C++ LinkedList using template

Question

I want to make my own linkedlist implementatin in c++ using templates. However, I run into several compiler errors. Here is the code:

template <class T>
class Node{
T datum;
Node _next = NULL;
 public:
 Node(T datum)
{
    this->datum = datum;
}
 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.
      size = 1;
  }
void add(T datum)
 {
   Node<T> *temp = new Node<T>(datum);
   (*currPtr).setNext((*temp));
   currPtr = temp;       
   size++;
 }
T next()
{
   next_pointer = node;
   T data = (*next_pointer).getDatum();
   next_pointer = (*next_pointer).getNext();
   return data;
}
int getSize()
{
   return size;
}   
};

when I try to instantiate this class, I got the following errors:

LinkedList.cpp: In instantiation of `Node<int>':
LinkedList.cpp:35:   instantiated from `LinkedList<T>::LinkedList(T) [with T = int]'
LinkedList.cpp:60:   instantiated from here
LinkedList.cpp:7: error: `Node<T>::_next' has incomplete type
LinkedList.cpp:5: error: declaration of `class Node<int>'
make[2]: *** [build/Debug/Cygwin-Windows/LinkedList.o] Error 1
make[1]: *** [.build-conf] Error 2
make: *** [.build-impl] Error 2

I'm using NetBeans IDE. Could anybody give me some suggestion to improve it? Thanks a lot!!

C++ isn't Java. A class can't contain a member of its own type. There are no "generics" in C++. C++ really isn't the same as Java. — Kerrek SB, Jun 19 '12 at 19:45
@Karthik, no, the class template's name injected into the scope, so `Node` with no template args refers to `Node`. Kerrek has the right answer. — Jonathan Wakely, Jun 19 '12 at 19:51

score 1 · Answer 1 · edited May 23 '17 at 12:12

1

Besides the syntax being wrong:

class Node{
T datum;
Node _next = NULL;
//....

you can't have a member of the same type as a member. You can use a pointer though:

template <class T>
class Node{
T datum;
Node<T>* _next;

and set it in the constructor (because =NULL wouldn't be legal there).

Related: Why is a class allowed to have a static member of itself, but not a non-static member?

edited May 23 '17 at 12:12

Community

1
1

answered Jun 19 '12 at 19:47

Luchian Grigore

253,575
64
457
625

The `=NULL` would be valid in C++11 – Jonathan Wakely Jun 19 '12 at 19:56

score 1 · Answer 2 · answered Jun 19 '12 at 19:48

1

Your Node class should contain a pointer to the next Node, not a Node directly. Here you've a recursive type which is impossible to handle for the compiler.

Another mistake is that you cannot initialize a member like this. You must do that in the constructor.

So, replace Node _next = NULL; by Node *_next; and initialize the pointer _next to nullptr in your constructor.

answered Jun 19 '12 at 19:48

Frédéric Terrazzoni

2,190
19
25

The compiler doesn't really matter. That's just a piece of software. What's important is that a type containing itself doesn't make *sense*. – Kerrek SB Jun 19 '12 at 20:52

score 1 · Answer 3 · answered Jun 19 '12 at 19:48

1

It might just be this line:

Node _next = NULL;

You really need that to be a pointer (Node<T>* _next). If a class contained an instance of its own type, then THAT instance would have its own instance of that type and so on ad memorum exhaustum.

answered Jun 19 '12 at 19:48

Matt K

13,370
2
32
51

score 1 · Accepted Answer · answered Jun 19 '12 at 19:51

1

Just like when you do Node<T> *node; in the LinkedList, you need to also do the same in your Node class

template <class T>
class Node{
    T datum;
    Node<T> *_next;//template parameter
                   //also, need to make this a pointer and initialized in constructor
public:

//...
    void setNext(Node<T> next) //template parameter
    {
        _next = next;
    }
    Node<T> getNext() //template parameter
    {
        return _next;
    }
//...
};

answered Jun 19 '12 at 19:51

crashmstr

28,043
9
61
79

Hi, Thanks a lot. Actually, when I change it to a pointer type, it passes. Even if I don't add T after the Node! Anyway, Thanks! – whileone Jun 19 '12 at 19:57
2

In the class' own scope `Node` is equivalent to `Node` – Jonathan Wakely Jun 19 '12 at 19:59
Yes, I found it does work without (as Jonathan Wakely points out in a comment). But I find it strange to not have them. – crashmstr Jun 19 '12 at 19:59

C++ LinkedList using template

4 Answers4

Linked