关于#c++#的问题：假如map类m的key值是一个Student类，现在已经重载比较运算符了，如果给出name或者id或者age进行查找

#include 
#include 
#include 
#include 
using namespace std;
class Student
{
public:
    string name;
    string id;
    int age;
    bool operator()(const Student& s1, const Student& s2)
    {
        if (s1.name == s2.name)
        {
            if (s1.id == s2.id)
            {
                return s1.age > s2.age;
            }
            else
            {
                return s1.id > s2.id;
            }
        }
        else
        {
            return s1.name > s2.name;
        }
    }
 
    // 新增函数，根据给出的关键字返回对应的Student对象
    Student getByKey(string key) const {
        Student s;
        if (key == "name") {
            s.name = this->name;
            s.id = "";
            s.age = 0;
        } else if (key == "id") {
            s.name = "";
            s.id = this->id;
            s.age = 0;
        } else if (key == "age") {
            s.name = "";
            s.id = "";
            s.age = this->age;
        }
        return s;
    }
};
int main()
{
    mapint>m;
    Student s1 = {"Tom", "001", 18};
    m[s1] = 100;
 
    // 根据name查找
    Student s2 = s1.getByKey("name");
    auto it = m.find(s2);
    if (it != m.end()) {
        cout << "Found, value is " << it->second << endl;
    } else {
        cout << "Not found" << endl;
    }
 
    // 根据id查找
    Student s3 = s1.getByKey("id");
    it = m.find(s3);
    if (it != m.end()) {
        cout << "Found, value is " << it->second << endl;
    } else {
        cout << "Not found" << endl;
    }
 
    // 根据age查找
    Student s4 = s1.getByKey("age");
    it = m.find(s4);
    if (it != m.end()) {
        cout << "Found, value is " << it->second << endl;
    } else {
        cout << "Not found" << endl;
    }
 
    return 0;
}

假如map类m的key值是一个Student类，现在已经重载比较运算符了，如果给出name或者id或者age进行查找，为什么这段代码报错？



#include <iostream>
#include <string>
#include <map>
#include <algorithm>
using namespace std;
class Student
{
public:
    string name;
    string id;
    int age;
    bool operator<(const Student& s1)const
    {
        if(s1.name==name)
        {
            if(s1.id==id)
            {
                return s1.age<age;
            }
            else
            {
                return s1.id<id;
            }
        }
        else
        {
            return s1.name<name;
        }
    }
 
    // 新增函数，根据给出的关键字返回对应的Student对象
    Student getByKey(string key) const 
    {
        Student s;
        if (key == "name") {
            s.name = this->name;
            s.id = "";
            s.age = 0;
        } else if (key == "id") {
            s.name = "";
            s.id = this->id;
            s.age = 0;
        } else if (key == "age") {
            s.name = "";
            s.id = "";
            s.age = this->age;
        }
        return s;
    }
};
int main()
{
    map<Student, int>m;
    Student s1 = {"Tom", "001", 18};
    m[s1] = 100;
 
    // 根据name查找
    Student s2 = s1.getByKey("name");
    auto it = m.find(s2);
    if (it != m.end()) {
        cout << "Found, value is " << it->second << endl;
    } else {
        cout << "Not found" << endl;
    }
 
    // 根据id查找
    Student s3 = s1.getByKey("id");
    it = m.find(s3);
    if (it != m.end()) {
        cout << "Found, value is " << it->second << endl;
    } else {
        cout << "Not found" << endl;
    }
 
    // 根据age查找
    Student s4 = s1.getByKey("age");
    it = m.find(s4);
    if (it != m.end()) {
        cout << "Found, value is " << it->second << endl;
    } else {
        cout << "Not found" << endl;
    }
 
    return 0;
}