LinkedList C++ implementation

I just created an implementation of LinkedList(just for self-education purpose.). I made it run, but the output result is kind of weird... Here is the code: ``` #include "stdafx.h" #include <iostream> #include <stdio.h> using namespace std; template <class T> class Node{ T datum; Node<T> *_next; public: Node(T datum) { this->datum = datum; _next = NULL; } void setNext(Node* next) { _next = next; } Node* getNext() { return _next; } T getDatum() { return datum; } }; template <class T> class LinkedList{ Node<T> *node; Node<T> *currPtr; Node<T> *next_pointer; int size; public: LinkedList(T datum) { node = new Node<T>(datum); currPtr = node; //assignment between two pointers. next_pointer = node; size = 1; } LinkedList* add(T datum) // return pointer type. { Node<T> *temp = new Node<T>(datum); currPtr->setNext(temp); currPtr = temp; size++; cout<<datum<<" is added."; return this; //pointer type specification } T next() { T data = (*next_pointer).getDatum(); cout<<data<<" is visited."; next_pointer = next_pointer->getNext(); return data; } int getSize() { return size; } }; ``` Now I tried to use LinkedList: ``` int main() { LinkedList<int> *list = new LinkedList<int>(1); list->add(2)->add(3)->add(4); cout<<endl; printf("%d %d %d %d",list->next(),list->next(),list->next(),list->next()); \\One cout<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<endl; \\Two cout<<list->next()<<endl;\\Three cout<<list->next()<<endl; cout<<list->next()<<endl; cout<<list->next()<<endl; } ``` The output One will display the data : 4 3 2 1. Two will display 4 3 2 1. Three will display 1 2 3 4. I don't know what happened during the runtime. All of them should output the data in 1 2 3 4 sequence... I'd appreciate your help! Thanks!