c++运算符重载习题

1、Task8-2/*1. 定义一个复数类 Complex ，重载运算符“ + ” ，使之能用于复数的加法运算。将运算符函数重载为非成员、非友元的普通函数。编写程序，求两个复数之和*/#include<iostream>using namespace std;class Complexpublic:Complex()real=0;imag=0;Complex(double r,double i)real=r;imag=i;void display();double real;double imag;void Complex:display()cout<<"(&qu

2、ot;<<real<<","<<imag<<"i)"-可编辑修改-Complex operator +(Complex &c1,Complex &c2) Complex p;p.real=c1.real+c2.real;p.imag=c1.imag+c2.imag;return p;int main()Complex c1(3,5),c2(2,5),c3;c1.display();cout<<"+"c2.display();cout<<"

3、="c3=c1+c2;c3.display();3 陈习章归便阳O,5i>+<2rGO=Ct,10L>Pr-ocess r«tujT-ried 04。冲sxscution time = 6.461 s Press an if key to continue.-可编辑修改-Task8-2/*2. 定义一个复数类 Complex ，重载运算符“ + ”、” - “、 ” *”、” /”，使之能用于复数的加、减、乘、除。运算符重载函数作为 Complex 类的成员函数，编程，分别求两个复数之和差积商。*/#include<iostream>using

4、 namespace std;class Complexpublic:Complex()real=0;imag=0;Complex(double r,double i)real=r;imag=i;Complex operator+(Complex &c2);Complex operator-(Complex &c2);Complex operator*(Complex &c2);Complex operator/(Complex &c2);void display();private:double real;double imag;Complex Complex

5、:operator +(Complex &c2) Complex c;-可编辑修改-c.real=real+c2.real;c.imag=imag+c2.imag;return c;Complex Complex:operator -(Complex &c2)Complex c;c.real=real-c2.real;c.imag=imag-c2.imag;return c;Complex Complex:operator *(Complex &c2)Complex c;c.real=real*c2.real;c.imag=imag*c2.imag;return c;C

6、omplex Complex:operator /(Complex &c2)Complex c;c.real=(real*c2.real+imag*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag);c.imag=(imag*c2.real-real*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag);return c;-可编辑修改-void Complex:display()cout<<"("<<real<<","<<imag&l

7、t;<"i)"<<endl;int main()Complex c1(3,4),c2(5,-10),c3;c3=c1+c2;cout<<"c1+c2="c3.display();c3=c1-c2;cout<<"c1-c2="c3.display();c3=c1*c2;cout<<"c1*c2="c3.display();c3=c1/c2;cout<<"c1/c2="c3.display();return 0;-可编辑修改-1 口1

8、0+翁习童作业 VaskW-ZexRcl+c2=<8,61l-c2=<-2,1413rl«c2=<15i.-40icl/c2-<-0.2,0.415Tracess returned 0 0x0 executlain time = 0.638 sFress any key to caintiriLiE_Task8-3/*-可编辑修改-3 .有两个矩阵a和b ,均为行m歹U （m、n的值自己给出），求两个矩阵之和、差、积、商,重载运算符“ +”、”，*， ，/，、/ ,使之能用于矩阵向加减乘除，如c=a+b 、 c=a*b 。*/#include <iost

9、ream>#define n 2#define m 3using namespace std;class Matrix/定义Matrix类public:Matrix。;/默认构造函数friend Matrix operator+(Matrix &,Matrix &);/重载运算符“ + ”friend Matrix operator-(Matrix &,Matrix &);friend Matrix operator*(Matrix &,Matrix &);friend Matrix operator/(Matrix &,Matri

10、x &);void input();/ 输入数据函数void display();private:/ 输出数据函数int matnm;Matrix:Matrix()for(int i=0;i<n;i+)for(int j=0;j<m;j+)matij=0;/ 定义构造函数Matrix operator+(Matrix &a,Matrix &b)函数Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.mati j=a.mati j+b.matij; return c;Matrix operator-(M

11、atrix &a,Matrix &b)/ 定义重载运算符“ +/ 定义重载运算符“ + ”函-可编辑修改-Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.mati j=a.mati j-b.matij;return c;Matrix operator*(Matrix &a,Matrix &b)/ 定义重载运算符“ + ”函数Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.mati j=a.mati j*b.matij;return c;Ma

12、trix operator/(Matrix &a,Matrix &b)/ 定义重载运算符“ + ”函数Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.mati j=a.mati j/b.mati j;return c;void Matrix:input()/ 定义输入数据函数cout<<"input value of matrix:"<<endl;int i,j;for(i=0;i<n;i+)for( j=0;j<m;j+)cin>>matij;vo

13、id Matrix:display()/ 定义输出数据函数for (int i=0;i<n;i+)for(int j=0;j<m;j+)cout<<mati j<<" "cout<<endl;int main()Matrix a,b,c;a.input();b.input();cout<<endl<<"Matrix a:"<<endl;a.display();cout<<endl<<"Matrix b:"<<endl

14、;b.display();c=a+b;/ 用重载运算符“ + ” 实现两个矩阵相加cout<<endl<<"Matrix c = Matrix a + Matrix b :"<<endl;c.display();c=a-b;/ 用重载运算符“ + ” 实现两个矩阵相加cout<<endl<<"Matrix c = Matrix a - Matrix b :"<<endl;c.display();c=a*b;/ 用重载运算符“ + ”实现两个矩阵相加cout<<endl&l

15、t;<"Matrix c = Matrix a * Matrix b :"<<endl;c.display();c=a/b;/用重载运算符“ + ”实现两个矩阵相加cout<<endl<<"Matrix c = Matrix a / Matrix b :"<<endl;c.display();return 0;E D:Cg酒;习茁8章作业g e8召田/ input value of matrix： 18 6 3 6 9 4 input v-alue of matrix： 1 2 3 4 3 2hat

16、1* lx a ：ks 6 3 8 9 4Matrix h： 12 3 4 3 2Mfi'ts'ZLx c =d + Ma.tr ix h19 B 6 12 12 6c = Hat=i耳 a - Ha.tr-Ik b k7 4 elatrix c - Matrix a * Matrix b ： 8 12 9 2 27 8latrix c - Matrix a / Matrix b * 8 3 1 ! 3 2irecess returned 0 (0x0) execution time e 46.069 s 电害意 anj Jtey to cont inu.e .Task8-4/

17、*4 . 在第三题的基础上，重载运算符>> 和 <<, 使之能用于该矩阵的输入和输出。# /#include <iostream># define n 2# define m 3using namespace std;class Matrix/ 定义 Matrix 类public:/ 默认构造函数/ 重载运算符“ +Matrix();friend Matrix operator+(Matrix &,Matrix &);friend Matrix operator-(Matrix &,Matrix &);friend Matri

18、x operator*(Matrix &,Matrix &);friend Matrix operator/(Matrix &,Matrix &);friend ostream& operator<<(ostream& ,Matrix&);friend istream& operator>>(istream& ,Matrix&);void input();/ 输入数据函数void display();/ 输出数据函数private:int matnm;-可编辑修改-;Matrix:Matri

19、x()for(int i=0;i<n;i+)for(int j=0;j<m;j+)matij=0;/ 定义构造函数Matrix operator+(Matrix &a,Matrix &b)函数Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.mati j=a.mati j+b.matij; return c;Matrix operator-(Matrix &a,Matrix &b)数Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.

20、mati j=a.mati j-b.matij;/ 定义重载运算符“ +/ 定义重载运算符“ + ”函return c;Matrix operator*(Matrix &a,Matrix &b)/ 定义重载运算符“ + ”函数Matrix c;for(int i=0;i<n;i+)for(int j=0;j<m;j+)c.mati j=a.mati j*b.matij;return c;Matrix operator/(Matrix &a,Matrix &b)/ 定义重载运算符“ + ”函数Matrix c;for(int i=0;i<n;i+

21、)for(int j=0;j<m;j+)c.mati j=a.mati j/b.mati j;return c;ostream& operator<<(ostream& out,Matrix& c)for(int i=0;i<n;i+)for(int j=0;j<m;j+)cout<<c.mati j<<'t'cout<<endl;return out;istream& operator>>(istream& in,Matrix& c)for(int i

22、=0;i<n;i+)for(int j=0;j<m;j+)cin>>c.mati j;cout<<endl;return in;/ 定义输入数据函数void Matrix:input() cout<<"input value of matrix:"<<endl;int i,j;for(i=0;i<n;i+)for( j=0;j<m;j+)(cin>>matij;)/定义输出数据函数void Matrix:display()for (int i=0;i<n;i+)for(int j=0;

23、j<m;j+)cout<<mati j<<" "cout<<endl;int main()Matrix a,b,c;cout<<" 请输入第一个n*m矩阵"<<endl;cin>>a;cout<<" 请输入第二个n*m矩阵"<<endl;cin>>b;cout<<endl<<"Matrix a:"<<endl;cout<<a;cout<<end

24、l<<"Matrix b:"<<endl;cout<<b;c=a+b;用重载运算符“+”实现两个矩阵相加cout<<endl<<"Matrix c = Matrix a + Matrix b :"<<endl;c=a-b;用重载运算符“+”实现两个矩阵相加cout<<endl<<"Matrix c = Matrix a - Matrix b :"<<endl;cout<<c;c=a*b;/用重载运算符“ + ”实现两

25、个矩阵相加cout<<endl<<"Matrix c = Matrix a * Matrix b :"<<endl;cout<<c;c=a/b;/用重载运算符“ + ”实现两个矩阵相加cout<<endl<<"Matrix c = Matrix a / Matrix b :"<<endl;cout<<c;return 0;B港习罩a受作曲3 5 kB-4.su请输入第一个口矩阵 IS B 6B 6 4请输入第二个矩阵 2 4 64 3 2latriy a ：

26、ISS6口64latrdx b ： 246432latrix c = tlatrix q * Flatrix b ： 1212121296latrdx c - Hatrlx a - Matrix L : B40432a -* Matrix b ： 20323632168latrlx c - Hati*ix a / Matrix h - 521222Process PEui'ned 3 <0x0>1ahPresE Any key +0 con± inue,t ime J 2G.7U t sTask8-5/* 5.实现分数类中的运算符重载，在分数类中可以完成分数的加减

27、乘除-可编辑修改-(运算后再化简) 、求反、比较( 6 种关系)的运算。*/#include<iostream>#include<stdlib.h>using namespace std;int gcd(int m,int n);class Fractionprivate:int nume; / 分子int deno; / 分母public:Fraction(int nu=0,int de=1);/ 构造函数，初始化用void Set(int nu=0,int de=1);/ 置值，改变值时用Fraction operator+(Fraction &c2);Fr

28、action operator-(Fraction &c2);Fraction operator*(Fraction &c2);Fraction operator/(Fraction &c2);bool operator>=(Fraction &c2);Fraction operator-();void display();void Simplify(); / 化简(使分子分母没有公因子);Fraction:Fraction(int nu,int de)if(de!=0)nume=nu;deno=de;elsecout<<" 格式错误

29、,程序退出n"exit(0);void Fraction: Set(int nu,int de)if(de!=0)nume=nu;deno=de;void Fraction:display()cout<<nume<<'/'<<deno<<endl;/ 求分子，分母最大公约数int gcd(int m,int n)int r;if(m<n)r=m;m=n;n=r;while(r=m%n)if(r=0)break;m=n;n=r;return n;/ 将分数化简void Fraction:Simplify()int n

30、=gcd(nume,deno);nume=nume/n;deno=deno/n;cout<<nume<<'/'<<deno;Fraction Fraction:operator+(Fraction &c2)Fraction c;c.deno=deno*c2.deno;c.nume=nume*c2.deno+c2.nume*deno;return c;Fraction Fraction:operator-(Fraction &c2)Fraction c;c.deno=deno*c2.deno;c.nume=nume*c2.deno-c2.nume*deno;return c;Fraction Fraction:operator*(Fraction &c2)Fraction c;c.den