A customer is acquiring a new IBM System Storage TS3500 Tape Library and requested to reuse four HP LTO-3 drives installed in a Sun L700 tape library to take advantage of the IBM multipath architecture. What action should the storage specialist take?()
第1题:
第2题:
阅读下列说明和C++代码,填补代码中的空缺,将解答填入答题纸的对应栏内。 【说明】 以下C++代码实现一个简单客户关系管理系统(CrM)中通过工厂(Customerfactory)对象来创建客户(Customer)对象的功能。客户分为创建成功的客户(realCustomer)和空客户(NullCustomer)。空客户对象是当不满足特定条件时创建或获取的对象。类间关系如图6-1所示。图6-1
【C++代码】 include<iostream> include<string> using namespace std; class Customer{ protected: string name; public: ( 1 ) boll isNil()=0; ( 2 ) string getName()=0; ﹜; class RealCustomer ( 3 ){ Public: realCustomer(string name){this->name=name;﹜ bool isNil(){ return false; ﹜ string getName(){ return name; ﹜ ﹜; class NullCustomer ( 4 ) { public: bool isNil(){ return true; ﹜ string getName(){ return 〝Not Available in Customer Database〞; ﹜ ﹜; class Customerfactory{ public: string names[3]={〝rob〞, 〝Joe〞,〝Julie〞﹜; public: Customer*getCustomer(string name){ for (int i=0;i<3;i++){ if (names[i].( 5 ) ){ return new realCustomer(name); ﹜ ﹜ return ( 6 ); ﹜ ﹜; class CRM{ public: void getCustomer(){ Customerfactory*( 7 ); Customer*customer1=cf->getCustomer(〝Rob〞); Customer*customer2=cf->getCustomer(〝Bob〞); Customer*customer3=cf->getCustomer(〝Julie〞); Customer*customer4=cf->getCustomer(〝Laura〞); cout<<〝Customers〞<<endl; cout<<Customer1->getName()<<endl; delete customer1; cout<<Customer2->getName()<<endl; delete customer2; cout<<Customer3->getName()<<endl; delete customer3; cout<<Customer4->getName()<<endl; delete customer4; delete cf; ﹜ ﹜; int main(){ CRM*crs=new CRM(); crs->getCustomer(); delete crs; return 0; ﹜ /*程序输出为: Customers rob Not Available in Customer Database Julie Not Available in Customer Database */
第3题:
请完成下列Java程序:假设某家银行,它可接受顾客的汇款,每做一次汇款,便可计算出汇款的总额。现有两个顾客,每人都分3次,每次50元将钱汇入。编写一个程序,模拟实际作业。要求实现2个类,一个是银行类,一个是顾客类。
注意:请勿改动main()主方法和其他已有语句内容,仅在下划线处填入适当的语句。
程序运行结果如下:
Customer2 :sum= 50
Customer1 :sum= 50
Customer1 :sum= 100
Customer2 :sum= 100
Customer1 :sum= 150
Customer2 :sum= 150
class bank{
private static int sum=0;
public static void add(int.n,char c){
int tmp=sum;
______________;
try{
Thread.sleep(((int) ((2000-500+1)*(Math.random())))+500);
}
catch(InterruptedException e){}
sum=tmp;
System.out.println("Customer"+c+" :sum= "+sum);
}
}
class customer extends Thread{
static char flag17_2 = '1';
public void run(){
char myflag17_2;
synchronized(this) {
myflag17_2 = flag17_2++;
}
for(int i=1;i<=3;i++)
______________________;
}
}
public class ex17_2{
public static void main(String args[]){
customer c1=new customer();
customer c2=new customer();
c1.start();
c2.start();
}
}
第4题:
Integer i = new Integer (42); Long 1 = new Long (42); Double d = new Double (42.0); Which two expressions evaluate to True?()
第5题:
The pSeries sales representative has done an excellent job in educating the customer on the benefits of working with IBM. The customer is planning to implement their first ERP application. This new application will be critical to running their business. The customer has only a competitor’s UNIX systemcurrently installed. They are interested in obtaining a proposal from IBM. In addition to the AIX hardware and HACMP software, what else should be included in the IBM proposal for the new ERP solution?()
第6题:
阅读以下说明和 Java程序,填补代码中的空缺,将解答填入答题纸的对应栏内。 【说明】 以下Java代码实现一个简单客户关系管理系统(CRM)中通过工厂(CustomerFactory )对象来创建客户(Customer)对象的功能。客户分为创建成功的客户(RealCustomer)和空客户 (NullCustomer)。空客户对象是当不满足特定条件时创建或获取的对象。类间关系如图 5-1 所示。图5-1 类图
【Java代码】 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(); ﹜ ﹜ /*程序输出为: Customers rob Not Available in Customer Database Julie Not Available in Customer Database */
第7题:
下面的XML片断中结构完整的是()。
第8题:
A.a new one
B.an new one
C.the new one
D.one new
第9题:
Given: Integer i = new Integer (42); Long l = new Long (42); Double d = new Double (42.0); Which two expression evaluate to true?()
第10题:
Organizations are now extending their () beyond the end customer to include the acceptance and disassembly of final products for reuse in the new products.