byte
float
String
long
第1题:
下列是定义一个接口ITF的程序,在横线处应填入的选项是 ( )public interface ITF{public static final double PI=3.14,public ______ double area(double a,double b);}
A.interfaee
B.static
C.final
D.abstract
第2题:
阅读以下函数说明和Java代码,将应填入(n)处的字句写在对应栏内。
【说明】
下面的程序先构造Point类,再顺序构造Ball类。由于在类Ball中不能直接存取类Point中的xCoordinate及yCoordinate属性值,Ball中的toString方法调用Point类中的toStrinS方法输出中心点的值。在MovingBsll类的toString方法中,super.toString调用父类Ball的toString方法输出类Ball中声明的属性值。
【Java代码】
//Point.java文件
public class Point{
private double xCoordinate;
private double yCoordinate;
public Point(){}
public Point(double x,double y){
xCoordinate=x;
yCoordinate=y;
}
public String toStrthg(){
return"("+Double.toString(xCoordinate)+","
+Double.toString(yCoordinate)+")";
}
//other methods
}
//Ball.java文件
public class Ball{
private (1);//中心点
private double radius;//半径
private String color;//颜色
public Ball(){}
public Ball(double xValue, double yValue, double r){
//具有中心点及其半径的构造方法
center=(2);//调用类Point中的构造方法
radius=r;
}
public Ball(double xValue, double yValue, double r, String c){
//具有中心点、半径和颜色的构造方法
(3);//调用3个参数的构造方法
color=c;
}
public String toString(){
return "A ball with center"+center.toString()
+",radius "+Double.toString(radius)+",color"+color;
}
//other methods
}
class MovingBall (4) {
private double speed;
public MovingBall(){}
public MoyingBall(double xValue, double yValue, double r, String c, double s){
(5);//调用父类Ball中具有4个参数的构造方法
speed=s;
}
public String toString(){
return super.toString()+",speed"+Double.toString(speed);
}
//other methods
}
public class test{
public static void main(String args[]){
MovingBall mb=new MovingBall(10,20,40,"green",25);
System.out.println(mb);
}
}
第3题:
下列是定义一个接口ITF的程序,在横线处应填入的选项是( )。 publid interface ITF { public static final double PI=3.14; public______double area(double a,double B) ; }
A.interface
B.static
C.final
D.abstract
第4题:
在下列方法的定义中,正确的是 ( )
A.public double x(){..;return false;}
B.public static int x(double y){...}
C.void x(doubled){...;return d}
D.public static x(double a){..}
第5题:
A.public double max(double x,double y)
B.publicintmax(intn,int k)
C.publicintmax(intx,int y, int z)
D.public double max(double n,double k)
第6题:
Which two are valid declarations within an interface definition?()
第7题:
在接口中以下哪条定义是正确的?()
第8题:
下列方法定义中,方法头不正确的是()。
第9题:
public class OuterClass { private double d1 1.0; //insert code here } You need to insert an inner class declaration at line2. Which two inner class declarations are valid?()
第10题:
Public double methoda();
Static void methoda (double d1) {}
Public native double methoda();
Abstract public void methoda();
Protected void methoda (double d1){}
第11题:
public static x(double a)
public static int x(double y)
void x(double d)
第12题:
static class InnerOne { public double methoda() { return d1; } }
static class InnerOne { static double methoda() { return d1; } }
private class InnerOne { public double methoda() { return d1; } }
protected class InnerOne { static double methoda() { return d1; } }
public abstract class InnerOne { public abstract double methoda(); }
第13题:
A.byte
B.float
C.String
D.long
第14题:
阅读以下说明和Java源程序,将应填入(n)处的字句写在答题纸的对应栏内。
说明
以下程序的功能是计算三角形、矩形和正方形的面积并输出。
程序由5个类组成:AreaTest是主类,类Triangle、Rectangle和Square分别表示三角形、矩形和正方形,抽象类Figure提供了一个计算面积的抽象方法。
程序
public class AreaTest{
public static void main(String args[]){
Figure[]figures={
new Triangle(2,3,3),new Rectangle(5,8), new Square(5)
};
for(int i=0;i<figures.1ength;i++){
System.out.println(figures[i]+"area="+figures[i].getArea());
}
}
}
public abstract class Figure{
public abstract double SetAJea();
public class Rectangle extends (1) {
double height;
double width;
public Rectangle(double height,double width){
this.height=height;
this.width=width;
}
public String toString(){
return "Rectangle:height="+height+",width="+width+":";
}
public double getArea() { return (2);
} } public class Square extends (3) {
public Square(double width) {
(4);
}
public String toString() {
return "Square:width="+width+":";
} } public class Triangle extends (5). {
double la;
double lb;
double lc;
public Triangle(double la,double lb,double lc) {
this.la=la; this.lb=lb; this.lc=lc;
public String toString(){
return "Triangle: sides="+la+","+lb+","+lc+":";
public double getArea() {
double s=(la+lb+lc)/2.0;
return Math.sqrt(s*(s-la)*(s-lb)*(s?1c));
}
}
第15题:
下列程序的执行结果为【 】。
include <iostream. h>
class Point
{
public:
Point(double i, double j) { x=i; y=j;}
double Area() const { return 0.0;}
private:
double x, y;
};
class Rectangle: public Point
{
public:
Rectangle(double i, double j, double k, double 1)
double Area() const {return w * h;}
private:
double w, h;
};
Rectangle: :Rectangle(double i, double j, double k. double 1): Point(i,j).
{
w=k, h=1
}
void fun(Point &s)
{
cout<<s. Area()<<end1;
}
void main( )
{
Rectangle rec(3.0, 5.2, 15.0. 25.0);
fun(rec)
}
第16题:
阅读以下说明和c++代码,将应填入(n)处的字句写在对应栏内。
【说明】
现要编写一个画矩形的程序,目前有两个画图程序:DP1和DP2,DP1用函数draw_a_line(x1, y1,x2,y2)画一条直线,DF2则用drawline(x1,x2,y1,y2)画一条直线。当实例画矩形时,确定使用DP1还是DP2。为了适应变化,包括“不同类型的形状”和“不同类型的画图程序”,将抽象部分与实现部分分离,使它们可以独立地变化。这里,“抽象部分”对应“形状”,“实现 部分”对应“画图”,与一般的接口(抽象方法)与具体实现不同。这种应用称为Bridge(桥接)模式。图9-7显示了各个类间的关系。

这样,系统始终只处理3个对象:Shape对象、Drawing对象、DP1或DP2对象。以下是 C++语言实现,能够正确编译通过。
【C++代码】
class DP1{
public:
static void draw_a_line(double x1, double y1,double x2, double y2){
//省略具体实现
}
);
class DP2{
public:
static void drawline(double x1, double x2,double y1, double y2){
//省略具体实现
}
};
class Drawing{
public:
(1) void drawLine(double x1,double y1,double x2,double y2)=0;
};
class V1Drawing:public Drawing{
public:
void drawLine(double x1, double y1,double x2, double y2){
DP1::draw_a_line(x1,y1,x2,y2);
}
};
class V2Drawing:public Drawing{
public:
void drawLine(double x1, double y1, double x2, double y2){
(2);
}
};
class Shape{
private:
(3) _dp;
public:
Shape(Drawing *dp);
virtual void draw()=0;
void drawLine(double x1, double y1, double x2, double y2);
};
Shape::Shape(Drawing *dp)
{
_dp = dp;
}
void Shape::drawLine(double x1, double y1, double x2, double y2)
{ //画一条直线
(4);
}
class Rectangle: public Shape{
private:
double _x1,_y1,_x2,_y2;
public:
Rectangle(Drawing *dp, double x1, double y1,
double x2, double y2);
void draw();
};
Rectangle::Rectangle(Drawing *dp, double x1, double y1, double x2, double y2)
:(5)
{
_x1=x1;_y1=y1;_x2=x2;_y2=y2;
}
void Rectangle::draw()
{
//省略具体实现
}
第17题:
Which will declare a method that is available to all members of the same package and can be referenced without an instance of the class?()
第18题:
Which will declare a method that forces a subclass to implement it? ()
第19题:
以下声明合法的是()
第20题:
在可变参数过程中的参数是()类型,因此可以把任何类型的实参传递给该过程。
第21题:
您已创建一个名为 CalcSalary,将确定 Certkiller.com 员工的薪酬类的责任。CalcSalary 类包括员工的薪酬递增和递减的方法。下面的代码包含在 CalcSalary 类中:()public class CalcSalary {// for promotionspublic static bool IncrementSalary (Employee Emp, double Amount){if (Emp.Status == QuarterlyReview.AboveGoals)Emp.Salary += Amount;return true;
第22题:
Abstract public void methoda();
Public abstract double methoda();
Static void methoda(double d1){}
Public native double methoda() {}
Protected void methoda(double d1) {}
第23题:
abstract public void methoda ();
public abstract double inethoda ();
static void methoda (double dl) {}
public native double methoda () {}
protected void methoda (double dl) {}