通过链表类从main中的另一个类调用函数

Question

Please, help me... I have Student class, LinkedList class and struct Node. I want to get Student's(object) name in main and I have so many errors. I'm not understanding typedef for calling the function.

There my code:

#include <iostream>
#include <string>

using namespace std;

class Student{
public:
string name;
int age;
Student(string n, int a){
    name = n;
    age = a;
}
Student(){};

void showname(){
    cout << "My name is: " << name << endl;
}

void showage(){
    cout << "My age is: " << age << endl;
}
};

template<class T>struct Node{
    T value;
    Node<T>* next;
};

template<class T>class LinkedList{
        typedef void (T::*fn)();
    public:
        Node<T>* first;
        Node<T>* temp;
        LinkedList(){
            first = NULL;
        }

    void insert(T a2){
        Node<T>* new_node = new Node<T>;
        new_node->value = a2;
        new_node->next = first;
        first = new_node;
    }

    void call(T b, fn op){
        (b.*op)();
    }

    void show(){
        temp = first;
        while(temp!=NULL){
            cout << temp->value;
            temp = temp->next;
        }
        cout << endl;
    }
};

int main(){
    Student s1("Nurbolat", 18);
    int a = 1;
    LinkedList<int> l1;
    LinkedList<Student> l2;
    l2.call(s1, &Student::showname);
    l2.call(s1, &Student::showage);
    return 0;
}

Answer 1

typedef void (T::*fn)();

create an alias fn as a member function of type T , receives no parameter and return void

Since int is a primitive type, it doesn't have any member function.

It's not required but it's allowed to instantiate all member function of LinkedList , then LinkedList<int> may give an error.

Remove that typedef and replace:

void call(T b, fn op){
    (b.*op)();
}

with:

template <typename F>
void call(T b, F op){
    (b.*op)();
}

then it should works