单选题public class Mycircle {   public double radius;   public double diameter;   public void setRadius(double radius)   this.radius = radius;   this.diameter= radius * 2;   }   public double getRadius() {  return radius;   }   Which statement is true?()AThe

题目
单选题
public class Mycircle {   public double radius;   public double diameter;   public void setRadius(double radius)   this.radius = radius;   this.diameter= radius * 2;   }   public double getRadius() {  return radius;   }   Which statement is true?()
A

 The Mycircle class is fully encapsulated.

B

 The diameter of a given MyCircle is guaranteed to be twice its radius.

C

 Lines 6 and 7 should be in a synchronized block to ensure encapsulation.

D

 The radius of a MyCircle object can be set without affecting its diameter.


相似考题

1.阅读下列Java程序和程序说明,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toString方法输出中心点的值。在MovingBall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。public class Point{private double xCoordinate;private double yCoordinate;public Point 0 }public Point(ouble x, double y){xCoordinate = x;yCoordinate = y;}public String toString(){return "( + Double.toString(Coordinate)+ ","+ Double.toString(Coordinate) + ");}//other methods}public class Ball{(1); //中心点private double radius; //半径private String colour; ///颜色public Ball() { }public Ball(double xValue, double yValue, double r)// 具有中心点及半径的构造方法{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、半径及颜色的构造方法{(3);//调用3个参数的构造方法colour = c;}public String toString(){return "A ball with center" + center, toString() + ", radius"+ Double.toString(radius) + ", colour" + colour;}//other methods}public class MovingBall. (4){private double speed;public MovingBall() { }public MovingBall(double xValue, double yValue, double r, String e, 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 Tester{public static void main(String args[]){MovingBall mb = new MovingBall(10,20,40,"green",25);System.out.println(mb);}}

2.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);}}

更多“单选题public class Mycircle {   public double radius;   public double diameter;   public void setRadius(double radius)   this.radius = radius;   this.diameter= radius * 2;   }   public double getRadius() {  return radius;   }   Which statement is true?()A  T”相关问题
  • 第1题:

    阅读以下说明和JAVA 2代码,填入(n)处的。

    [说明]

    以下JAVA程序实现了在接口interface iShape2D的定义和应用,仔细阅读代码和相关注释,将程序补充完整。

    [代码6-1]

    interface iShape2D //定义接口

    {

    (1)

    (2)

    }

    (3)//实现CRectangle类

    {

    int width, height;

    (4) CRectangle (int w,int h) {

    width=w;

    height=h;

    }

    public void area ( ){ //定义area( )的处理方式

    System. out.println ("area="+width*height);

    }

    }

    (5)//实现CCircle类

    {

    double radius;

    (6) CCircle (double r) {

    radius=r;

    }

    public void area ( ) { //定义area( )的处理方式

    System.out.println ("area="+pi*radius*radius);

    }

    }

    [代码6-2]

    public class app10_4

    {

    public static void main(String args[])

    {

    CRectangle rect=new CRectangle (5,10);

    rect.area ( ); //调用CRectangle类里的area ( ) method

    CCircle cir=new CCircle (2.0);

    cir.area ( ); //调用CCircl类里的area ( ) method

    }

    }


    正确答案:(1)final double pi=3.14; (2)abstract void area(); (3)class CRectangle implements iShape2D (4)public (5)class CCircle implements iShape2D (6)public
    (1)final double pi=3.14; (2)abstract void area(); (3)class CRectangle implements iShape2D (4)public (5)class CCircle implements iShape2D (6)public 解析:本题JAVA程序实现了接口interface iShape2D的定义和应用。(1)和(2)定义pi和面积函数area(),可从下文得到,它们位置可以互换。(3)定义Crectangle,继承iShape2D。(4)应该为public关键字。(5)定义CCircle,继承iShape2D。(6)应该为public关键字。

  • 第2题:

    下列程序中,先声明一个圆类circle和一个桌子类table,另外声明一个圆桌类roundtable,它是由 circle和table两个类派生的,要求声明一个圆桌类对象,并输出圆桌的高度,面积和颜色。请填空完成程序

    include<iostream.h>

    include<string.h>

    class circle{

    double radius;

    public:

    circle(double r){radius=r;}

    double get_area(){return 3.416*radius*radius;}

    };

    class table{

    double height;

    public:

    table(double h)<height=h;}

    double get_height(){return height;}

    };

    class roundtable:public table,public circle{

    char *color;

    public:

    roundtable(double h,double r,char c[]): 【 】 {

    color=new char[strlen(c) +1];

    【 】;

    };

    char*get_color(){return color;}

    }:

    void main(){

    roundtable rt(0.8,1.0,“白色”);

    cout<<"圆桌的高:"<<rt. get_height()<<end1;

    cout<<"圆桌面积:"<<rt.get_area()<<end1;

    cout<<"圆桌颜色:"<<n.get color()<<end1;

    }


    正确答案:circle(r) table(h) strcpy(color c)
    circle(r), table(h) strcpy(color, c)

  • 第3题:

    本题的功能是用文本框来设定表盘中指针的位置。窗口中有一个画板和两个文本框,画板中绘制了一个表盘和时针、分针,通过文本框分别设定“时”和“分”,表盘中的时针和分针就会指到对应的位置上。

    import java.awt.*;

    import java.awt.event*;

    import java.awt.geom.*;

    import javax.swing.*;

    import javax.swing.event.*;

    public class java3

    {

    public static void main(String[]args)

    {

    TextTestFrame. frame=new TextTestFrame():

    frame.setDefauhCloseOperation(JFrame.EXIT_

    0N_CLOSE);

    frame.show();

    }

    }

    class TextTestFrame. extends JFrame

    {

    public TextTestFrame()

    {

    setTitle("java3"):

    setSize(DEFAULT_WIDTH,DEFAULT_

    HElGHT);

    Container contentPane=getContentPane();

    DocumentListener listener=new DoeumentListen-

    er();

    JPanel panel=new JPanel();

    hourField=new JTextField("12",3);

    panel.add(hourField);

    hourField.getDocument().addDocumentListener

    (this);

    minuteField=new JTextField("00",3):

    panel.add(minuteField);

    minuteField.getDocument().addDocumentListener

    (listener);

    contentPane.add(panel,BorderLayout.S()UTH);

    clock=new ClockPanel();

    contentPane.add(clock,BorderLayout.CEN-

    TER);

    }

    public void setClock()

    {

    try

    {

    int hours

    =Integer.parseInt(hourField.getText().trim

    ()):

    int minutes

    =Integer.parseInt(minuteField.getText().trim

    ());

    clock.setTime(hours,minutes);

    }

    catch(NumberFormatExcepfion e){}

    }

    public static final int DEFAULT_WIDTH=300;

    public static final int DEFAULT_HEIGHT

    =300;

    private J TextField hourField;

    private JTextField minuteField;

    private ClockPanel clock;

    private class clockFieldListener extends Docu-

    mentListener

    {

    public void insertUpdate(DocumentEvent e){ set-

    Clock();}

    public void removeUpdate(DocumentEvent e){

    setClock();}

    public void changedUpdate(DocumentEvent e){}

    }

    }

    class ClockPanel extends JPanel

    {

    pubhc void paintComponent(Graphics g)

    {

    super.paintComponent(g);

    Graphies2D g2=(Graphics2D)g;

    Ellipse2D circle

    =new Ellipse2D.Double(0,0,2* RADIUS,2

    *RADIUS);

    g2.draw(circle);

    double hourAngle

    =Math.toRadians(90-360*minutes/(12

    *60));

    drawHand(92,hourAngle,HOUR_HAND_

    LENGTH);

    double minuteAngle

    =Math.toRadians(90-360*minutes/60);

    drawHand(g2,minuteAngle,MINUTE_HAND_

    LENGTH):

    }

    punic void drawHand(Graphics2D g2,

    double angle,double handLength)

    {

    Point2D end=new Point2D.Double(

    RADIUS+handLength*Math.cos(angle),

    RADIUS-handLength*Math.sin(angle));

    Point2D center=new Point2D.Double(RADIUS,

    RADIUS):

    g2.draw(new Line2D.Double(center,end));

    }

    public void setTime(int h,int m)

    {

    minutes=h*60+m;

    repaint();

    }

    private double minutes=0;

    private double RADIUS=100;

    private double MINUTE_HAND_LENGTH=

    0.8*RADIUS;

    private double HOUR_HAND_LENGTH=0.6

    *RADIUS:

    }


    正确答案:
    第1处:DocumentListenerlistener=newClockField-Listener()第2处:hourField.getDocument().addDocumentLis-tener(listener)第3处:privateclassClockFieldListenerimplementsDocu-mentListener【解析】第1处从后面程序可以看出ClockFieldListener类扩展了DocumentListener,此处应使用继承后的子类;第2处注册窗体的监听器,参数应为事件源。第3处实现的是接口,应使用implements。

  • 第4题:

    Which will declare a method that is available to all members of the same package and be referenced without an instance of the class?()

    • A、 abstract public void methoda ();
    • B、 public abstract double inethoda ();
    • C、 static void methoda (double dl) {}
    • D、 public native double methoda () {}
    • E、 protected void methoda (double dl) {}

    正确答案:C

  • 第5题:

    class super {   public float getNum() {return 3.0f;}   }   public class Sub extends Super {   }   Which method, placed at line 6, will cause a compiler error?()

    • A、 Public float getNum() {return 4.0f; }
    • B、 Public void getNum (){}
    • C、 Public void getNum (double d){}
    • D、 Public double getNum (float d) {retrun 4.0f; }

    正确答案:B

  • 第6题:

    public class Mycircle {  public double radius;  public double diameter;  public void setRadius(double radius)  this.radius = radius;  this.diameter= radius * 2;  }  public double getRadius()   {  return radius;  }  Which statement is true?()  

    • A、 The Mycircle class is fully encapsulated.
    • B、 The diameter of a given MyCircle is guaranteed to be twice its radius.
    • C、 Lines 6 and 7 should be in a synchronized block to ensure encapsulation.
    • D、 The radius of a MyCircle object can be set without affecting its diameter.

    正确答案:B

  • 第7题:

    1. class super {  2. public float getNum() {return 3.0f;}  3. }  4.    5. public class Sub extends Super { 6.   7. }   Which method, placed at line 6, will cause a compiler error?()  

    • A、  Public float getNum()   {return 4.0f; }
    • B、  Public void getNum ()  { }
    • C、  Public void getNum (double d)   { }
    • D、  Public double getNum (float d) {retrun 4.0f; }

    正确答案:B

  • 第8题:

    public abstract class Shape {  private int x;  private int y;  public abstract void draw();  public void setAnchor(int x, int y) {  this.x = x;  this.y = y;  }  }  Which two classes use the Shape class correctly?()

    • A、 public class Circle implements Shape { private int radius; }
    • B、 public abstract class Circle extends Shape { private int radius; }
    • C、 public class Circle extends Shape { private int radius; public void draw(); }
    • D、 public abstract class Circle implements Shape { private int radius; public void draw(); }
    • E、 public class Circle extends Shape { private int radius;public void draw() {/* code here */} }
    • F、 public abstract class Circle implements Shape { private int radius;public void draw() { / code here */ } }

    正确答案:B,E

  • 第9题:

    单选题
    class super {   public float getNum() {return 3.0f;}   }   public class Sub extends Super {   }   Which method, placed at line 6, will cause a compiler error?()
    A

     Public float getNum() {return 4.0f; }

    B

     Public void getNum (){}

    C

     Public void getNum (double d){}

    D

     Public double getNum (float d) {retrun 4.0f; }


    正确答案: C
    解析: 暂无解析

  • 第10题:

    单选题
    public class Mycircle {  public double radius;  public double diameter;  public void setRadius(double radius)  this.radius = radius;  this.diameter= radius * 2;  }  public double getRadius()   {  return radius;  }  Which statement is true?()
    A

     The Mycircle class is fully encapsulated.

    B

     The diameter of a given MyCircle is guaranteed to be twice its radius.

    C

     Lines 6 and 7 should be in a synchronized block to ensure encapsulation.

    D

     The radius of a MyCircle object can be set without affecting its diameter.


    正确答案: B
    解析: 暂无解析

  • 第11题:

    多选题
    public abstract class Shape {  private int x;  private int y;  public abstract void draw();  public void setAnchor(int x, int y) {  this.x = x;  this.y = y;  }  }  Which two classes use the Shape class correctly?()
    A

    public class Circle implements Shape { private int radius; }

    B

    public abstract class Circle extends Shape { private int radius; }

    C

    public class Circle extends Shape { private int radius; public void draw(); }

    D

    public abstract class Circle implements Shape { private int radius; public void draw(); }

    E

    public class Circle extends Shape { private int radius;public void draw() {/* code here */} }

    F

    public abstract class Circle implements Shape { private int radius;public void draw() { / code here */ } }


    正确答案: B,F
    解析: 暂无解析

  • 第12题:

    单选题
    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?()
    A

     Abstract public void methoda();

    B

     Public abstract double methoda();

    C

     Static void methoda(double d1){}

    D

     Public native double methoda(){}

    E

     Protected void methoda(double d1){}


    正确答案: C
    解析: 暂无解析

  • 第13题:

    阅读以下说明和C++程序,将应填入(n)处的字句写在对应栏内。

    【说明】

    以下程序的功能是计算正方体、球体和圆柱体的表面积和体积并输出。

    程序由4个类组成:类cube、sphere和cylinder分别表示正方体、球体和圆柱体;抽象类 container为抽象类,提供了两个纯虚拟函数surface_area()和volum(),作为通用接口。

    【C++程序】

    include<iostream.h>

    define pi 3.1416

    class container{

    protected:

    double radius;

    public:

    container(double radius) {container::radius=radius;}

    virtual double surface_area()=0;

    virtual double velum()=0;

    };

    class cube:(1){ //定义正方体类

    public:

    cube(double radius):container(radius){};

    double surface_area () {return 6 * radius * radius;}

    double volum() {return radius * radius * radius;}

    };

    class sphere:(2){ //定义球体类

    public:

    sphere(double radius): container(radius){};

    double surface_area() { return (3);}

    double volum() {return pi * radius * radius * radius * 4/3;}

    };

    class cylinder:(4){ //定义圆柱体类

    double height;

    public:

    cylinder(double radius,double height):container(radius)

    {

    container::height=height;

    }

    double surface_are a () { return 2 * pi * radius * (height+radius); }

    double volum () {return (5);}

    };

    void main()

    {

    container * p;

    cube obj1 (5);

    sphere obj2(5);

    cylinder obj3(5,5);

    p=&obj1;

    cout<<“正方体表面积”(<<p->surface_area()<<end1;

    cont<<“正方体体积”<<p->volume()<<end1;

    p=&obj2;

    cout<<“球体表面积”<<p->surface_area()<<end1;

    cout<<“球体体积”<<p->volume()<<end1;

    p=&obj3;

    cout<<“球体表面积”<<p->surface_area()<<end1;

    cout<<“球体体积”<<p->volume()<<end1;

    }


    正确答案:(1)public container (2)public container (3) 4 * pi * radius * radius (4) public container (5)pi * radius * radius * height
    (1)public container (2)public container (3) 4 * pi * radius * radius (4) public container (5)pi * radius * radius * height 解析:类cube、sphere和cylinder分别表示正方体、球体和圆柱体,它们都需要求各自的表面积和体积,而抽象类container提供纯虚拟函数 surface_area()和velum(),所以类cube、sphere和cylinder都以类contain为基类,公有继承,所以(1)、(2)和(4)空应填入“public container”。
    (3)空处为类sphere中求表面积函数的返回值,所以根据球体表面积公式应填入“4*pi*radius*radius”。
    (5)空处为类cylinder中求圆柱体体积函数的返回值,所以根据圆柱体体积公式应填入“pi*radius*radius*height”。

  • 第14题:

    阅读以下说明和JAVA 2代码,将应填入(n)处的字句写在对应栏内。

    [说明]

    以下程序为类类型的变量应用实例,通过异常处理检验了类CCircle的变量的合法性,即参数半径应为非负值。仔细阅读代码和相关注释,将程序补充完整。

    [JAVA代码]

    //定义自己的异常类

    class CCircleException extends Exception

    {

    }

    // 定义类 CCircle

    class CCircle

    {

    private double radius;

    public void setRadius ( double r ) (1)

    {

    if ( r<0 ) {

    (2)

    }

    else

    (3)

    }

    Public void show ( ) {

    System. out. println ( "area="+3.14*radius*radius );

    }

    }

    public class ciusample

    {

    public static void main ( String args[] )

    {

    CCircle cir=new CCircle( );

    (4) {

    cir. setRadius ( -2.0 )

    }

    (5)

    {

    System. out. println ( e+" throwed" ) ;

    }

    cir. show( ) ;

    }

    }


    正确答案:(1)throws CCircleException (2)throw new CCircleException(); //抛出异常 (3)radius=r; (4)try (5)catch(CCircleException e) //捕捉由setRadius()抛出的异常
    (1)throws CCircleException (2)throw new CCircleException(); //抛出异常 (3)radius=r; (4)try (5)catch(CCircleException e) //捕捉由setRadius()抛出的异常 解析:本题主要考查JAVA语言中Class类型的变量应用。本段代码中对于类Ccircle的半径变量进行合法性检验,如果圆Ccircle的半径为负值,则抛出异常处理。

  • 第15题:

    阅读以下说明和Java程序,填写程序中的空(1)~(6),将解答写入答题纸的对应栏内。
    【说明】
    以下Java代码实现一个简单绘图工具,绘制不同形状以及不同颜色的图形。部分接口、类及其关系如图5-1所示。




    【Java代码】
    interface?DrawCircle?{? //绘制圆形 public(1) ;}class?RedCircle?implements?DrawCircle?{? ?//绘制红色圆形???????public?void?drawCircle(int?radius,intx,?int?y)??{????????????System.out.println("Drawing?Circle[red,radius:"?+?radius?+",x:"?+?x?+?",y:"?+y+?"]");???????}}class?GreenCircle?implements?DrawCircle?{????//绘制绿色圆形??????public?void?drawCircle(int?radius,?int?x,int?y)?{???????????System.out.println("Drawing?Circle[green,radius:"?+radius+",x:?"?+x+?",y:?"?+y+?"]");??????}}abstract?class?Shape?{????//形状? protected? ? (2)???;? ? public?Shape(DrawCircle?drawCircle)?{? ?this.drawCircle=?drawCircle;? ? ? public?abstract?void?draw();}class?Circle?extends?Shape?{? //圆形? ?private?int?x,y,radius;? public?Circle(int?x,int?y,intradius,DrawCircle?drawCircle)?{? ?(3)???;? this.x?=?x;? ? ? this.y?=?y;? ?this.radius?=radius;? }? ? ?public?void?draw()?{? ? drawCircle.? ?(4)? ?;? ? ? }}public?class?DrawCircleMain?{? public?static?void?main(String[]?args)?{? Shape?redCircle=new?Circle(?100,100,10,? (5) );//绘制红色圆形? Shape?greenCircle=new?Circle(200,200,10,(6) );//绘制绿色圆形? ?redCircle.draw(); greenCircle.draw();? ?}}


    答案:
    解析:
    (1)void drawCircle (int radius,int x,int y)
    (2)DrawCircle drawCircle
    (3)super.drawcircle=drawcircle
    (4)drawCircle(radius,x,y)
    (5)new RedCircle()
    (6)new GreenCircle()【解析】
    第一空是填接口里面的方法,在接口的实现里面找,可以发现应该填void drawCircle (int radius,int x,int y)。
    第二空可以根据后面this drawCircle=drawCircle判断,这里应该有一个drawCircle属性,因此应该填)DrawCircle drawCircle。
    第三空这里用super,用super. drawcircle来引用父类的成员。
    第四空调用drawCircle(radius,x,y)方法。
    第五、六空分别创建一个红色圆形对象和一个绿色圆形对象作为Circle里面的实参。

  • 第16题:

    1. class Super {  2. public float getNum() { return 3.0f; }  3. }  4.   5. public class Sub extends Super {  6.   7. }  Which method, placed at line6, causes compilation to fail?()  

    • A、 public void getNum(){}
    • B、 public void getNum(double d){}
    • C、 public float getNum() { return 4.0f; }
    • D、 public double getNum(float d) { return 4.0d; }

    正确答案:A

  • 第17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?()

    • A、 static class InnerOne {  public double methoda() {return d1;}  }
    • B、 static class InnerOne {  static double methoda() {return d1;}  }
    • C、 private class InnerOne {  public double methoda() {return d1;}  }
    • D、 protected class InnerOne {  static double methoda() {return d1;}  }
    • E、 public abstract class InnerOne {  public abstract double methoda();  }

    正确答案:C,E

  • 第18题:

    1. public class OuterClass {  2. private double d1 = 1.0;  3. // insert code here  4. }  Which two are valid if inserted at line 3?()  

    • A、 static class InnerOne { public double methoda() { return d1; } }
    • B、 static class InnerOne { static double methoda() { return d1; } }
    • C、 private class InnerOne { public double methoda() { return d1; } }
    • D、 protected class InnerOne { static double methoda() { return d1; } }
    • E、 public abstract class InnerOne { public abstract double methoda(); }

    正确答案:C,E

  • 第19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?()  

    • A、 Abstract public void methoda();
    • B、 Public abstract double methoda();
    • C、 Static void methoda(double d1){}
    • D、 Public native double methoda()  {}
    • E、 Protected void methoda(double d1)  {}

    正确答案:C

  • 第20题:

    多选题
    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?()
    A

    static class InnerOne {  public double methoda() {return d1;}  }

    B

    static class InnerOne {  static double methoda() {return d1;} }

    C

    private class InnerOne {  public double methoda() {return d1;} }

    D

    protected class InnerOne {  static double methoda() {return d1;} }

    E

    public abstract class InnerOne {  public abstract double methoda();  }


    正确答案: D,C
    解析: 暂无解析

  • 第21题:

    单选题
    1. class Super {  2. public float getNum() { return 3.0f; }  3. }  4.   5. public class Sub extends Super {  6.   7. }  Which method, placed at line6, causes compilation to fail?()
    A

     public void getNum(){}

    B

     public void getNum(double d){}

    C

     public float getNum() { return 4.0f; }

    D

     public double getNum(float d) { return 4.0d; }


    正确答案: A
    解析: 暂无解析

  • 第22题:

    多选题
    1. public class OuterClass {  2. private double d1 = 1.0;  3. // insert code here  4. }  Which two are valid if inserted at line 3?()
    A

    static class InnerOne { public double methoda() { return d1; } }

    B

    static class InnerOne { static double methoda() { return d1; } }

    C

    private class InnerOne { public double methoda() { return d1; } }

    D

    protected class InnerOne { static double methoda() { return d1; } }

    E

    public abstract class InnerOne { public abstract double methoda(); }


    正确答案: C,A
    解析: 暂无解析

  • 第23题:

    单选题
    Which will declare a method that is available to all members of the same package and be referenced without an instance of the class?()
    A

     abstract public void methoda ();

    B

     public abstract double inethoda ();

    C

     static void methoda (double dl) {}

    D

     public native double methoda () {}

    E

     protected void methoda (double dl) {}


    正确答案: A
    解析: 暂无解析

  • 第24题:

    单选题
    1. class super {  2. public float getNum() {return 3.0f;}  3. }  4.    5. public class Sub extends Super { 6.   7. }   Which method, placed at line 6, will cause a compiler error?()
    A

      Public float getNum()   {return 4.0f; }

    B

      Public void getNum ()  { }

    C

      Public void getNum (double d)   { }

    D

      Public double getNum (float d) {retrun 4.0f; }


    正确答案: C
    解析: 暂无解析