2015 第9周專案1 複數類中的運算子過載（續）

【專案1-複數類中的運算子過載（續）】

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。

#include using namespace std;

class complex

complex(double r,double i)

complex operator-();

friend complex operator+(complex &c1, complex &c2);

friend complex operator+(double d1, complex &c2);

friend complex operator+(complex &c1, double d2);

friend complex operator-(complex &c1, complex &c2);

friend complex operator-(double d1, complex &c2);

friend complex operator-(complex &c1, double d2);

friend complex operator*(complex &c1, complex &c2);

friend complex operator*(double d1, complex &c2);

friend complex operator*(complex &c1, double d2);

friend complex operator/(complex &c1, complex &c2);

friend complex operator/(double d1, complex &c2);

friend complex operator/(complex &c1, double d2);

friend ostream& operator << (ostream& output,const complex &c);

friend istream& operator >> (istream& input, complex &c);

private:

double real;

double imag;

};ostream &operator<<(ostream &output , const complex &c)

while(!((sign=='+'||sign=='-')&&i=='i'));

c.real=a;

c.imag=(sign=='+')?b:-b;

return input;

}complex complex::operator-()

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

complex operator+(double d1, complex &c2)

complex operator+(complex &c1, double d2)

complex operator-(complex &c1, complex &c2)

complex operator-(double d1, complex &c2)

complex operator-(complex &c1, double d2)

//複數相乘：(a+bi)(c+di)=(ac－bd)+(bc+ad)i.

complex operator*(complex &c1, complex &c2)

complex operator*(double d1, complex &c2)

complex operator*(complex &c1, double d2)

//複數相除：(a+bi)/(c+di)=(ac+bd)/(c^2+d^2) +(bc-ad)/(c^2+d^2)i

complex operator/(complex &c1, complex &c2)

complex operator/(double d1, complex &c2)

complex operator/(complex &c1, double d2)

int main()

{ complex c1,c2,c3;

double d=11;

cout<<"c1: "<>c1;

cout<<"c2: "<>c2;

cout<<"c1="<

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