请问cpp中为什么不能输出重载后的后置递增运算符的值？

Question

请问cpp中为什么不能输出重载后的后置递增运算符的值？

# include<iostream>

using namespace std;


class myInt
{
    friend ostream& operator<<(ostream& cout,myInt& a);
private:
    int m_num;
public:
    myInt(){
        m_num = 0;
    };
    //前置递增运算符重载
    myInt& operator++(){
        m_num++;
        return *this;
    }
    //后置递增运算符重载
    myInt operator++(int){//int是占位符
        myInt temp = *this;
        m_num++;
        return temp;
    }
};

//<<重载
ostream& operator<<(ostream& cout,myInt& a){
    cout << a.m_num;
    return cout;
}

ostream& operator<<(ostream& cout,myInt& a){
    cout << a.m_num;
    return cout;
}



int main(){
    myInt m_int;
    cout << m_int << endl;
    cout << ++m_int << endl;
    cout << m_int << endl;
    cout << m_int++ << endl;

    
}

Answer 1

右值引用

#include <iostream>

using namespace std;

class myInt
{
    friend ostream &operator<<(ostream &cout, myInt &a);

    friend ostream &operator<<(ostream &cout, myInt &&a);

  private:
    int m_num;

  public:
    myInt()
    {
        m_num = 0;
    };
    // 前置递增运算符重载
    myInt &operator++()
    {
        m_num++;
        return *this;
    }
    // 后置递增运算符重载
    myInt operator++(int)
    { // int是占位符
        myInt temp = *this;
        m_num++;
        return temp;
    }
};

//<<重载
ostream &operator<<(ostream &cout, myInt &a)
{
    cout << a.m_num;
    return cout;
}

ostream &operator<<(ostream &cout, myInt &&a)
{
    cout << a.m_num;
    return cout;
}

int main()
{
    myInt m_int;
    cout << m_int << endl;
    cout << ++m_int << endl;
    cout << m_int << endl;
    cout << m_int++ << endl;
}