一个类中含有const数据成员，这个类的2个对象之间如何才能用=号赋值？

C/C++ code

#include <iostream>
using namespace std;

class complex
{
private:
    //定义实数部分和虚拟部分
    double    real,image;
    //5_5_2题 定义一个常成员
    const double consum;
public:
    //complex();
    complex(double, double);
    ~complex();
    //复制构造函数
    complex( const complex &);

    double GetReal();
    double GetImage();

    void Show();

    //声明友元函数add
    friend complex add(complex, complex);

    //5_5_2题 声明一个常成员函数;const关键字必须放在函数参数表之后
    void ConShow  ( ) const ;
};

//complex::complex()
//{
//}

complex::complex(  const complex &ccp):consum(0) //有没有这个复制构造函数，都报error: non-static const member 'const double complex::consum', can't use default assignment operator
{
    real = ccp.real;
    image = ccp.image;
}

complex::complex(double x, double y):consum(x)
{
    real = x;
    image = y;
}

complex::~complex()
{
}

double complex::GetReal()
{
    return real;
}

double complex::GetImage()
{
    return image;
}

//5_5_2题 定义一个常成员函数，Page114上说定义const成员函数时，const关键字可以放在参数列表前，也可以放在参数列表后
void complex::ConShow  () const
{
    cout << consum << endl;
}

complex add(complex a, complex b)
{
    complex sum(0,0);
    sum.real = a.real + b.real;
    sum.image = a.image + b.image;
    return sum;
}

int main( ) {
    complex one(3.3,5.6), two(1.5,-0.2), three(0,0);
    three = add(one, two); //该如何写才能让这句话可以正确执行呢?
    //three(add(one, two)); 
    cout << "a + b 的实数部分 =  " << three.GetReal() << endl;
    cout << "a + b 的虚数部分 =  " << three.GetImage() << endl;
    return 0;
}

//运行结果：

编译结果：
g++ -O0 -g3 -Wall -c -fmessage-length=0 -o Practise_5_5_1.o ..\Practise_5_5_1.cpp
..\Practise_5_5_1.cpp: In member function 'complex& complex::operator=(const complex&)':
..\Practise_5_5_1.cpp:13: error: non-static const member 'const double complex::consum', can't use default assignment operator
..\Practise_5_5_1.cpp: In function 'int main()':
..\Practise_5_5_1.cpp:84: note: synthesized method 'complex& complex::operator=(const complex&)' first required here 

cast_const

你需要自己定义complex类的赋值运算符，原型如下：

complex& complex::operator=(const complex&);

一般编译器会自动为你生成这个函数，但是当类中有const成员或引用成员时，它就不会生成了，因为它不知道该如何做。你必须自己定义这个函数，在其中忽略该const成员的赋值。

C/C++ code

class complex
{
// ...
    complex& operator=(complex& c)
    {
        real=c.real;
        image=c.image;
        double *temp=const_cast<double*>(&consum);
        *temp=c.consum;
        return *this;
    }
};