在前缀递增时，若想要实现 ++(++a) 这种连续自加，就要返回其对象的引用，这样才能保证操作的是同一块内存空间，否则就只是单纯的赋值操作，原来的对象并未被修改。

#include <iostream>
using namespace std;
class Complex {
private:
    double i;
    double j;
public:
    Complex(int = 0, int = 0);
    void display();
    Complex& operator ++();//前缀自增
    Complex operator ++(int);//后缀自增，参数需要加int
};

Complex::Complex(int a, int b) {
    i = a;
    j = b;
}

void Complex::display() {
    cout << "i="<< i <<"\tj="<< j << endl;
}

Complex& Complex::operator ++() {
    ++i;
    ++j;
    return *this;
}

Complex Complex::operator ++(int) {
    Complex temp = *this;
    ++* this;
    return temp;
}

int main()
{
    Complex comnum1(2, 2), comnum2, comnum3;
    cout << "自增计算前:" << endl;
    cout << "comnum1:";
    comnum1.display();
    cout << "comnum2:";
    comnum2.display();
    cout << "comnum3:";
    comnum3.display();
    cout << endl;

    cout << "前缀自增计算后：" << endl;
    comnum2 = ++comnum1;
    cout << "comnum1:";
    comnum1.display();
    cout << "comnum2:";
    comnum2.display();
    cout << endl;

    cout << "后缀自增计算后:" << endl;
    comnum3 = comnum1++;
    cout << "comnum1:";
    comnum1.display();
    cout << "comnum3:";
    comnum3.display();

    cout << "前缀递增加引用是为了连续自加++(++a)" << endl;
    Complex comcum4;
    ++(++comcum4);
    cout << "comcum4:";
    comcum4.display();

    return 0;
}