递增运算符怎么回事？

Question

#include<iostream>
using namespace std;

class myinteger {
    friend ostream& operator<<(ostream& cout, myinteger& myint);
public:
    myinteger() {
        m_num = 0;
    }
    myinteger& operator++() {
        m_num++;
        return *this;
    }
private:
    int m_num;
};

ostream& operator<<(ostream& cout, myinteger& myint) {
    cout << myint.m_num;
    return cout;
}

int main() {
    myinteger myint;
    cout << ++myint << endl;//myint.operator++()
}

这里的myinteger的对象myint调用operator不是myint.operator++()这样的吗，那简化之后应该是myint++吧，为什么是++myint呢？

Answer 1

++有两种不同的形式++i或者i++，operator++()实现的是++i，operator++(int)实现的是i++
可以参考https://blog.csdn.net/qq_24447809/article/details/111184133
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