複數的共軛函數

Question

我試圖創建一個函數，複合號碼爲 ，例如A（2，3）將變成A（2，-3），通過輸入〜A 我已經完成了一些代碼下面但我想這是錯誤的，我希望你能幫我解決這個問題。 我已經引用了我在下面的代碼中做錯了的部分。

#include <iostream> 
    using namespace std; 
    class Complex 
    { 
    private: 
     double real; 
     double imaginenary; 
    public: 
     Complex(); 
     Complex(double r, double i = 0); 

     // Declaration 
     Complex operator~(const Complex & c) const; 


     Complex operator+(const Complex & c) const; 
     Complex operator-(const Complex & c) const; 
     Complex operator*(const Complex & c) const; 
     Complex operator*(double n) const; 
     friend Complex operator*(double m, const Complex & c) 
       { return c * m; }  
     friend ostream & operator<<(ostream & os, const Complex & c); 
    }; 
    Complex::Complex() 
    { 
     real = imaginenary = 0; 
    } 

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



    // Definition 
    Complex Complex::operator~(const Complex & c) const 
    { 
     Complex conj; 
     conj.imaginenary = -1 * imaginenary; 
     conj.real = real; 
    } 


    Complex Complex::operator+(const Complex & c) const 
    { 
     Complex sum; 
     sum.imaginenary = imaginenary + c.imaginenary; 
     sum.real = real + c.real; 
     return sum; 
    } 

    Complex Complex::operator-(const Complex & c) const 
    { 
     Complex diff; 
     diff.imaginenary = imaginenary - c.imaginenary; 
     diff.real = real - c.real; 
     return diff; 
    } 
    Complex Complex::operator*(const Complex & c) const 
    { 
     Complex mult; 
     mult.imaginenary = imaginenary * c.imaginenary; 
     mult.real = real * c.real; 
     return mult; 
    } 

    Complex Complex::operator*(double mult) const 
    { 
     Complex result; 
     result.real = real * mult; 
     result.imaginenary = imaginenary * mult; 
     return result; 
    } 
    ostream & operator<<(ostream & os, const Complex & c) 
    { 
     os << "(" << c.real <<"," << c.imaginenary<<"i)"; 
     return os; 
    } 
    int main() 
    { 
     Complex A; 
     Complex B(5, 40); 
     Complex C(2, 55); 
     cout << "A, B, and C:\n"; 
     cout << A <<"; " << B << ": " << C << endl; 
     cout << " complex conjugate is" << ~C << endl; 
     cout << "B + C: " << B+C << endl; 
     cout << "B * C: " << B*C << endl; 
     cout << "10 * B: " << 10*B << endl; 
     cout << "B - C: " << B - C << endl; 
     return 0; 
    } 

Answer 1

嘗試

Complex Complex::operator~() const  
{ 
    Complex conj; 
    conj.imaginenary = -1 * imaginenary; 
    conj.real = real; 
    return conj; 
} 

但它可能是明智的，從類定義中刪除運營商和創建（朋友）函數來代替。它與隱式類型轉換效果更好。

Answer 2

的符號（〜）運算符是一元運算符所以它不應該接受一個參數（它的工作原理上*this）。你也忘了從operator~返回一個值。

Complex operator~() const 
{ 
    return Complex(real, -1 * imaginenary); 
} 

請參閱您的固定碼here

BTW：它的拼寫虛。

Answer 3

Complex Complex::operator~(const Complex & c) const 
{ 
    Complex conj; 
    conj.imaginenary = -1 * c.imaginenary; 
    conj.real = c.real; 
    return conj; 
}

這應該做。

儘管這不是回到你非全局範圍內分配什麼好主意，因爲在內存區域可以隨時改寫。順便說一句和它的想象，不是imaginenary :)

Answer 4

大量的代碼，但如果你會比你的operator~到例如operator+，你會發現一個細節 - 沒有返回聲明。

Answer 5

friend Complex operator*(double m, const Complex & c) { return c * m; } 
friend ostream & operator<<(ostream & os, const Complex & c); 

反正有沒有在這個問題上使用朋友？因爲我從書中讀到了這些，但是不太明白。