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試卷名稱 2017年上半年程序員考試下午真題試題(案例分析)
考試中心《2017年上半年程序員考試下午真題試題(案例分析)》在線考試
試卷年份2017年上半年
試題題型【分析簡(jiǎn)答題】
試題內(nèi)容

閱讀以下說明和 Java 程序,填補(bǔ)代碼中的空缺,將解答填入答題紙的對(duì)應(yīng)欄內(nèi)。
【說明】
以下 Jave 代碼實(shí)現(xiàn)一個(gè)簡(jiǎn)單客戶關(guān)系管理系統(tǒng)(CrM)中通過工廠(Customerrfactory )對(duì)象來創(chuàng)建客戶(Customer)對(duì)象的功能??蛻舴譃閯?chuàng)建成功的客戶(realCustomer)和空客戶(NullCustomer)??湛蛻魧?duì)象是當(dāng)不滿足特定條件時(shí)創(chuàng)建或獲取的對(duì)象。類間關(guān)系如圖 5-1 所示。

【Java 代碼】
Abstract class Customer﹛
Protected String name;
()boolean isNil()
()String getName();


Class realCustomer ()Customer﹛
Public realCustomer(String name )﹛ return false; ﹜


Class NullCustomer()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].())﹛
return new realCusωmer(name);


return ()



Public class CrM﹛
Public viod get Customer()﹛
Customerfactory()
Customer customer1-cf.getCustomer(″rob″);
Customer customer2=cf.getCustomer(″rob″);
Customer customer3= cf.getCustomer(″Julie″);
Customer customer4= cf.getCustomer(″Laura″);
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:
Customer
rob
Not Available in Customer Database
Julie
Not Available in Customer Datable
*/
int main()﹛
CrM*crs=newCrM();
Crs->getCustomer();
Crs->getCustomer();
Delete crs;
return();

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


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