C++笔记 第三十一课 完善的复数类---狄泰学院

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/weixin_42187898/article/details/83902410

如果在阅读过程中发现有错误，望评论指正，希望大家一起学习，一起进步。
学习C++编译环境：Linux

第三十一课 完善的复数类

1.完善的复数类

复数类应该具有的操作
运算：+，-，*，/
比较：==，！=
赋值：=
求模：modulus
利用操作符重载
统一复数与实数的运算方式
统一复数与实数的比较方式

Complex 复数类的实现

Complex.h

#ifndef _COMPLEX_H_ //定义预处理宏
#define _COMPLEX_H_
class Complex
{
    double a;// 实部
    double b;//虚部
public:
//定义功能函数
    Complex(double a = 0, double b = 0);
    double getA();//获取实部
    double getB();
    double getModulus();
//定义操作符重载函数
    Complex operator + (const Complex& c);
    Complex operator - (const Complex& c);
    Complex operator * (const Complex& c);
    Complex operator / (const Complex& c);
//比较运算
    bool operator == (const Complex& c);
    bool operator != (const Complex& c);
//赋值操作符的重载，特殊之处在于只能当做成员函数实现
    Complex& operator = (const Complex& c);
};
#endif

Complex.cpp

#include "Complex.h"
#include "math.h" //数学头文件
Complex::Complex(double a, double b)
{
    this->a = a;
    this->b = b;
}
double Complex::getA()
{
    return a;
}
double Complex::getB()
{
    return b;
}
double Complex::getModulus()
{
    return sqrt(a * a + b * b);
}
Complex Complex::operator + (const Complex& c)
{
    double na = a + c.a;
    double nb = b + c.b;
    Complex ret(na, nb);
    
    return ret;
}
Complex Complex::operator - (const Complex& c)
{
    double na = a - c.a;
    double nb = b - c.b;
    Complex ret(na, nb);
    
    return ret;
}
Complex Complex::operator * (const Complex& c)
{
    double na = a * c.a - b * c.b;
    double nb = a * c.b + b * c.a;
    Complex ret(na, nb);
    
    return ret;
}
Complex Complex::operator / (const Complex& c)
{
    double cm = c.a * c.a + c.b * c.b;
    double na = (a * c.a + b * c.b) / cm;
    double nb = (b * c.a - a * c.b) / cm;
    Complex ret(na, nb);
    
    return ret;
}
    
bool Complex::operator == (const Complex& c)
{
    return (a == c.a) && (b == c.b);
}
bool Complex::operator != (const Complex& c)
{
    return !(*this == c);//特殊的计算
}
    
Complex& Complex::operator = (const Complex& c)
{
    if( this != &c )//查看是否为同一个复数
    {
        a = c.a;
        b = c.b;
    }
    
    return *this;
}

工程进行阶段性编译，更好排除编译错误

31-1.cpp

#include <stdio.h>
#include "Complex.h"
int main()
{
    Complex c1(1, 2);
    Complex c2(3, 6);
    Complex c3 = c2 - c1;
    Complex c4 = c1 * c3;
    Complex c5 = c2 / c1;
    
    printf("c3.a = %f, c3.b = %f\n", c3.getA(), c3.getB());
    printf("c4.a = %f, c4.b = %f\n", c4.getA(), c4.getB());
    printf("c5.a = %f, c5.b = %f\n", c5.getA(), c5.getB());
    
    Complex c6(2, 4);
    
    printf("c3 == c6 : %d\n", c3 == c6);
    printf("c3 != c4 : %d\n", c3 != c4);
    
    (c3 = c2) = c1;
    
    printf("c1.a = %f, c1.b = %f\n", c1.getA(), c1.getB());
    printf("c2.a = %f, c2.b = %f\n", c2.getA(), c2.getB());
    printf("c3.a = %f, c3.b = %f\n", c3.getA(), c3.getB());
    
    return 0;
}
运行结果：
c3.a = 2.000000, c3.b = 4.000000
c4.a = -6.000000, c4.b = 8.000000
c5.a = 3.000000, c5.b = 0.000000
c3 == c6 : 1
c3 != c4 : 1
c1.a = 1.000000, c1.b = 2.000000
c2.a = 3.000000, c2.b = 6.000000
c3.a = 1.000000, c3.b = 2.000000

2.注意事项

C++规定赋值操作符（=）只能重载为成员函数
操作符重载不能改变原操作符的优先级
操作符重载不能改变操作数的个数
操作符重载不能改变操作符的原有语义
小结
复数的概念可以通过自定义类实现
复数中的运算操作可以通过操作符重载实现
赋值操作符只能通过成员函数实现
操作符重载的本质为函数定义—扩展原有的功能