第九周專案一:複數類中的運算子過載(續)
阿新 • • 發佈:2019-02-08
問題及程式碼:
在複數類中的運算子過載基礎上
(1)再定義一目運算子 -,-c相當於0-c。
(2)定義Complex類中的<<和>>運算子的過載,實現輸入和輸出,改造原程式中對運算結果顯示方式,使程式讀起來更自然。
解決程式碼:
#include <iostream> using namespace std; class Complex { public: Complex(){real=0;imag=0;} Complex(double r,double i){real=r; imag=i;} Complex operator+(const Complex &c2); Complex operator-(const Complex &c2); Complex operator*(const Complex &c2); Complex operator/(const Complex &c2); Complex operator-(); friend ostream& operator << (ostream&,const Complex&); friend istream& operator >> (istream&,Complex&); private: double real; double imag; }; //下面定義成員函式 //複數相加: (a+bi)+(c+di)=(a+c)+(b+d)i. ostream& operator << (ostream& output,const Complex& c) { output<<"("<<c.real<<","<<c.imag<<"i)"; return output; } istream& operator >> (istream& input,Complex& c) { cout<<"please input real part and imaginary part:"; input>>c.real>>c.imag; return input; } Complex Complex::operator+(const Complex &c2) { Complex c; c.real=real+c2.real; c.imag=imag+c2.imag; return c; } //複數相減:(a+bi)-(c+di)=(a-c)+(b-d)i. Complex Complex::operator-(const Complex &c2) { Complex c; c.real=real-c2.real; c.imag=imag-c2.imag; return c; } //複數相乘:(a+bi)(c+di)=(ac-bd)+(bc+ad)i. Complex Complex::operator*(const Complex &c2) { Complex c; c.real=real*c2.real-imag*c2.imag; c.imag=imag*c2.real+real*c2.imag; return c; } //複數相除:(a+bi)/(c+di)=(ac+bd)/(c^2+d^2) +(bc-ad)/(c^2+d^2)i Complex Complex::operator/(const 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; } int main() { Complex c1,c2,c3; cin>>c1>>c2; cout<<"c1="<<c1<<endl; cout<<"c2="<<c2<<endl; c3=c1+c2; cout<<"c1+c2="<<c3<<endl; c3=c1-c2; cout<<"c1-c2="<<c3<<endl; c3=c1*c2; cout<<"c1*c2="<<c3<<endl; c3=c1/c2; cout<<"c1/c2="<<c3<<endl; return 0; }
執行結果:
學習心得:運用到了過載單目運算子,還有流插入運算子和流提取運算子,感覺過載了流插入運算子和流提取運算子後程序更加簡明清晰了!