Sorted Linked List in C++ - c++

Question

Here is my problem.
I have a Linked List class as follows:
#include <iostream>
#include "List.h"
template <class Elem>
class LList : public List<Elem> { //List is a virtual base class
protected:
Node <Elem> *head;
Node <Elem> *fence;
Node <Elem> *tail;
int leftCount;
int rightCount;
void init();
void removeAll();
public:
LList();
~LList();
//Rest of methods overridden from List class
//////
};
Then I have a class called SortedLList which inherits from LList as follows:
#include "LinkedList.h"
#include "Helper.h"
template <class Elem>
class SortedLList : public LList<Elem> {
protected:
Helper *helper;
public:
SortedLList();
~SortedLList();
bool insert(const Elem&); //Override insertion method from LList class
};
In the implementation of SortedLList (SortledLList.cpp):
#include "SortedLList.h"
template <class Elem>
SortedLList<Elem>::~SortedLList() {
removeAll();
}
template <class Elem>
bool SortedLList<Elem>::insert(const Elem &_e) {
fence = head;
//Rest of Code..
}
I am having a compiler error that says : Use of undeclared identifier removeAll(). Same thing for fence and head pointers are not being recognized. What did I do wrong?
Thank You.

Answer 1

Because your class is a template there are certain issues that can happen to confuse the compiler. You may think your code is straight forward and easy to understand, and in this case it is. Older compilers used to do their best to guess and compile this code.
However, newer compilers are more strict and fail on all versions of this type of code, in order to prevent programmers from relying on it.
What you need to do is use the this pointer when calling base class functions. That makes the call unambiguous and clear. That would look like this->removeAll().
Another option would be to use a full name qualification like LList<Elem>::removeAll(). I prefer using this because it is easier to read.