A.look over
B.look after
C.look for
D.look forward to
第1题:
A.write, calls
B. wrote, called
C.was writing, calls
D. was writing, called
第2题:
下列程序的输出结果是______。
include<iostream.h>
class base
{
int x,y;
public:
base(int i,int j){x=i;y=j;}
virtual int add( ){return x+y;}
};
class three:public base
{
int z;
public:
three(int i,int j,int k):base(i,j){z=k;)
int add( ){return(base::add( )+z);}
};
void main( )
{
three*q=new three(10,20,30);
cout<<q->add( )<<endl;
}
第3题:
请看一下代码,《插入代码》处应填入的代码是()Map map=new HashMap(); map.put(“tom”,123.6); map.put(“jim”,234.5); map.put(“terry”,45.3); 《插入代码》 其中《插入代码》处要实现的功能是把key为“jim”的value值在原有数字的基础上添加100。
A.map.put(“jim”,map.get(“jim”)+100);
B.map.set(“jim”,map.get(“jim”)+100);
C.map.put(“jim”,234.5);
D.map.set(“jim”,234.5);
第4题:
A.written
B.write
C.wrote
第5题:
第6题:
通过 SQL,在表 Actor 中选择 FirstName 等于 Jim 而 LastName 等于 Carter 的所有记录的表述正确的是()。
A.SELECT * FROM Actor WHERE FirstName LIKE 'Jim' AND LastName LIKE 'Carter'
B.SELECT FirstName='Jim', LastName='Carter' FROM Actor
C.SELECT * FROM Actor WHERE FirstName='Jim' AND LastName='Carter'
D.SELECT * FROM Actor WHERE FirstName='Jim' OR LastName='Carter'