Related
I don't know where I am wrong, when I debugged the code I found out that the 'new node' address is 'new node' address, basically the new node is referring to itself
void insertend(struct node *parent, int item)
{
while (parent->addr != NULL)
parent = parent->addr;
struct node new_node;
new_node.a = item;
parent->addr = &new_node;
parent->addr->addr = NULL;
}
void insertend(struct node *parent, int item)
{
while (parent->addr != NULL)
parent = parent->addr;
struct node new_node;
new_node.a = item;
parent->addr = &new_node;
parent->addr->addr = NULL;
}
The lifetime of new_node is limited to the function. Once that function returns, it is no longer valid.
In order to circumvent this, it is necessary to dynamically allocate memory for new_node. Of course, as already pointed out, this means explicitly deallocating the memory eventually.
Note: as this is C++ rather than C, we do not need to add struct to the front of the type in use, and NULL is better spelled nullptr.
void insertend(node *parent, int item)
{
while (parent->addr != nullptr)
parent = parent->addr;
node *new_node = new node;
new_node.a = item;
parent->addr = new_node;
parent->addr->addr = nullptr;
}
As C++ structs are just classes with default public access, it's also worth noting we could implement this as a member function. Something like:
template <typename T>
struct Node {
T value;
Node<T> *next;
void append(T val) {
Node<T> * temp = this;
while (temp->next != nullptr) {
temp = temp->next;
}
temp->next = new Node<T>;
temp->next->value = val;
temp->next->next = nullptr;
}
};
int main() {
auto n = Node<int>();
n.value = 27;
n.append(42);
n.append(34);
for (Node<int> *t = &n; t != nullptr; t = t->next) {
std::cout << t->value << std::endl;
}
return 0;
}
The next step would be implementing a constructor and destructor.
One more thing to keep in mind is that getting to the end of a list this way is O(n) time complexity. Doing it over and over again is costly. If you have Node::append return a pointer to the new Node, then you can call append on that.
template <typename T>
struct Node {
T value;
Node<T> *next;
Node<T> *append(T val) {
Node<T> * temp = this;
while (temp->next != nullptr) {
temp = temp->next;
}
temp->next = new Node<T>;
temp->next->value = val;
temp->next->next = nullptr;
return temp->next;
}
};
int main() {
auto n = Node<int>();
n.value = 27;
auto n2 = n.append(42);
n2 = n2->append(34);
n2 = n2->append(15);
for (Node<int> *t = &n; t != nullptr; t = t->next) {
std::cout << t->value << std::endl;
}
return 0;
}
I’m not sure what the addr refers to, but I would assume it points to the next node in the linked list. In your code, there needs to be another node to use as a temporary node, because the parent node keeps being reassigned, and the previous items are lost. We also need to check that the parent node is not NULL.
void insertend(struct node **parent, int item)
{
node *new_node = new node;
new_node->a = item;
new_node->addr = NULL;
if(*parent == NULL)
{
*parent = new_node;
}
else
{
node *temp = *parent;
while (temp->addr != NULL)
temp = temp->addr;
temp->addr = new_node;
}
}
The if statement checks whether the parent node is NULL. If true, then we assign parent node to new_node. If false, then we go to the end of the list and assign the next node in the list to new_node.
As Igor correctly pointed out, your new node gets destroyed when the function finishes. So, in this context you could just allocate memory. You can use the new operator for this. However, you would need to explicitly free the memory, eventually.
void insertend(struct node *parent, int item)
{
while (parent->addr != NULL)
parent = parent->addr;
struct * new_node = new node;
new_node->a = item;
parent->addr = new_node;
parent->addr->addr = NULL;
}
I am implementing a link list with cpp,what is wrong with the following code?
Every time i step into the function---AddToTail, the "list" can't get correct value. It changes it value to the new constructed node.
#include <iostream>
using namespace std;
struct Node
{
int value;
Node * next;
};
void AddToTail(Node* &list, int value)
{
Node newnode;
newnode.value = value;
newnode.next = NULL;
if (list == NULL)
list = &newnode;
else
{
Node * list1 = list;
while (list1->next != NULL)
{
list1 = list1->next;
}
list1->next = &newnode;
int a = 1;
}
}
int main()
{
Node *list=NULL;
AddToTail(list, 1);
AddToTail(list, 2);
AddToTail(list, 3);
while (list->next != NULL)
{
cout << list->value << endl;
list = list->next;
}
system("pause");
}
void AddToTail(Node* &list, int value)
{
Node newnode;
// Set up fields of newnode.
// Store address of newnode into some other data structure.
}
This is your issue. You are creating a node on the stack and this node will go out of scope at the end of the function. The reason it seems to be interfering with later node creations is because re-entering the function will almost certainly create newnode at exactly the same address as in the previous call.
If you want objects to survive function scope, you're going to need to allocate them dynamically, something like:
void AddToTail (Node *&list, int value) {
Node *newnode = new Node(); // create on heap.
newnode->value = value; // set up node.
newnode->next = nullptr;
if (list == nullptr) { // list empty,
list = newnode; // just create.
return;
}
Node *lastNode = list; // find last item.
while (lastNode->next != nullptr)
lastNode = lastNode->next;
lastNode->next = newnode; // append to that.
}
I'm trying to write an insert function for string values for a circular doubly linked list. I saw that creating a dummy node is beneficial in doing this so I can eliminate special cases like when the list is empty. The problem is I'm not finding alot of good information on dummy head nodes. I understand their purpose, but I don't understand how I create/implement it.
appreciate all the code examples guys, tried to figure it out on my own getting a little stuck though if someone can look at it.
#include <iostream>
#include <string>
using namespace std;
typedef string ListItemType;
struct node {
node * next;
node * prev;
ListItemType value;
};
node * head;
node * dummyHead = new node;
void insert(const ListItemType input, node * & within);
void main(){
insert("bob",dummyHead);
}
void insert( const ListItemType input, node * &ListHead){
node *newPtr = new node;
node *curr;
newPtr->value = input;
curr = ListHead->next; //point to first node;
while (curr != ListHead && input < curr->value){
curr = curr->next;
}
//insert the new node pointed to by the newPTr before
// the node pointed to by curr
newPtr->next = curr;
newPtr->prev = curr->prev;
curr->prev = newPtr;
newPtr->prev->next = newPtr;
}
For a circular doubly linked list, you can setup 1 sentinel node where both "next" and "prev" points to itself when list is empty. When list is not empty, sentinel->next points to first element and sentinel->prev points to last element. With this knowledge, your insert and remove function would look something like this.
This is very basic and your LinkedList and Node class maybe implemented differently. That is OK. The main thing is the insert() and remove() function implementation that shows how sentinel node(s) removes the need for checking for NULL values.
Hope this helps.
class DoublyLinkedList
{
Node *sentinel;
int size = 0;
public DoublyLinkedList() {
sentinel = new Node(null);
}
// Insert to the end of the list
public void insert(Node *node) {
// being the last node, point next to sentinel
node->next = sentinel;
// previous would be whatever sentinel->prev is pointing previously
node->prev = sentinel->prev;
// setup previous node->next to point to newly inserted node
node->prev->next = node;
// sentinel previous points to new current last node
sentinel->prev = node;
size++;
}
public Node* remove(int index) {
if(index<0 || index>=size) throw new NoSuchElementException();
Node *retval = sentinel->next;
while(index!=0) {
retval=retval->next;
index--;
}
retval->prev->next = retval->next;
retval->next->prev = retval->prev;
size--;
return retval;
}
}
class Node
{
friend class DoublyLinkedList;
string *value;
Node *next;
Node *prev;
public Node(string *value) {
this->value = value;
next = this;
prev = this;
}
public string* value() { return value; }
}
Why are you trying to use dummy node?
I hope you can handle it without a dummy node.
Eg:
void AddNode(Node node)
{
if(ptrHead == NULL)
{
ptrHead = node;
}else
{
Node* itr = ptrHead;
for(int i=1; i<listSize; i++)
{
itr = itr->next;
}
itr->next = node;
}
listSize++;
}
The above one is an example to handle the linked list without dummy node.
For a circular double linked list without a dummy node, the first node previous pointer points to the last node, and the last node next pointer points to the first node. The list itself has a head pointer to first node and optionally a tail pointer to last node and/or a count.
With a dummy node, the first node previous pointer points to the dummy node and the last node next pointer points to the dummy node. The dummy nodes pointers can point to the dummy node itself or be null.
The HP / Microsoft STL list function uses a dummy node as a list head node with the next pointer used as a head pointer to the first real node, and the previous pointer used as a tail pointer to the last real node.
#include <iostream>
#include <string>
using namespace std;
typedef string ElementType;
struct Node
{
Node(){}
Node(ElementType element, Node* prev = NULL, Node* next = NULL):element(element){}
ElementType element;
Node* prev;
Node* next;
};
class LinkList
{
public:
LinkList()
{
head = tail = dummyHead = new Node("Dummy Head", NULL, NULL);
dummyHead->next = dummyHead;
dummyHead->prev = dummyHead;
numberOfElement = 0;
}
void insert(ElementType element)
{
Node* temp = new Node(element, NULL, NULL);
if (0 == numberOfElement)
{
head = tail = temp;
head->prev = dummyHead;
dummyHead->next = head;
tail->next = dummyHead;
dummyHead->prev = tail;
}
else
{
tail->next = temp;
temp->prev = dummyHead->next;
temp->next = dummyHead;
dummyHead->next = temp;
tail = temp;
}
numberOfElement += 1;
}
int length() const { return numberOfElement; }
bool empty() const { return head == dummyHead; }
friend ostream& operator<< (ostream& out, const LinkList& List);
private:
Node* head;
Node* tail;
Node* dummyHead;
int numberOfElement;
};
ostream& operator<< (ostream& out, const LinkList& List)
{
Node* current = List.head;
while (current != List.dummyHead)
{
out<<current->element<<" ";
current = current->next;
}
out<<endl;
return out;
}
int main()
{
string arr[] = {"one", "two", "three", "four", "five"};
LinkList list;
int len = sizeof(arr) / sizeof(arr[0]);
for (int i = 0; i < len; ++i)
{
list.insert(arr[i]);
}
cout<<list<<endl;
}
I think this code can help you. When you want to implement some data structure, you must have a clear blueprint about it.
Do the following inside the constructor
ptrHead = new Node();
listSize = 1;
if you have tail also,
ptrHead->next = ptrTail;
The above code will create dummy node.
Make sure you implementation should not affected by this dummy node.
eg:
int getSize()
{
return listSize-1;
}
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:
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)
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;
}
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
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();
}
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);
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.
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;
}
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);
}
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();
}
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;
}
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;
}
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.stack.imgur.com/C2D9y.jpg
I was writing a simple function to insert at the end of a linked list on C++, but finally it only shows the first data. I can't figure what's wrong. This is the function:
void InsertAtEnd (node* &firstNode, string name){
node* temp=firstNode;
while(temp!=NULL) temp=temp->next;
temp = new node;
temp->data=name;
temp->next=NULL;
if(firstNode==NULL) firstNode=temp;
}
What you wrote is:
if firstNode is null, it's replaced with the single node temp which
has no next node (and nobody's next is temp)
Else, if firstNode is not null, nothing happens, except that the temp
node is allocated and leaked.
Below is a more correct code:
void insertAtEnd(node* &first, string name) {
// create node
node* temp = new node;
temp->data = name;
temp->next = NULL;
if(!first) { // empty list becomes the new node
first = temp;
return;
} else { // find last and link the new node
node* last = first;
while(last->next) last=last->next;
last->next = temp;
}
}
Also, I would suggest adding a constructor to node:
struct node {
std::string data;
node* next;
node(const std::string & val, node* n = 0) : data(val), next(n) {}
node(node* n = 0) : next(n) {}
};
Which enables you to create the temp node like this:
node* temp = new node(name);
You've made two fundamental mistakes:
As you scroll through the list, you roll off the last element and start constructing in the void behind it. Finding the first NULL past the last element is useless. You must find the last element itself (one that has its 'next' equal NULL). Iterate over temp->next, not temp.
If you want to append the element at the end, you must overwrite the last pointer's NULL with its address. Instead, you write the new element at the beginning of the list.
void InsertAtEnd (node* &firstNode, string name)
{
node* newnode = new node;
newnode->data=name;
newnode->next=NULL;
if(firstNode == NULL)
{
firstNode=newnode;
}
else
{
node* last=firstNode;
while(last->next != NULL) last=last->next;
last->next = newnode;
}
}
Note, this gets a bit neater if you make sure never to feed NULL but have all lists always initialized with at least one element. Also, inserting at the beginning of list is much easier than appending at the end: newnode->next=firstNode; firstNode=newnode.
The last element in your list never has it's next pointer set to the new element in the list.
The problem is that you are replacing the head of the linked list with the new element, and in the process losing the reference to the actual list.
To insert at the end, you want to change the while condition to:
while(temp->next != null)
After the loop, temp will point to the last element in the list. Then create a new node:
node* newNode = new node;
newNode->data = name;
newNode->next = NULL;
Then change temps next to this new node:
temp->next = newNode;
You also do not need to pass firstNode as a reference, unless you want NULL to be treated as a linked list with length 0. In that case, you will need to significantly modify your method so it can handle the case where firstNode is NULL separately, as in that case you cannot evaluate firstNode->next without a segmentation fault.
If you don't want to use reference pointer, you could use pointer to pointer. My complete code goes like below:
void insertAtEnd(struct node **p,int new_data)
{
struct node *new_node=(struct node *)malloc(sizeof(struct node));
new_node->data=new_data;
new_node->next=NULL;
if((*p)==NULL)//if list is empty
{
*p=new_node;
return;
}
struct node* last=*p;//initailly points to the 1st node
while((last)->next != NULL)//traverse till the last node
last=last->next;
last->next=new_node;
}
void printlist(struct node *node)
{
while(node != NULL);
{
printf("%d->",node->data);
node=node->next;
}
}
int main()
{
struct node *root=NULL;
insertAtEnd(&root,1);
insertAtEnd(&root,2);
insertAtEnd(&root,3);
insertAtEnd(&root,4);
insertAtEnd(&root,5);
printlist(root);
return 0;
}
Understanding the need of the below two variables is key to understanding the problem:
struct node **p: Because we need to link it from the root node created in the main.
struct node* last: Because if not used, the original content will be changed with the contents of the next node inside the while loop. In the end only 2 elements will be printed, the last 2 nodes, which is not desired.
void addlast ( int a)
{
node* temp = new node;
temp->data = a;
temp->next = NULL;
temp->prev=NULL;
if(count == maxnum)
{
top = temp;
count++;
}
else
{
node* last = top;
while(last->next)
last=last->next;
last->next = temp;
}
}
#include <bits/stdc++.h>
using namespace std;
class Node
{
public:
int data;
Node *next;
};
void append(Node *first, int n)
{
Node *foo = new Node();
foo->data = n;
foo->next = NULL;
if (first == NULL)
{
first = foo;
}
else
{
Node *last = first;
while (last->next)
last = last->next;
last->next = foo;
}
}
void printList(Node *first)
{
while (first->next != NULL)
{
first = first->next;
cout << first->data << ' ';
}
}
int main()
{
Node *node = new Node();
append(node, 4);
append(node, 10);
append(node, 7);
printList(node);
return 0;
}
Output: 4 10 7
You can use this code:
void insertAtEnd(Node* firstNode, string name)
{
Node* newn = new Node; //create new node
while( firstNode->next != NULL ) //find the last element in yur list
firstNode = firstNode->next; //he is the one that points to NULL
firstNode->next = newn; //make it to point to the new element
newn->next = NULL; //make your new element to be the last (NULL)
newn->data = name; //assign data.
}
void InsertAtEnd (node* &firstNode, string name){
node* temp=firstNode;
while(temp && temp->next!=NULL) temp=temp->next;
node * temp1 = new node;
temp1->data=name;
temp1->next=NULL;
if(temp==NULL)
firstNode=temp1;
else
temp->next= temp1;
}
while loop will return at temp==null in your code instead you need to return last node pointer from while loop like this
while(temp && temp->next!=NULL) temp=temp->next;
and assign a new node to next pointer of the returned temp node will add the data to the tail of linked list.