Friend class declaration error [duplicate] - c++

Question

This question already has answers here:
Understanding templates in c++
(2 answers)
Closed 7 years ago.
I'm implementing a singly linked list. Here's the node class
template <typename T> class node{
friend class slist<T>;
public:
explicit node():data(0), next(NULL){}
private:
T data;
node<T> *next;
};
And here's the list class.
template<typename T> class slist{
public:
slist():head(NULL){}
bool empty(){
return head == NULL;
}
void add_front(const T& data){
if(head == NULL){
node<T>* n = new node<T>();
n->data = data;
n->next = NULL;
head = n;
return;
}
node<T>* n = new node<T>();
n->data = data;
n->next = head;
head = n;
}
void remove_front(){
if(head == NULL) return;
node<T>* old = head;
head = old->next;
delete old;
}
void print_list(){
if(head == NULL) return;
while(!empty()){
std::cout<<head->data<<std::endl;
head = head->next;
}
}
~slist(){
while(!empty()) remove_front();
}
private:
node<T>* head;
};
The implementation works perfectly fine if I declare the members in the node as public. However the moment I declare them private and make the slist a friend class, I get the following error.
In file included from src.cpp:3:
./slist.hpp:5:28: error: redefinition of 'slist' as different kind of symbol
template<typename T> class slist{
^
./node.hpp:4:15: note: previous definition is here
friend class slist;
I can obviously find other implementation of single linked list, but I'm trying to understand what's going wrong here. So please refrain from unsolicited advice as "google it". Thanks.

Answer 1

You need to forward declare the List class as a template class:
#include <iostream>
template<typename T> class List; //<< fwd decl
template<typename T> class Node {
int i;
public:
friend class List<T>; //<< usage same as you have now
Node() : i(123) {}
int value() { return i; }
};
template<typename T> class List { //<< definition
public:
List(Node<T>* node) { node->i++; }
};
int main() {
Node<int> node{};
List<int> list{&node};
std::cout << node.value() << "\n";
}
http://ideone.com/2RUWgj

Answer 2

The friend class declaration needs it's own template parameter declaration without shadowing the enclosing classes template parameters:
template<typename U> class slist; // Forward declare slist
template <typename T>
class node{
template<typename U> // <<<<<
friend class slist<U>;
// ^ <<<<<
// ...
};