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試卷年份2017年上半年
試題題型【分析簡答題】
試題內(nèi)容

閱讀以下說明和 Java程序,填補代碼中的空缺,將解答填入答題紙的對應欄內(nèi)。
【說明】
以下Java代碼實現(xiàn)一個簡單客戶關系管理系統(tǒng)(CRM)中通過工廠(CustomerFactory )對象來創(chuàng)建客戶(Customer)對象的功能??蛻舴譃閯?chuàng)建成功的客戶(RealCustomer)和空客戶 (NullCustomer)??湛蛻魧ο笫钱敳粷M足特定條件時創(chuàng)建或獲取的對象。類間關系如圖 5-1 所示。

圖5-1 類圖
【Java代碼】
Abstract class Customer﹛
Protected String name;
(  1  )boolean isNil();
(  2  )String getName();

Class RealCustomer (   3   )Customer{
Public RealCustomer(String name){  this.name=name;   }
Public String getName(){   return name ;   }
Public boolean is Nil() {   return false;   }


Class NullCustomer( 4  )Customer﹛
Public String getName()﹛   return "Not Available in Customer Database";   ﹜
Public boolean isNil() ﹛    return true;    ﹜

class Customerfactory {
public String[] names = {"Rob","Joe","Julie"};
public Customer getCustomer(String name) {
for (int i = 0; i < names.length;i++) {
if (names[i].(   5   ))﹛
return new RealCustomer(name);


return (   6   );


Public class CrM﹛
Public viod get Customer()﹛
Customerfactory(  7   );
Customer customer1-cf.getCustomer("Rob");
Customer customer2=cf.getCustomer("Bob");
Customer customer3= cf.getCustomer("Julie");
Customer customer4= cf.getCustomer("Laura");
System.out.println("customers”)
System.out.println(customer1.getName());
System.out.println(customer2getName());
System.out.println(customer3.getName());
System.out.println(customer4.getName());

Public static viod main (String[]arge)﹛
CRM crm =new CRM();
Crm.getCustomer();


/*程序輸出為:
Customers
rob
Not Available in Customer Database
Julie
Not Available in Customer Database
*/


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