7: OBJECT-ORIENTED PROGRAMMING in Java

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CHAPTER 7: OBJECT-ORIENTED PROGRAMMING
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713 Inheritance Versus Aggregation
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Figure 75 is a UML class diagram showing several aggregation relationships and one inheritance relationship The class diagram shows a queue defined by aggregation and a stack defined by inheritance Both are based on linked lists A linked list is defined by aggregation A non-generic implementation of these data structures is shown in Example 715 The purpose of the example is to illustrate inheritance and aggregation, not industrial-strength implementation of queues and stacks The class Node at (1) is straightforward, defining two fields: one denoting the data and the other denoting the next node in the list The class LinkedList at (2) keeps track of the list by managing a head and a tail reference Nodes can be inserted in the front or back, but deleted only from the front of the list
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Implementing Data Structures by Inheritance and Aggregation
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next has QueueByAggregation has 1 LinkedList has 01 head 01 tail
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has data StackByInheritance 1
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Example 715 Implementing Data Structures by Inheritance and Aggregation class Node { private Object data; private Node next; // (1) // Data // Next node
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// Constructor for initializing data and reference to the next node Node(Object data, Node next) { thisdata = data; thisnext = next; }
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713: INHERITANCE VERSUS AGGREGATION // Methods: public void public Object public void public Node
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setData(Object obj) getData() setNext(Node node) getNext()
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data = return next = return
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obj; } data; } node; } next; }
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} //______________________________________________________________________________ class LinkedList { // (2) protected Node head = null; protected Node tail = null; // Methods: public boolean isEmpty() { return head == null; } public void insertInFront(Object dataObj) { if (isEmpty()) head = tail = new Node(dataObj, null); else head = new Node(dataObj, head); } public void insertAtBack(Object dataObj) { if (isEmpty()) head = tail = new Node(dataObj, null); else { tailsetNext(new Node(dataObj, null)); tail = tailgetNext(); } } public Object deleteFromFront() { if (isEmpty()) return null; Node removed = head; if (head == tail) head = tail = null; else head = headgetNext(); return removedgetData(); } } //______________________________________________________________________________ class QueueByAggregation { // (3) private LinkedList qList; // Constructor QueueByAggregation() { qList = new LinkedList(); } // Methods: public boolean isEmpty() { return qListisEmpty(); } public void enqueue(Object item) { qListinsertAtBack(item); } public Object dequeue() { if (qListisEmpty()) return null; return qListdeleteFromFront(); } public Object peek() { return (qListisEmpty() null : qListheadgetData()); } } //______________________________________________________________________________
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CHAPTER 7: OBJECT-ORIENTED PROGRAMMING class StackByInheritance extends LinkedList { // (4) public void push(Object item) { insertInFront(item); } public Object pop() { if (isEmpty()) return null; return deleteFromFront(); } public Object peek() { return (isEmpty() null : headgetData()); } } //______________________________________________________________________________ public class Client { // (5) public static void main(String[] args) { String string1 = "Queues are boring to stand in!"; int length1 = string1length(); QueueByAggregation queue = new QueueByAggregation(); for (int i = 0; i<length1; i++) queueenqueue(new Character(string1charAt(i))); while (!queueisEmpty()) Systemoutprint(queuedequeue()); Systemoutprintln(); String string2 = "!no tis ot nuf era skcatS"; int length2 = string2length(); StackByInheritance stack = new StackByInheritance(); for (int i = 0; i<length2; i++) stackpush(new Character(string2charAt(i))); stackinsertAtBack(new Character( ! )); while (!stackisEmpty()) Systemoutprint(stackpop()); Systemoutprintln(); } }
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// (6)
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Choosing between inheritance and aggregation to model relationships can be a crucial design decision A good design strategy advocates that inheritance should be used only if the relationship is-a is unequivocally maintained throughout the lifetime of the objects involved; otherwise, aggregation is the best choice A role is often confused with an is-a relationship For example, given the class Employee, it would not be a good idea to model the roles an employee can play (such as a manager or a cashier) by inheritance if these roles change intermittently Changing roles would involve a new object to represent the new role every time this happens Code reuse is also best achieved by aggregation when there is no is-a relationship Enforcing an artificial is-a relationship that is not naturally present is usually not a good idea This is illustrated in Example 715 at (6) Since the class StackByInheritance at (4) is a subclass of the class LinkedList at (2), any inherited method from the superclass can be invoked on an instance of the subclass Also,