by performing a cast is not usually a good idea. In our example, you do not need to cast an Employee object to a Manager object for most purposes. The getSalary method will work correctly on both objects of both classes. The dynamic binding that makes polymorphism work locates the correct method automatically.

The only reason to make the cast is to use a method that is unique to managers, such as setBonus. If for some reason you find yourself wanting to call setBonus on Employee objects, ask yourself whether this is an indication of a design flaw in the superclass. It may make sense to redesign the superclass and add a setBonus method. Remember, it takes only one uncaught ClassCastException to terminate your program. In general, it is best to minimize the use of casts and the instanceof operator.

C++ NOTE

Java uses the cast syntax from the "bad old days" of C, but it works like the safe dynamic_cast operation of C++. For example,

 Manager boss = (Manager) staff[1]; // Java 

is the same as

 Manager* boss = dynamic_cast<Manager*>(staff[1]); // C++ 

with one important difference. If the cast fails, it does not yield a null object but throws an exception. In this sense, it is like a C++ cast of references. This is a pain in the neck. In C++, you can take care of the type test and type conversion in one operation.

 Manager* boss = dynamic_cast<Manager*>(staff[1]); // C++ if (boss != NULL) . . . 

In Java, you use a combination of the instanceof operator and a cast.

 if (staff[1] instanceof Manager) {    Manager boss = (Manager) staff[1];    . . . } 

Abstract Classes

As you move up the inheritance hierarchy, classes become more general and probably more abstract. At some point, the ancestor class becomes so general that you think of it more as a basis for other classes than as a class with specific instances you want to use. Consider, for example, an extension of our Employee class hierarchy. An employee is a person, and so is a student. Let us extend our class hierarchy to include classes Person and Student. Figure 5-2 shows the inheritance relationships between these classes.

Figure 5-2. Inheritance diagram for Person and its subclasses

Why bother with so high a level of abstraction? There are some attributes that make sense for every person, such as the name. Both students and employees have names, and introducing a common superclass lets us factor out the getName method to a higher level in the inheritance hierarchy.

Now let's add another method, getdescription, whose purpose is to return a brief description of the person, such as

 an employee with a salary of $50,000.00 a student majoring in computer science 

It is easy to implement this method for the Employee and Student classes. But what information can you provide in the Person class? The Person class knows nothing about the person except the name. Of course, you could implement Person.getDescription() to return an empty string. But there is a better way. If you use the abstract keyword, you do not need to implement the method at all.

 public abstract String getDescription();    // no implementation required 

For added clarity, a class with one or more abstract methods must itself be declared abstract.

 abstract class Person {  . . .    public abstract String getDescription(); } 

In addition to abstract methods, abstract classes can have concrete data and methods. For example, the Person class stores the name of the person and has a concrete method that returns it.

 abstract class Person {    public Person(String n)    {       name = n;    }    public abstract String getDescription();    public String getName()    {        return name;    }    private String name; } 

TIP

Many programmers think that abstract classes should have only abstract methods. However, this is not true. It always makes sense to move as much functionality as possible into a superclass, whether or not it is abstract. In particular, move common fields and methods (whether abstract or not) to the abstract superclass.

Abstract methods act as placeholders for methods that are implemented in the subclasses. When you extend an abstract class, you have two choices. You can leave some or all of the abstract methods undefined. Then you must tag the subclass as abstract as well. Or you can define all methods. Then the subclass is no longer abstract.

For example, we will define a Student class that extends the abstract Person class and implements the geTDescription method. Because none of the methods of the Student class are abstract, it does not need to be declared as an abstract class.

A class can even be declared as abstract even though it has no abstract methods.

Abstract classes cannot be instantiated. That is, if a class is declared as abstract, no objects of that class can be created. For example, the expression

 new Person("Vince Vu") 

is an error. However, you can create objects of concrete subclasses.

Note that you can still create object variables of an abstract class, but such a variable must refer to an object of a nonabstract subclass. For example,

 Person p = new Student("Vince Vu", "Economics"); 

Here p is a variable of the abstract type Person that refers to an instance of the nonabstract subclass Student.

C++ NOTE

In C++, an abstract method is called a pure virtual function and is tagged with a trailing = 0, such as in

 class Person // C++ { public:    virtual string getDescription() = 0;    . . . }; 

A C++ class is abstract if it has at least one pure virtual function. In C++, there is no special keyword to denote abstract classes.

Let us define a concrete subclass Student that extends the abstract Person class:

 class Student extends Person {    public Student(String n, String m)    {       super(n);       major = m;    }    public String getDescription()    {       return "a student majoring in " + major;    }    private String major; } 

The Student class defines the getdescription method. Therefore, all methods in the Student class are concrete, and the class is no longer an abstract class.

The program shown in Example 5-2 defines the abstract superclass Person and two concrete subclasses, Employee and Student. We fill an array of Person references with employee and student objects.

 Person[] people = new Person[2]; people[0] = new Employee(. . .); people[1] = new Student(. . .); 

We then print the names and descriptions of these objects:

 for (Person p : people)    System.out.println(p.getName() + ", " + p.getDescription()); 

Some people are baffled by the call

 p.getDescription() 

Isn't this call an undefined method? Keep in mind that the variable p never refers to a Person object because it is impossible to construct an object of the abstract Person class. The variable p always refers to an object of a concrete subclass such as Employee or Student. For these objects, the geTDescription method is defined.

Could you have omitted the abstract method altogether from the Person superclass and simply defined the geTDescription methods in the Employee and Student subclasses? If you did that, then you wouldn't have been able to invoke the geTDescription method on the variable p. The compiler ensures that you invoke only methods that are declared in the class.

Abstract methods are an important concept in the Java programming language. You will encounter them most commonly inside interfaces. For more information about interfaces, turn to Chapter 6.

Example 5-2. PersonTest.java

  1. import java.text.*;  2. import java.util.*;  3.  4. public class PersonTest  5. {  6.    public static void main(String[] args)  7.    {  8.       Person[] people = new Person[2];  9. 10.       // fill the people array with Student and Employee objects 11.       people[0] = new Employee("Harry Hacker", 50000, 1989, 10, 1); 12.       people[1] = new Student("Maria Morris", "computer science"); 13. 14.       // print out names and descriptions of all Person objects 15.       for (Person p : people) 16.          System.out.println(p.getName() + ", " + p.getDescription()); 17.    } 18. } 19. 20. abstract class Person 21. { 22.    public Person(String n) 23.    { 24.       name = n; 25.    } 26. 27.    public abstract String getDescription(); 28. 29.    public String getName() 30.    { 31.       return name; 32.    } 33. 34.    private String name; 35. } 36. 37. class Employee extends Person 38. { 39.    public Employee(String n, double s, 40.       int year, int month, int day) 41.    { 42.       super(n); 43.       salary = s; 44.       GregorianCalendar calendar = new GregorianCalendar(year, month - 1, day); 45.       hireDay = calendar.getTime(); 46.    } 47. 48.    public double getSalary() 49.    { 50.       return salary; 51.    } 52. 53.    public Date getHireDay() 54.    { 55.       return hireDay; 56.    } 57. 58.    public String getDescription() 59.    { 60.       return String.format("an employee with a salary of $%.2f", salary); 61.    } 62. 63.    public void raiseSalary(double byPercent) 64.    { 65.       double raise = salary * byPercent / 100; 66.       salary += raise; 67.    } 68. 69.    private double salary; 70.    private Date hireDay; 71. } 72. 73. 74. class Student extends Person 75. { 76.    /** 77.       @param n the student's name 78.       @param m the student's major 79.    */ 80.    public Student(String n, String m) 81.    { 82.       // pass n to superclass constructor 83.       super(n); 84.       major = m; 85.    } 86. 87.    public String getDescription() 88.    { 89.       return "a student majoring in " + major; 90.    } 91. 92.    private String major; 93. } 

Protected Access

As you know, fields in a class are best tagged as private, and methods are usually tagged as public. Any features declared private won't be visible to other classes. As we said at the beginning of this chapter, this is also true for subclasses: a subclass cannot access the private fields of its superclass.

There are times, however, when you want to restrict a method to subclasses only or, less commonly, to allow subclass methods to access a superclass field. In that case, you declare a class feature as protected. For example, if the superclass Employee declares the hireDay field as protected instead of private, then the Manager methods can access it directly.

However, the Manager class methods can peek inside the hireDay field of Manager objects only, not of other Employee objects. This restriction is made so that you can't abuse the protected mechanism and form subclasses just to gain access to the protected fields.

In practice, use the protected attribute with caution. Suppose your class is used by other programmers and you designed it with protected fields. Unknown to you, other programmers may inherit classes from your class and then start accessing your protected fields. In this case, you can no longer change the implementation of your class without upsetting the other programmers. That is against the spirit of OOP, which encourages data encapsulation.

Protected methods make more sense. A class may declare a method as protected if it is tricky to use. This indicates that the subclasses (which, presumably, know their ancestors well) can be trusted to use the method correctly, but other classes cannot.

A good example of this kind of method is the clone method of the Object class see Chapter 6 for more details.

C++ NOTE

As it happens, protected features in Java are visible to all subclasses as well as to all other classes in the same package. This is slightly different from the C++ meaning of protected, and it makes the notion of protected in Java even less safe than in C++.

Here is a summary of the four access modifiers in Java that control visibility:

1. Visible to the class only (private).

2. Visible to the world (public).

3. Visible to the package and all subclasses (protected).

4. Visible to the package the (unfortunate) default. No modifiers are needed.

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