单选题public class Score implements Comparable {  private int wins, losses;  public Score(int w, int 1) { wins = w; losses = 1; }  public int getWins() { return wins; }  public int getLosses() { return losses; }  public String toString() {  return “”; }  // 

题目
单选题
public class Score implements Comparable {  private int wins, losses;  public Score(int w, int 1) { wins = w; losses = 1; }  public int getWins() { return wins; }  public int getLosses() { return losses; }  public String toString() {  return “”; }  // insert code here  }  Which method will complete this class?()
A

 public int compareTo(Object o) {/*mode code here*/}

B

 public int compareTo(Score other) {/*more code here*/}

C

 public int compare(Score s1,Score s2){/*more code here*/}

D

 public int compare(Object o1,Object o2){/*more code here*/}


相似考题
参考答案和解析
正确答案: C
解析: 暂无解析
更多“public class Score implements Comparable {  private int wins”相关问题
  • 第1题:

    下面这个程序的结果是includeclass A{private:int a;public:void seta();int geta(

    下面这个程序的结果是 #include<iostream.h> class A { private: int a; public: void seta();int geta();}; void A∷seta() {a=1;} int A∷geta() {return a;} class B { pnvate: int a; publiC: void sera();int geta();}; void B∷seta() {a = 2;} int B∷geta() {return a;} class C:public A,public B { private: int b; public: void display();}; void C∷display() { int b=geta(); cout<<b;} void main() { C c; c.seta(); c.display();}

    A.1

    B.2

    C.随机输出1或2

    D.程序有错


    正确答案:D
    解析:在类A中有geta()函数,在类B中也有geta()函数,类C继承了类A和类B,这样就产生了二义性,所以程序会出错。

  • 第2题:

    Given:Which method will complete this class?()

    A.public int compareTo(Object o){/*more code here*/}

    B.public int compareTo(Score other){/*more code here*/}

    C.public int compare(Score s1,Score s2){/*more code here*/}

    D.public int compare(Object o1,Object o2){/*more code here*/}


    参考答案:B

  • 第3题:

    有以下程序:include using namespace std;class A{private:int a;public:A (int i){a

    有以下程序: #include <iostream> using namespace std; class A { private: int a; public: A (int i) { a=i; } void disp() { cout<<a<<","; } }; class B { private: int b; public: B(int j) { b=j; } void disp() { cout<<b<<","; } }; class C : public B,public A { private: int c; public: C(int k) :A(k-2),B(k+2) { c=k; } void disp () { A::disp (); B::disp (); cout<<c<<endl; } }; int main () { C obi (10); obj.disp (); return 0; } 程序执行后的输出结果是

    A.10,10,10

    B.10,12,14

    C.8,10,12

    D.8,12,10


    正确答案:D
    解析:本题考核派生类构造函数的使用。本题中,派生类C具有多重继承关系,所以在派生类C的构造函数中应该包含基类A和B的成员初始化列表。

  • 第4题:

    interface A{

    int x = 0;

    }

    class B{

    int x =1;

    }

    class C extends B implements A {

    public void pX(){

    System.out.println(x);

    }

    public static void main(String[] args) {

    new C().pX();

    }

    }


    正确答案:

     

    错误。在编译时会发生错误(错误描述不同的JVM 有不同的信息,意思就是未明确的

    x 调用,两个x 都匹配(就象在同时import java.util 和java.sql 两个包时直接声明Date 一样)。

    对于父类的变量,可以用super.x 来明确,而接口的属性默认隐含为 public static final.所以可

    以通过A.x 来明确。

  • 第5题:

    若有以下程序:includeusing namespace std;class A{private:inta;public:voidseta(in

    若有以下程序: #include <iostream> using namespace std; class A { private: int a; public: void seta(int x) { a=x; } void showa() { cout<<a<<","; } }; class B { private: int b; public: void setb (int x) { b=x; } void showb() { cout<<b<<","; } }; class C :public A,private B { private: int c; public: void setc(int x, inc y, int z) { c=z; seta (x); setb (y); } void showc() { showa (); showb (); cout<<c<<end1; } }; int main () { C c; c. setc(1,2,3); c.showc(); return 0; } 程序执行后的输出结果是

    A.1,2,3

    B.1,1,1

    C.2,2,2

    D.3,3,3


    正确答案:A
    解析:本题考核派生类的应用。本题中类A和类B都是基类。而类C从类A公有派生,从类B处私有派生.所以类C中的函数成员可以访问类A和类B中的公有成员。在类C的函数成员setc中,调用基类A的函数成员seta对A的数据成员a赋值,还调用了基类B的函数成员setb对类B的数据成员b赋值,然后对类C自己的数据成员c赋值。在类C的函数成员showc中,调用基类A的函数成员showa显示数据成员a的值,还调用基类B的函数成员showb显示数据成员b的值,然后输出类C自己的数据成员c的值.在主函数main中,先定义派生类的对象c,然后调用setc对c中的数据成员赋值,然后输出赋值结果。所以程序最后输出应该为:1,2,3。

  • 第6题:

    下列类的定义中,有( )处语法错误。 class Base { public: Base(){} Base(int i) { data=i; } private: int data; }; class Derive : public Base { public: Derive() : Base(O) {} Derive (int x) { d=x; } void setvalue(int i) { data=i; } private: int d; };

    A.1

    B.2

    C.3

    D.4


    正确答案:B
    解析:本题考核派生类的定义和成员的访问权限。第①处错误:在派生类的构造函数Derive(intx)中没有调用基类的构造函数对基类对象初始化。第②处错误:数据data是基类Base的私有成员,派生类Derive不能访问,所以在函数setvalue中对data的赋值是错误的。

  • 第7题:

    1. public class Target {  2. private int i = 0;  3. public int addOne() {  4. return ++i;  5. }  6. }  And:  1. public class Client {  2. public static void main(String[] args) {  3. System.out.println(new Target().addOne());  4. }  5. }  Which change can you make to Target without affecting Client?() 

    • A、 Line 4 of class Target can be changed to return i++;
    • B、 Line 2 of class Target can be changed to private int i = 1;
    • C、 Line 3 of class Target can be changed to private int addOne() {
    • D、 Line 2 of class Target can be changed to private Integer i = 0;

    正确答案:D

  • 第8题:

    public class Score implements Comparable {  private int wins, losses;  public Score(int w, int 1) { wins = w; losses = 1; }  public int getWins() { return wins; }  public int getLosses() { return losses; }  public String toString() {  return “<“ + wins + “,“ + losses + “>”; }  // insert code here  }  Which method will complete this class?() 

    • A、 public int compareTo(Object o) {/*mode code here*/}
    • B、 public int compareTo(Score other) {/*more code here*/}
    • C、 public int compare(Score s1,Score s2){/*more code here*/}
    • D、 public int compare(Object o1,Object o2){/*more code here*/}

    正确答案:B

  • 第9题:

    Which two demonstrate an “is a” relationship?()   

    • A、 public interface Person { }  public class Employee extends Person { }
    • B、 public interface Shape { }  public class Employee extends Shape { }
    • C、 public interface Color { }  public class Employee extends Color { }
    • D、 public class Species { }  public class Animal (private Species species;)
    • E、 interface Component { }  Class Container implements Component ( Private Component[ ] children;  )

    正确答案:D,E

  • 第10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,E
    解析: 暂无解析

  • 第12题:

    多选题
    Which the two demonstrate an “is a” relationship?()
    A

    public interface Person {}  Public class Employee extends Person {}

    B

    public interface Shape {}  public interface Rectangle extends Shape {}

    C

    public interface Color {}  public class Shape { private Color color; }

    D

    public class Species {}  public class Animal { private Species species; }

    E

    interface Component {} Class Container implements Component {private Component [] children;


    正确答案: B,A
    解析: 暂无解析

  • 第13题:

    若有以下程序 include using namespace std; class A {private:int a; public: A(in

    若有以下程序 #include <iostream> using namespace std; class A { private: int a; public: A(int i) { a=i; } void disp() { cout<<a<<","; } }; class B { private: int b; public: B(int j) { b=j; } void disp() { cout<<b<<","; } }; class C: public B,public A { private: int c; public: C(int k):A(k-2),B(k+2) { c=k; } void disp() { A::disp(); B::disp(); cout<<c<<end1; } }; int main() { C obj(10); obj.disp(); return 0; } 程序执行后的输出结果是

    A.10,10,10

    B.10,12,14

    C.8,10,12

    D.8,12,10


    正确答案:D
    解析:本题考核派生类构造函数的使用。本题中,派生C具有多重继承关系,所以在派生类C的构造函数中应该包含基类A和B的成员初始化列表。

  • 第14题:

    下面这个程序的结果是 include class A { private: int a; public: void seta( ) ;

    下面这个程序的结果是

    #include<iostream.h>

    class A

    {

    private:

    int a;

    public:

    void seta( ) ;int geta( ) ;};

    void A: :seta( )

    { a=1;}

    int A: :geta( )

    { retum a;}

    class B

    { private:

    int a;

    public:

    void seta( ) ;int geta( ) ;};

    void B: :seta( )

    {a=2;}

    int B: :geta( )

    {return a;}

    class C: public A,public B

    { private:

    int b;

    public:

    void display( ) ;};

    void C: :display( )

    { int b=geta( ) ;

    cout < < b;}

    void main( )

    { C c;

    c. seta( ) ;

    c. display( ) ;}

    A.1

    B.2

    C.随机输出1或2

    D.程序有错


    正确答案:D
    解析:在类A中有geta( ) 函数,在类B中也有geta( ) 函数,类C继承了类A和类B,这样就产生了二义性,所以程序会出错。

  • 第15题:

    若有以下程序:include using namespace std;class A{private:int a;public:A(im i){a

    若有以下程序: #include <iostream> using namespace std; class A { private: int a; public: A(im i) { a=i; } void disp() cout<<a<<","; } }; class B { private: int b; public: B(int j) { b=j; } void disp0 { cout<<b<<","; } }; class C: public B,public A { private: int c; public: C(int k):A(k-2),B(k+2) { c=k; } void disp0 { A::disp(); B::disp(); cout<<c<<endl; } }; int main() { C obj(10); obj.disp(); return 0; } 程序执行后的输出结果是( )。

    A.10,10,10

    B.10,12,14

    C.8,10,12

    D.8,12,10


    正确答案:D

  • 第16题:

    若有以下程序:includeusing namespace std;class A{private:int a; public:void seta

    若有以下程序: #include<iostream> using namespace std; class A { private: int a; public: void seta(int x) { a=x; } void showa() { cout<<a<<","; } }; class B { private: int b; public: void setb(int x) { b=x; } void showb() { cout<<b<<",”; } }; class C:pUblic A,private B { private: int c; public: void setc(int x,int y,int z) { c=z; seta(x); setb(y); } void showc() { showa(); showb(); cout<<c<<end1; } }; int main() { Cc; c.setc(1,2,3); c.showc(); retrun 0; } 程序执行后的输出结果是

    A.1,2,3

    B.1,1,1

    C.2,2,2

    D.3,3,3


    正确答案:A
    解析:本题考核派生类的应用。本题中类A和类B都是基类。而类C从类A公有派生,从类B处私有派生。所以类C中的函数成员可以访问类A和类B中的公有成员。在类C的函数成员setc中,调用基类A的函数成员seta对A的数据成员a赋值,还调用了基类B的函数成员setb对类B的数据成员b赋值,然后对类C自己的数据成员c赋值。在类C的函数成员showc中,调用基类A的函数成员showa显示数据成员a的值,还调用基类B的函数成员showb显示数据成员b的值,然后输出类C自己的数据成员c的值。在主函数main中,先定义派生类的对象c,然后调用setc对c中的数据成员赋值,然后输出赋值结果。所以程序最后输出应该为:1,2,3。

  • 第17题:

    若有以下程序:include using namespace std;class A{private:int a;public:void seta

    若有以下程序:#include <iostream>using namespace std;class A{private: int a;public: void seta(int x) { a=x; } void showa() { cout<<a<<","; }};class B{private: int b;public: void setb(int x) { b=x; } void showb() { cout<<b<<","; }};class C: public A, private B{private: int c;public: void setc(int x, int y, int z) { c=z; seta(x); setb(y); } void showc() { showa(); showb(); cout<<c<<end1; }};int main(){ C c; c.setc(1,2,3); c.showc(); return 0;}程序执行后的输出结果是( )。

    A.1,2,3

    B.1,1,1

    C.2,2,2

    D.3,3,3


    正确答案:A

  • 第18题:

    若有以下程序:include using namespace std;class A{private: int a;public: A(int

    若有以下程序: #include <iost ream> using namespace std; class A { private: int a; public: A(int i) { a=i; } void disp () { cout<<a<<", "; } }; class B { private:

    A.10,10,10

    B.10,12,14

    C.8,10,12

    D.8,12,10


    正确答案:D
    解析:本题考核派生类构造函数的使用。本题中,派生C具有多重继承关系,所以在派生类C的构造函数中应该包含基类A和B的成员初始化列表。而且派生类C中的成员函数disp分别调用了基类A、B的公有成员函数disp()用于输出基类中私有数据变量初始化后的值。所以由类C构造函数的初始化形式可知,程序最后输出为8,12,10,即D选项。

  • 第19题:

    Which three demonstrate an “is a” relationship?() 

    • A、 public class X {  }     public class Y extends X { }
    • B、 public interface Shape { }     public interface Rectangle extends Shape{ }
    • C、 public interface Color { }     public class Shape { private Color color; }
    • D、 public interface Species { }     public class Animal { private Species species; }
    • E、 public class Person { }    public class Employee {      public Employee(Person person) { }
    • F、 interface Component { }     class Container implements Component {   private Component[] children; }

    正确答案:A,B,F

  • 第20题:

    Which the two demonstrate an “is a” relationship?()

    • A、 public interface Person {}  Public class Employee extends Person {}
    • B、 public interface Shape {}  public interface Rectangle extends Shape {}
    • C、 public interface Color {}  public class Shape { private Color color; }
    • D、 public class Species {}  public class Animal { private Species species; }
    • E、 interface Component {} Class Container implements Component {private Component [] children;

    正确答案:B,E

  • 第21题:

    public class Parent {     int change() {…}     }  class Child extends Parent {     }  Which methods can be added into class Child?()    

    • A、 public int change(){}
    • B、 int chang(int i){}
    • C、 private int change(){}
    • D、 abstract int chang(){}

    正确答案:A,B

  • 第22题:

    多选题
    Which three demonstrate an “is a” relationship?()
    A

    public class X {  }     public class Y extends X { }

    B

    public interface Shape { }     public interface Rectangle extends Shape{ }

    C

    public interface Color { }     public class Shape { private Color color; }

    D

    public interface Species { }     public class Animal { private Species species; }

    E

    public class Person { }    public class Employee {      public Employee(Person person) { }

    F

    interface Component { }     class Container implements Component {   private Component[] children; }


    正确答案: A,F
    解析: 暂无解析

  • 第23题:

    多选题
    Which two demonstrate an “is a” relationship?()
    A

    public interface Person { }  public class Employee extends Person { }

    B

    public interface Shape { }  public class Employee extends Shape { }

    C

    public interface Color { }  public class Employee extends Color { }

    D

    public class Species { }  public class Animal (private Species species;)

    E

    interface Component { }  Class Container implements Component ( Private Component[ ] children;  )


    正确答案: D,B
    解析: 暂无解析

  • 第24题:

    单选题
    public class Score implements Comparable {  private int wins, losses;  public Score(int w, int 1) { wins = w; losses = 1; }  public int getWins() { return wins; }  public int getLosses() { return losses; }  public String toString() {  return “”; }  // insert code here  }  Which method will complete this class?()
    A

     public int compareTo(Object o) {/*mode code here*/}

    B

     public int compareTo(Score other) {/*more code here*/}

    C

     public int compare(Score s1,Score s2){/*more code here*/}

    D

     public int compare(Object o1,Object o2){/*more code here*/}


    正确答案: B
    解析: 暂无解析