friend Complex operator + (Complex c1,Complex c2);
事先定义了转换构造函数，但是接下来在这个友员函数定义里面，参数不能用引用Complex &c1，Complex &c2),用引用的话则会出现以下错误

binary '+' : no operator defined which takes a left-hand operand of type 'double' (or there is no acceptable conversion)
这个我想c4=i+c1; 这时i不是复数类型，调用转换构造函数，但是引用应该没关系的啊？原因实部是只能用值传递才能使得转换构造函数初始化？
我发现也可以把Complex(double r){real=r;imag=0;}修改成Complex(double &r){real=r;imag=0;}加一个引用符号，尽量减少时空开销。
我觉得尽管用了转换构造函数，可以使得运算符重载后不许重载更多就能适应交换律，但是这样，反而增加了时空开销，虽然不那么麻烦了。

    #include <iostream.h>
class Complex
{
public:
    Complex(){real=0;imag=0;}
    Complex(double r,double i){real=r;imag=i;}
    Complex(double r){real=r;imag=0;}
    friend Complex operator + (Complex c1,Complex c2);
    friend Complex operator - (Complex c1,Complex c2);
    friend Complex operator * (Complex c1,Complex c2);
    friend Complex operator / (Complex c1,Complex c2);
    void display();
private:
    double real;
    double imag;
};

Complex operator + (Complex c1,Complex c2)
{return Complex(c1.real+c2.real,c1.imag+c2.imag);}

Complex operator - (Complex c1,Complex c2)
{return Complex(c1.real-c2.real,c1.imag-c2.imag);}

Complex operator * (Complex c1,Complex c2)
{return Complex(c1.real*c2.real-c1.imag*c2.imag,c1.imag*c2.real+c1.real*c2.imag);}

Complex operator / (Complex c1,Complex c2)
{return Complex((c1.real*c2.real+c1.imag*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag),
                (c1.imag*c2.real-c1.real*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag));}

void Complex::display()
{cout<<"("<<real<<","<<imag<<"i)"<<endl;}

int main()
{
    Complex c1(3,4),c2(5,-10),c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14;
    double i=5;
    c3=c1+c2;
    cout<<"c1=";c1.display();cout<<"c2=";c2.display();
    cout<<"c1+c2=";c3.display();

    c4=i+c1;
    cout<<"i="<<i;cout<<",c1=";c1.display();
    cout<<"i+c1=";c4.display();

    c5=c2+i;
    cout<<"i="<<i;cout<<",c2=";c2.display();
    cout<<"c2+i=";c5.display();

    c6=c1-c2;
    cout<<"c1=";c1.display();cout<<"c2=";c2.display();
    cout<<"c6=c1-c2=";c6.display();

    c7=c1-i;
    cout<<"i="<<i;cout<<",c1=";c1.display();
    cout<<"c7=c1-i=";c7.display();

    c8=i-c2;
    cout<<"i="<<i;cout<<"c2=";c2.display();
    cout<<"c8=i-c2=";c8.display();

    c9=c1*c2;
    cout<<"c1=";c1.display();cout<<"c2=";c2.display();
    cout<<"c9=c1*c2=";c9.display();

    c10=c1*i;
    cout<<"i="<<i;cout<<",c1=";c1.display();
    cout<<"c10=c1*i=";c10.display();

    c11=i*c2;
    cout<<"i="<<i;cout<<",c2=";c2.display();
    cout<<"c11=i*c2=";c11.display();

    c12=c1/c2;
    cout<<"c1=";c1.display();
    cout<<"c2=";c2.display();
    cout<<"c12=c1/c2=";c12.display();

    c13=c1/i;
    cout<<"i="<<i;cout<<",c1=";c1.display();
    cout<<"c13=c1/i=";c13.display();

    return 0;
}

哎，这个程序是书上的一个例题。

我想问的是那个重载操作符的函数参数为什么不能用引用型？ 抱歉，前面我应该问清楚。

+操作符已经被重载了，乃觉得怎莫就八能用咧。