String[]names={zhang, wang, li);
String names[] =new String [3];names [0] =zhang; names [1] =wang; names [2] =li;
String[3] names={zhang, wang, li};
以上皆正确
第1题:
String s="zhang san,li si,wang wu"; 按字母顺序对姓名
第2题:
定义结构体数组: struct stu { int num; char nameE20]; }X[5]={1,"LI",2,"ZHAO",3,"WANG",4," ZHANG",5,"LIU"); for(i=1;i<5;i++) printf("%d%c",x[i].num,x[i].name[2]); 以上程序段的输出结果为( )。
A.2A3N4A5U
B.112A3H4I
C.1A2N3A4U
D.2H3A4H5I
第3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4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++代码实现一个简单的聊天室系统(ChatRoomSystem),多个用户(User)可以向聊天室(ChatRoom)发送消息,聊天室将消息展示给所有用户。类图如图6-1所表示。
图6-1 类图
【C++代码】 include<iostream> include <string> using namespace std; class User { private: string name; public: User(string name){ (1) =name; } ~User(){} void setName(string name) { this->name=name; } string getName(){ return name; } void sendMessage(string message); }; class ChatRoom { . public: static void showMessage(User* user, string message) { cout<<"["<<user->getName()<<"] : "<<message<<endl; } }; void User::sendMessage(string message) { (2) (this,message); } class ChatRoomSystem{ public: . . void startup() { User* zhang = new User(“John"); User* li = new User("Leo"); zhang->sendMessage("Hi! Leo!"); li_>sendMessage("Hi! John!"); } void join(User* user) { (3) ("HeIIo Everyone! l am"+user->getName()); } . }; int main(){ ChatRoomSystem*crs= (4) ; crs->startup(); crs->join( (5) ("Wayne")); delete crs; } /* 程序运行结果: [John]:Hi! Leo! [Leo]:Hi! John! [Wayne]:Hello Everyone! I am Wayne /*
第5题:
Wang Li is()a new marketing campaign at the moment.
第6题:
Which of the following statements are true when creating NETBIOS names? ()
第7题:
以下哪种初始化数组的方式是错误的?()
第8题:
以下哪种初始化数组的方式是错误的?()
第9题:
String[]names={zhang, wang, li);
String names[] =new String [3];names [0] =zhang; names [1] =wang; names [2] =li;
String[3] names={zhang, wang, li};
以上皆正确
第10题:
An exception may be thrown at runtime.
The code may run with no output, without exiting.
The code may rum with output “A B A B C C “, then exit.
The code may ruin with output “A A A B C A B C C “, then exit.
The code may rum with output “A B C A B C A B C “, then exit.
The code may ruin with output “A B C A A B C A B C “, then exit.
第11题:
NETBIOS names can only use alphanumeric characters.
You can use a ’.’ in a NETBIOS name.
You can use an ’_’ (underscore) in a NETBIOS name.
NETBIOS names must be UPPERCASE
NETBIOS names can be a maximum of 32 characters
第12题:
The list of column names in the FOREIGN KEY clause can be a subset of the list of column names in the primary key of T2 or a UNIQUE constraint that exists on T2.
The list of column names in the FOREIGN KEY clause can be a subset of the list of column names in the primary key of T1 or a UNIQUE constraint that exists on T1.
The list of column names in the FOREIGN KEY clause must be identical to the list of column names in the primary key of T2 or a UNIQUE constraint that exists on T2.
The list of column names in the FOREIGN KEY clause must be identical to the list of column names in the primary key of T1 or a UNIQUE constraint that exists on T1.
第13题:
定义结构体数组: struct stu { int num; char name[20]; }x[5]={1,"LI",2,"ZHAO",3"WANG",4,"ZHANG",5"LIU"}; for(i=1;i<5;i++) printf("%d %c", x[i].num, x[i].name[2]); 以上程序段的输出结果为( ).
A.2A3N4A5U
B.1I2a3h4I
C.1A2N3A4U
D.2H3A4H5I
第14题:
定义结构体数组:
struct stu
{ int num;
char nameE20];
}X[5]={1,"LI",2,"ZHAO",3,"WANG",4," ZHANG",5,"LIU");
for(i=1;i<5;i++)
printf("%d%c",x[i].num,x[i].name[2]);
以上程序段的输出结果为( )。
A.2A3N4A5U
B.112A3H4I
C.1A2N3A4U
D.2H3A4H5I
第15题:
A.使用len(列表名)测量元素的个数names_list=["zhangsan","lisi","wangwu"]print(len(names_list))
B.使用列表名[下标]获取列表的某个元素,例如:names_list=["zhangsan","lisi","wangwu"]print(names_list[2])
C.向列表中添加新元素有三个方法:append、extend、insert,例如:names_list=["zhangsan","lisi","wangwu"]names_list.append("zhaoliu")names_list.extend(["zhaoliu","liqi"])names_list.insert(1,"zhaoliu")print(names_list)
D.已有列表nums=[11,22,33,44,55],使用while循环遍历列表nums=[11,22,33,44,55]i=0 whilei
print(nums[i])i+=1
第16题:
Wang Li is()a new marketing campaign at the moment.
Aplan
Bplaning
Cplanning
第17题:
下面关于数组声明和初始化的语句哪个有语法错误()
第18题:
When defining a referential constraint between the parent table T2 and the dependent table T1, which of the following is true?()
第19题:
import java.util.*; public class NameList { private List names = new ArrayList(); public synchronized void add(String name) { names.add(name); } public synchronized void printAll() { for (int i = 0; i
第20题:
String[] name =,“zhang”,”wang”,”li”-
String*3+ names=,“zhang”,”wang”,”li”-
String names[] =new String[3] names*0+=”wang”
names*1+=”wang”
names*2+=”li”
以上皆正确
第21题:
plan
planing
planning
第22题:
1
2
3
4
5
第23题:
int a1[]={3,4,5};
String a2[]={string1,string1,string1};
String a3[]=new String(3);
int[][] a4=new int[3][3];
第24题:
the names of all persons on board
only the names of the crew members on board
only the names of passengers on board
information on emergency training drills