Simple linked list in C++

Question

I am about to create a linked that can insert and display until now:

struct Node {
    int x;
    Node *next;
};

This is my initialisation function which only will be called for the first Node:

void initNode(struct Node *head, int n){
    head->x = n;
    head->next = NULL;
}

To add the Node, and I think the reason why my linked list isn't working correct is in this function:

void addNode(struct Node *head, int n){
    struct Node *NewNode = new Node;
    NewNode-> x = n;
    NewNode -> next = head;
    head = NewNode;
}

My main function:

int _tmain(int argc, _TCHAR* argv[])
{
    struct Node *head = new Node;

    initNode(head, 5);
    addNode(head, 10);
    addNode(head, 20);
    return 0;
}

Let me run the program as I think it works. First I initialise the head Node as a Node like this:

head = [ 5 |  NULL ]

Then I add a new node with n = 10 and pass head as my argument.

NewNode = [ x | next ] where next points at head. And then I change the place where head is pointing to NewNode, since NewNode is the first Node in LinkedList now.

Why isn't this working? I would appreciate any hints that could make me move in the right direction. I think LinkedList is a bit hard to understand.

When I'm printing this, it only returns 5:

Thats more C rather than C++. You should encapsulate the linked list inside a class. And the pointer Node *head should be a private member variable inside the class pointing directly on the first Node (The actual way you have to allocate memory for one dummy node pointing on the next element. So you waste memory and the logic is more complex and does not represent the model's idea). Tell me if you need some example code. — exilit
– exilit, Commented Mar 3, 2014 at 8:05

Community · Accepted Answer · 2020-06-20 09:12:55Z

This is the most simple example I can think of in this case and is not tested. Please consider that this uses some bad practices and does not go the way you normally would go with C++ (initialize lists, separation of declaration and definition, and so on). But that are topics I can't cover here.

#include <iostream>
using namespace std;

class LinkedList{
    // Struct inside the class LinkedList
    // This is one node which is not needed by the caller. It is just
    // for internal work.
    struct Node {
        int x;
        Node *next;
    };

// public member
public:
    // constructor
    LinkedList(){
        head = NULL; // set head to NULL
    }

    // destructor
    ~LinkedList(){
        Node *next = head;
        
        while(next) {              // iterate over all elements
            Node *deleteMe = next;
            next = next->next;     // save pointer to the next element
            delete deleteMe;       // delete the current entry
        }
    }
    
    // This prepends a new value at the beginning of the list
    void addValue(int val){
        Node *n = new Node();   // create new Node
        n->x = val;             // set value
        n->next = head;         // make the node point to the next node.
                                //  If the list is empty, this is NULL, so the end of the list --> OK
        head = n;               // last but not least, make the head point at the new node.
    }

    // returns the first element in the list and deletes the Node.
    // caution, no error-checking here!
    int popValue(){
        Node *n = head;
        int ret = n->x;

        head = head->next;
        delete n;
        return ret;
    }

// private member
private:
    Node *head; // this is the private member variable. It is just a pointer to the first Node
};

int main() {
    LinkedList list;

    list.addValue(5);
    list.addValue(10);
    list.addValue(20);

    cout << list.popValue() << endl;
    cout << list.popValue() << endl;
    cout << list.popValue() << endl;
    // because there is no error checking in popValue(), the following
    // is undefined behavior. Probably the program will crash, because
    // there are no more values in the list.
    // cout << list.popValue() << endl;
    return 0;
}

I would strongly suggest you to read a little bit about C++ and Object oriented programming. A good starting point could be this: http://www.galileocomputing.de/1278?GPP=opoo

EDIT: added a pop function and some output. As you can see the program pushes 3 values 5, 10, 20 and afterwards pops them. The order is reversed afterwards because this list works in stack mode (LIFO, Last in First out)

bah, an example without even rudimentary tests, that's no example!
It compiles and exits normally. Isn't that enough? I think he needs a very simple example because he has no knowledge about OO C++. If he is interested in that topic (and he should be, if he wants to become a C++ programmer) he has to learn a lot more than this 30 loc.
Possibly, but in case he needs more, this: pastebin.com/yGh8hjnx can be compiled with g++ -o ll+ ll.cc, proves that there aren't bugs (that I could find,easily :) ) in your code, introduces the rudimentary concept of a cursor (kind of necessary to make lists at all useful :) ) and seems to work.
Yes of course there is much more needed to be a real list and to make it kind of useful (Your example is more useful in that case, +1). But maybe it is more easy to understand for beginners.
Im not sure what answer I should accept as "correct", because every answer helped me. How is the approach on this forum when something like this happens?

Blaz Bratanic · Accepted Answer · 2014-03-03 08:08:27Z

5

You should take reference of a head pointer. Otherwise the pointer modification is not visible outside of the function.

void addNode(struct Node *&head, int n){
    struct Node *NewNode = new Node;
 NewNode-> x = n;
 NewNode -> next = head;
 head = NewNode;
}

answered Mar 3, 2014 at 8:08

Blaz Bratanic

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Comments

erik258 · Accepted Answer · 2014-03-03 08:32:49Z

I'll join the fray. It's been too long since I've written C. Besides, there's no complete examples here anyway. The OP's code is basically C, so I went ahead and made it work with GCC.

The problems were covered before; the next pointer wasn't being advanced. That was the crux of the issue.

I also took the opportunity to make a suggested edit; instead of having two funcitons to malloc, I put it in initNode() and then used initNode() to malloc both (malloc is "the C new" if you will). I changed initNode() to return a pointer.

#include <stdlib.h>
#include <stdio.h>

// required to be declared before self-referential definition
struct Node;

struct Node {
    int x;
    struct Node *next;
};

struct Node* initNode( int n){
    struct Node *head = malloc(sizeof(struct Node));
    head->x = n;
    head->next = NULL;
    return head;
}

void addNode(struct Node **head, int n){
 struct Node *NewNode = initNode( n );
 NewNode -> next = *head;
 *head = NewNode;
}

int main(int argc, char* argv[])
{
    struct Node* head = initNode(5);
    addNode(&head,10);
    addNode(&head,20);
    struct Node* cur  = head;
    do {
        printf("Node @ %p : %i\n",(void*)cur, cur->x );
    } while ( ( cur = cur->next ) != NULL );

}

compilation: gcc -o ll ll.c

output:

Node @ 0x9e0050 : 20
Node @ 0x9e0030 : 10
Node @ 0x9e0010 : 5

Saurabh Raoot · Accepted Answer · 2016-07-21 11:08:48Z

Below is a sample linkedlist

    #include <string>
    #include <iostream>

    using namespace std;


    template<class T>
    class Node
    {
    public:
        Node();
        Node(const T& item, Node<T>* ptrnext = NULL);
        T value;
        Node<T> * next;
    };

    template<class T>
    Node<T>::Node()
    {
        value = NULL;
        next = NULL;
    }
    template<class T>
    Node<T>::Node(const T& item, Node<T>* ptrnext = NULL)
    {
        this->value = item;
        this->next = ptrnext;
    }

    template<class T>
    class LinkedListClass
    {
    private:
        Node<T> * Front;
        Node<T> * Rear;
        int Count;
    public:
        LinkedListClass();
        ~LinkedListClass();
        void InsertFront(const T Item);
        void InsertRear(const T Item);
        void PrintList();
    };
    template<class T>
    LinkedListClass<T>::LinkedListClass()
    {
        Front = NULL;
        Rear = NULL;
    }

    template<class T>
    void LinkedListClass<T>::InsertFront(const T  Item)
    {
        if (Front == NULL)
        {
            Front = new Node<T>();
            Front->value = Item;
            Front->next = NULL;
            Rear = new Node<T>();
            Rear = Front;
        }
        else
        {
            Node<T> * newNode = new Node<T>();
            newNode->value = Item;
            newNode->next = Front;
            Front = newNode;
        }
    }

    template<class T>
    void LinkedListClass<T>::InsertRear(const T  Item)
    {
        if (Rear == NULL)
        {
            Rear = new Node<T>();
            Rear->value = Item;
            Rear->next = NULL;
            Front = new Node<T>();
            Front = Rear;
        }
        else
        {
            Node<T> * newNode = new Node<T>();
            newNode->value = Item;
            Rear->next = newNode;
            Rear = newNode;
        }
    }
    template<class T>
    void LinkedListClass<T>::PrintList()
    {
        Node<T> *  temp = Front;
        while (temp->next != NULL)
        {
            cout << " " << temp->value << "";
            if (temp != NULL)
            {
                temp = (temp->next);
            }
            else
            {
                break;
            }
        }
    }

    int main()
    {
        LinkedListClass<int> * LList = new LinkedListClass<int>();
        LList->InsertFront(40);
        LList->InsertFront(30);
        LList->InsertFront(20);
        LList->InsertFront(10);
        LList->InsertRear(50);
        LList->InsertRear(60);
        LList->InsertRear(70);
        LList->PrintList();
    }

Peter Mortensen · Accepted Answer · 2017-12-02 01:55:24Z

Both functions are wrong. First of all function initNode has a confusing name. It should be named as for example initList and should not do the task of addNode. That is, it should not add a value to the list.

In fact, there is not any sense in function initNode, because the initialization of the list can be done when the head is defined:

Node *head = nullptr;

or

Node *head = NULL;

So you can exclude function initNode from your design of the list.

Also in your code there is no need to specify the elaborated type name for the structure Node that is to specify keyword struct before name Node.

Function addNode shall change the original value of head. In your function realization you change only the copy of head passed as argument to the function.

The function could look as:

void addNode(Node **head, int n)
{
    Node *NewNode = new Node {n, *head};
    *head = NewNode;
}

Or if your compiler does not support the new syntax of initialization then you could write

void addNode(Node **head, int n)
{
    Node *NewNode = new Node;
    NewNode->x = n;
    NewNode->next = *head;
    *head = NewNode;
}

Or instead of using a pointer to pointer you could use a reference to pointer to Node. For example,

void addNode(Node * &head, int n)
{
    Node *NewNode = new Node {n, head};
    head = NewNode;
}

Or you could return an updated head from the function:

Node * addNode(Node *head, int n)
{
    Node *NewNode = new Node {n, head};
    head = NewNode;
    return head;
}

And in main write:

head = addNode(head, 5);

6502 · Accepted Answer · 2014-03-03 08:10:20Z

2

The addNode function needs to be able to change head. As it's written now simply changes the local variable head (a parameter).

Changing the code to

void addNode(struct Node *& head, int n){
    ...
}

would solve this problem because now the head parameter is passed by reference and the called function can mutate it.

answered Mar 3, 2014 at 8:10

6502

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Comments

Peter Mortensen · Accepted Answer · 2017-12-02 01:52:44Z

2

head is defined inside the main as follows.

struct Node *head = new Node;

But you are changing the head in addNode() and initNode() functions only. The changes are not reflected back on the main.

Make the declaration of the head as global and do not pass it to functions.

The functions should be as follows.

void initNode(int n){
    head->x = n;
    head->next = NULL;
}

void addNode(int n){
    struct Node *NewNode = new Node;
    NewNode-> x = n;
    NewNode->next = head;
    head = NewNode;
}

edited Dec 2, 2017 at 1:52

Peter Mortensen

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answered Mar 3, 2014 at 8:11

Deepu

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Comments

Paul Roub · Accepted Answer · 2017-06-03 00:40:41Z

1

I think that, to make sure the indeep linkage of each node in the list, the addNode method must be like this:

void addNode(struct node *head, int n) {
  if (head->Next == NULL) {
    struct node *NewNode = new node;
    NewNode->value = n;
    NewNode->Next = NULL;
    head->Next = NewNode;
  }
  else 
    addNode(head->Next, n);
}

edited Jun 3, 2017 at 0:40

Paul Roub

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answered Jun 3, 2017 at 0:25

JOSMAR BARBOSA - M4NOV3Y

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Comments

Peter Mortensen · Accepted Answer · 2017-12-02 01:57:58Z

Use:

#include<iostream>

using namespace std;

struct Node
{
    int num;
    Node *next;
};

Node *head = NULL;
Node *tail = NULL;

void AddnodeAtbeggining(){
    Node *temp = new Node;
    cout << "Enter the item";
    cin >> temp->num;
    temp->next = NULL;
    if (head == NULL)
    {
        head = temp;
        tail = temp;
    }
    else
    {
        temp->next = head;
        head = temp;
    }
}

void addnodeAtend()
{
    Node *temp = new Node;
    cout << "Enter the item";
    cin >> temp->num;
    temp->next = NULL;
    if (head == NULL){
        head = temp;
        tail = temp;
    }
    else{
        tail->next = temp;
        tail = temp;
    }
}

void displayNode()
{
    cout << "\nDisplay Function\n";
    Node *temp = head;
    for(Node *temp = head; temp != NULL; temp = temp->next)
        cout << temp->num << ",";
}

void deleteNode ()
{
    for (Node *temp = head; temp != NULL; temp = temp->next)
        delete head;
}

int main ()
{
    AddnodeAtbeggining();
    addnodeAtend();
    displayNode();
    deleteNode();
    displayNode();
}

While this code snippet may solve the question, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.

Vladimir Popov · Accepted Answer · 2018-02-17 16:45:25Z

1

In a code there is a mistake:

void deleteNode ()
{
    for (Node * temp = head; temp! = NULL; temp = temp-> next)
        delete head;
}

It is necessary so:

for (; head != NULL; )
{
    Node *temp = head;
    head = temp->next;

    delete temp;
}

edited Feb 17, 2018 at 16:45

answered Feb 17, 2018 at 16:33

Vladimir Popov

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Comments

JustWe · Accepted Answer · 2019-03-04 08:23:59Z

Here is my implementation.

#include <iostream>

using namespace std;

template< class T>
struct node{
    T m_data;
    node* m_next_node;

    node(T t_data, node* t_node) :
        m_data(t_data), m_next_node(t_node){}

    ~node(){
        std::cout << "Address :" << this << " Destroyed" << std::endl;
    }
};

template<class T>
class linked_list {
public:
    node<T>* m_list;

    linked_list(): m_list(nullptr){}

    void add_node(T t_data) {
        node<T>* _new_node = new node<T>(t_data, nullptr);
        _new_node->m_next_node = m_list;
        m_list = _new_node;
    }


    void populate_nodes(node<T>* t_node) {
        if  (t_node != nullptr) {
            std::cout << "Data =" << t_node->m_data
                      << ", Address =" << t_node->m_next_node
                      << std::endl;
            populate_nodes(t_node->m_next_node);
        }
    }

    void delete_nodes(node<T>* t_node) {
        if (t_node != nullptr) {
            delete_nodes(t_node->m_next_node);
        }
        delete(t_node);
    }

};


int main()
{
    linked_list<float>* _ll = new linked_list<float>();

    _ll->add_node(1.3);
    _ll->add_node(5.5);
    _ll->add_node(10.1);
    _ll->add_node(123);
    _ll->add_node(4.5);
    _ll->add_node(23.6);
    _ll->add_node(2);

    _ll->populate_nodes(_ll->m_list);

    _ll->delete_nodes(_ll->m_list);

    delete(_ll);

    return 0;
}

Parteek Sharma · Accepted Answer · 2020-12-06 14:48:00Z

link list by using node class and linked list class

this is just an example not the complete functionality of linklist, append function and printing a linklist is explained in the code

code :

#include<iostream>
using namespace std;

Node class

class Node{
    public:
    int data;
    Node* next=NULL;
    Node(int data)
        {
            this->data=data;


        }   
    };

link list class named as ll

class ll{
    public:
        Node* head;

ll(Node* node)
    {
        this->head=node;
    }

void append(int data)
    {
        Node* temp=this->head;
        while(temp->next!=NULL)
            {
                temp=temp->next;
            }
        Node* newnode= new Node(data);
        // newnode->data=data;
        temp->next=newnode;
    }
void print_list()
    {   cout<<endl<<"printing entire link list"<<endl;
        Node* temp= this->head;
        while(temp->next!=NULL)
            {
                cout<<temp->data<<endl;
                temp=temp->next;
            }
        cout<<temp->data<<endl;;

    }
};

main function

int main()
{
  cout<<"hello this is an example of link list in cpp using classes"<<endl;
  ll list1(new Node(1));
  list1.append(2);
  list1.append(3);
  list1.print_list();
  }

thanks ❤❤❤

screenshot https://i.sstatic.net/C2D9y.jpg

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