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1题:
It's very interesting to study names of different countries.Chinese names are different (1) foreign names.Once an English lady came to visit me.When I was introduced to her she said, “ Glad to meet you, Miss Ping.” Then she gave me her name card with three words on it:"Betty J.Black.So I said, “Thank you, Miss Betty.” We looked at each other and laughed heartily.Later I found that the English people (2)their family names last and the given names first, while their middle names are not used very much.I explained to her that the Chinese family name comes first, the given name last, so she(3) never call me Miss Ping.She asked if we Chinese had a middle name.I told her we didn't.but people may often find three words
on a Chinese name card.In this case the family name still come first, and the other words after it(4)a two-word given name.it is quite usual in China.My sister is Li Xiaofang.She has two words in her given name instead (5) just one like mine.(完型填空)
A.Put
B.From
C.Should
D.of
E.are
答案:B A C E D
解析:本段意思:研究不同国家的名称是非常有趣的。中国人的名字不同于外国人的名字。有一次,一位英国女士来看我。当我被介绍给她时,她说:“很高兴认识你,萍小姐。然后她给了我她的名片,上面有三个字:“贝蒂·j·布莱克。”所以我说:“谢谢你,贝蒂小姐。我们面面相觑,开怀大笑。后来我发现英国人把姓放在最后,名放在前面,中间的名字用得不多。我向她解释中国姓在前,名在后,所以她永远不要叫我萍小姐。她问我们中国人有没有中间名。我告诉她我们没有。但是人们经常会在一张中文名片上发现三个字。在这种情况下,姓仍然在前面,在它后面的其他单词是两个单词组成的名字。这在中国很常见。我妹妹是李小芳。她有两个字在她的名字,而不像我一样只有一个。
第2题:
String s="zhang san,li si,wang wu"; 按字母顺序对姓名
第3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4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5题:
以下正确的电子邮箱地址的格式是
A.wang.163.com
B.wang@163.com
C.wang#163.com
D.www.wang.163.com
第6题:
Wang Li is()a new marketing campaign at the moment.
Aplan
Bplaning
Cplanning
第7题:
下面关于数组声明和初始化的语句哪个有语法错误()
第8题:
以下哪种初始化数组的方式是错误的?()
第9题:
以下哪种初始化数组的方式是错误的?()
第10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11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12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13题:
importjava.util.*;
publicclassNameList{
privateListnames=newArrayList();
publicsynchronizedvoidadd(Stringname){names.add(name);}
publicsynchronizedvoidprintAll(){
for(inti=0;iSystem.out.print(names.get(i)+);
}
}
publicstaticvoidmain(String[]args){
finalNameListsl=newNameList();
for(inti=0;i<2;i++){
newThread(){
publicvoidruin(){
sl.add(”A”);
sl.add(”B”);
sl.add(”C”);
sl.printAll();
}
}.start();
}
}
}
Whichtwostatementsaretrueifthisclassiscompiledandrun?()
第14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15题:
include<iostream>
include<string>
using namespace std;
class MyClass
{
public:
MyClass(int i=0){member=i;}
void SetMember(const string m){member="big"+m;m=member;}
string GetMember(){return member;}
void print()const{cout<<"Stu:"<<member<<endl;}
private:
string member;
};
void main()
{
MyClass obj1,*obj2;
string i="wang lin";
obj2=&obj1;
obj2->SetMember(i);
obj1.print();
}
第16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17题:
第18题:
Wang Li is()a new marketing campaign at the moment.
第19题:
Which of the following statements are true when creating NETBIOS names? ()
第20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21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22题:
plan
planing
planning
第23题:
Wang:19
Wang:17
Li:20
Li:19
第24题:
Select name as’姓名’from table
Select name as姓名from table where id=1
Sleect name=姓名from table姓名=names(正确答案)
Select names姓名from table