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學習C++編譯環境：Linux

第三十一課 完善的複數類

1.完善的複數類

複數類應該具有的操作
運算：+，-，*，/
比較：==，！=
賦值：=
求模：modulus
利用操作符過載
統一複數與實數的運算方式
統一複數與實數的比較方式

Complex 複數類的實現

Complex.h

#ifndef _COMPLEX_H_ //定義預處理巨集
#define _COMPLEX_H_
class Complex
{
    double a;// 實部
    double b;//虛部
public:
//定義功能函式
    Complex(double a = 0, double b = 0);
    double getA();//獲取實部
    double getB();
    double getModulus();
//定義操作符過載函式
    Complex operator + (const Complex& c);
    Complex operator - (const Complex& c);
    Complex operator * (const Complex& c);
    Complex operator / (const Complex& c);
//比較運算
    bool operator == (const Complex& c);
    bool operator != (const Complex& c);
//賦值操作符的過載，特殊之處在於只能當做成員函式實現
    Complex& operator = (const Complex& c);
};
#endif
 

Complex.cpp

#include "Complex.h"
#include "math.h" //數學標頭檔案
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);//特殊的計算
}
    
Complex& Complex::operator = (const Complex& c)
{
    if( this != &c )//檢視是否為同一個複數
    {
        a = c.a;
        b = c.b;
    }
    
    return *this;
}
 

工程進行階段性編譯，更好排除編譯錯誤

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;
}
執行結果：
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.注意事項

C++規定賦值操作符（=）只能過載為成員函式
操作符過載不能改變原操作符的優先順序
操作符過載不能改變運算元的個數
操作符過載不能改變操作符的原有語義
小結
複數的概念可以通過自定義類實現
複數中的運算操作可以通過操作符過載實現
賦值操作符只能通過成員函式實現
操作符過載的本質為函式定義—擴充套件原有的功能