C++程序设计与应用基础类和对象习题答案

1、C+程序设计与应用基础第二章类和对象习题答案1、请完成下列填空题1)在c语言中，编程的单位是函数_:在C+语言中，编程的单位是类_O2)在下面程序的横线处填上适当的语句,使类型该程序执行结果为10o#include class MyClasspublic:- MyClass (int a)x=a; /为x置值- int GetNumO return x;/取x值private:int x;void main()Myclass my(10);Coutmy. Get Num () end;3)在类中定义和实现的函数称为_内联函数_。4)对类中的成员函数和属性的访问是通过_ublic_、_rotec

2、ted_和_rivate_这3个关键字来控制的。5) _new运算符对指定类型对象动态分配内存并返回该类型的指针6）假定用户没有给一个名MyClass的类定义构造函数,则系统对其定义的默认构造函数形式为_MyClassO_。2、编程题1）定义一个复数类imaginaryo用友元函数实现该类的加、减、乘和除运算。答案：class Imaginarypublic：Imaginary(float a, float b);private:float x, y;Imaginary:Imaginary(float a, float b)y=b；Imaginary Add(Imaginary& imgl,

3、Imaginary& img2) float a, b;a=img 1. x+img2 x;friendImaginary Add(Imaginary&imgl, Imaginary&img2);friendImaginary Sub(Imaginary&imgl, Imaginary&img2);friendImaginary Mui(Imaginary&imgl, Imaginary&img2);friendImaginary Div(Imaginary&imgl, Imaginary&img2);b=img1. y+img2 y;return Imaginary(a, b):Imagin

4、ary Sub(Imaginaryft imgl, Imaginaryft img2) float a, b;a=img 1. ximg2 x;b=img 1 y-img2 y;return Imaginary(a, b):Imaginary Mui(Imaginaryft imgl, Imaginaryft img2) float a, b;a=img 1. x*img2 x-img 1. y*img2 y;b=img 1. x*img2 y+img 1 y*img2 x;return Imaginary(a, b):Imaginary Div(Imaginary& imgl, Imagin

5、aryft img2) float a, b;辻(img2 x*img2. x+img2. y*img2. y) !=0)a=(img 1. x*img2 x+img 1. y*img2 y)/ (img2 x*img2 x+img2 y*img2 y); b=(img1. y*img2 x-img 1. x*img2 y)/ (img2 x*img2 x+img2 y*img2 y); returnImaginary(a, b):elsereturn Imaginary (0. 0, 0. 0);2)定义类Rationale该类存放分数形式的有理数。要求如下所述。(1)定义带默认参数值的构造

6、函数，默认分数为lo(2)定义私有变量X和Y分别存放分子和分母。同时分子分母要以最简形式存 放。例如:分数3/9应该以1/3形式存放。(3)定义成员函数Add、sub、Mui和Div。计算结果仍然以最简形式存放。(4)以X/Y的形式打印分数。(5)以浮点数形式打印分数。3)定义整数集合intset，该类包括以下函数成员答案：include include class Rationalpublic:Rational(int a=l,int b二1)int c;c=Common_Divisor(a, b);X=a/c;Y二b/c;Rational Add(Rational& r);Rational

7、 Sub(Rational& r);Rational Mui(Rational& r):Rational Div(Rational& r):void Print () cout，X/Y=，X，/，/Yendl;void PrintPoint () if (Y=0) exit (0) : cout，X/Y=，double (X) /Yendl;int Common_Divisor(int a, int b) ; private:int X, Y;；Rational Rational:Add(Rational& r) int a, b, c;a=X*r. Y +Y*r. X;b=Y*r. Y;c=

8、Common_Divisor(a, b);return Rational(a/c, b/c);Rational Rational:Sub(Rational& r) int a, b, c;a=X*r. Y-Y*r. X;b=Y*r. Y;c=Common_Divisor(a, b);return Rational(a/c, b/c);Rational Rational:Mui(Rational& r)int a, b, c;a=X*r X;b=Y*r. Y;c=Common_Divisor(a, b);return Rational(a/c, b/c): Rational Rational:D

9、iv(Rational& r)int a, b, c;a=X*r. Y;b二Y*r. X;c=Common_Divisor(a, b);return Rational(a/c, b/c);int Rational:Common_Divisor(int a, int b)int t, c;if(ab)t=a;ab;b二t;doc=a%b;a=b;b二c;while (c!=0);return a;#include class intSetpublic：intSet ();intSet(int a, int size):void Empty ();bool IsEmpty ():bool IsMe

10、mberOf(int a):bool Add(int a);bool Sub(int a):bool IsEqual(intSet& Set);intSet Intersection(intSet& Set);intSet Merge(intSet& set);void Copy (intSet& Set);void Print ();private:int elementE100:public：int EndPosition;intSet:intSet ()/初始化整数集合元素for (int I=0;I=100)EndPosition=99;elseEndPosition二sizeT;fo

11、r(int I二0;I二EndPosition;I+)elementIj=aI;for(I=size;K100;I+)element L=0;void intSet:Empty()清空整数集合元素for (int I=0;I100;I+)element I =0;EndPosition二 T;bool intSet:IsEmpty() 判断整数集合是否为空if(EndPo s i t i on=-1)return true;elsereturn false;bool intSet:IsMemberOf(int a) for(int 1=0;I=EndPosition;I+)if(element

12、I=a)return true;return false;bool intSet:Add(int a) /能否增加一个元素，需要知道：1.该元素是否已经存在:2.整数集合是否已经 存满辻(IsMemberOf (a)return true;else辻(EndPosition=99)return false;elseEndPosition+;e1ementEndPosition=a;return true;bool intSet:Sub(int a) 能否删除一个元素，需要知道整数集合是否已存在该元素int pos;if (!IsMemberOf(a)return false;for (int

13、1=0;I=EndPosition;I+)if(elementI=a)pos二1;break;for (I=pos; KEndPosition; I+)element匸二element 1+1;EndPosition一一；return true;bool intSet:IsEqual(intSet& set)/判断两个整数集合是否相等bool Equal;if(EndPosition!=set EndPosition)return false;for (int 1=0;I=EndPosition;I+)Equal二false;for (int j=0;j=set EndPosition;j+)

14、if(elementI=set elementj)Equal二true;break;if (!Equal)return false;return true;intSet intSet:Intersection(intSetft set)求两个整数集合的交集int a100,size二0;for (int 1=0;I=EndPosition;I+)for (int j二0;j二set EndPosition;j+)if(elementI=set elementj)asize=elementIZ;size+;break;return intSet(a, size):intSet intSet:Me

15、rge(intSet& set) 求两个整数集合的并集int aFlOOZ,size;for (int 1=0;I=EndPosition;I+)aI二elementI;size二EndPosition+1;for (1=0;I=set EndPosition;I+)if(IsMemberOf(set elementI) continue;else辻(size=99)break;&size二set elementI:size+;return intSet(a,size): void intSet:Copy(intSet& set) for (int 1=0;I=set EndPosition;

16、I+)elementI=set elementI;EndPosition=set EndPosition; void intSet:Print()for (int 1=0;I二EndPosition;I+)cout element ，I，=，element Iendl;4)定义和实现栈类stack,栈的数据数据结果如图所示。该类包括类的常用操 作如下列。答案：#include const int MaxSize=100;class Stackpublic:Stack ();void Push(const int a);int Pop();void Clear ();int Peek() con

17、st;bool IsEmpty() const;bool IsFull () const;private:int ListMaxSizeZ;int Top;Stack: :Stack()/Top表示栈顶，初始化时栈为空;Top =1;void Stack:Push(const int a)if (Top二二MaxSize-1)return;Top卄;ListTop二a;int Stack: :Pop()int element;/如果栈为空,则返回0if (Top=T) return 0;element二ListTop_;Top;return element;void Stack:Clear() Top =1;int Stack:Peek() const /如果栈为空,则返回0if (Top = -1)return 0;return ListTop: bool Stack:IsEmpty() const if (Top = -1)return true;elsereturn false;bool St