c++运算符重载习题

Task8-1

/*

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

using namespace std;

class Complex

{

public:

|

Complex(){real=0;imag=0;}

Complex(double r,double i){real=r;imag=i;}

void display();

double real;

double imag;

};

void Complex::display()

{

cout<<"("<

*

}

Complex operator +(Complex &c1,Complex &c2)

{

Complex p;

=+;

=+;

return p;

}

int main()

】

{

Complex c1(3,5),c2(2,5),c3;

();

cout<<"+";

();

cout<<"=";

c3=c1+c2;

();

}

)

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

using namespace std;

class Complex

{public:

（

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::operator +(Complex &c2)

{Complex c;

=real+;

=imag+;

return c;

}

Complex Complex::operator -(Complex &c2)

{Complex c;

=;

$

=;

return c;

}

Complex Complex::operator *(Complex &c2)

{Complex c;

=real*;

=imag*;

return c;

}

Complex Complex::operator /(Complex &c2)

"

{Complex c;

=(real*+imag*/*+*;

=(imag**/*+*;

return c;

}

void Complex::display()

{cout<<"("<

}

int main()

{Complex c1(3,4),c2(5,-10),c3;

|

c3=c1+c2;

cout<<"c1+c2=";

();

c3=c1-c2;

cout<<"c1-c2=";

();

c3=c1*c2;

cout<<"c1*c2=";

();

c3=c1/c2;

[

cout<<"c1/c2=";

();

return 0;

}

Task8-3 /*

3. 有两个矩阵a和b，均为n行m列（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

#define n 2

#define m 3

using namespace std;

class Matrix

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

#define n 2

#define m 3

~

using namespace std;

class Matrix

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乘除

（运算后再化简）、求反、比较（6种关系）的运算。

*/

#include

#include<>

using namespace std;

int gcd(int m,int n);

class Fraction

{

》

private:

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);

Fraction 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;

}

else

{

cout<<"格式错误,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<

}

//求分子，分母最大公约数

int gcd(int m,int n)

{

int r;

if(m

{

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=gcd(nume,deno);

nume=nume/n;

deno=deno/n;

cout<

?

}

Fraction Fraction::operator+(Fraction &c2) {

Fraction c;

=deno*;

=nume*+*deno;

}

Fraction Fraction::operator-(Fraction &c2) {

》

Fraction c;

=deno*;

=nume* return c;

}

Fraction Fraction::operator*(Fraction &c2) {

Fraction c;

=deno*;

=nume*;

return c;

'

}

Fraction Fraction::operator/(Fraction &c2) {

Fraction c;

=deno*;

=nume*;

return c;

}

（

bool Fraction::operator>=(Fraction &c2) {

nume=nume*;

=*deno;

if(nume>=

return true;

else

return false;

^

}

Fraction Fraction::operator-()

{

Fraction c;

=-nume;

=deno;

}

int main()

{

\

Fraction c1(1,2),c2(5,8),c;

cout<<"c1=";

();

cout<<"c2=";

();

c=c1+c2;

cout<<"c1+c2=";

();

cout<

c=c1-c2;

cout<<"c1-c2=";

();

cout<

c=c1*c2;

cout<<"c1*c2=";

();

cout<

c=c1/c2;

cout<<"c1/c2=";

();

cout<

c=-c1;

cout<<"-c1=";

();

cout<

if(c1>=c2) cout<<"c1>=c2"<

}