I got some problem with the linked list I've written. I dunno if it's either my insert function that the problem, or if it's my traverse function that's not correct. I hope for some input. A side note, I'm initalising the list in main now since I don't know if my initNode function is correct.
#include <iostream>
using namespace std;
typedef struct Node
{
int data;
Node *next;
};
void initNode(Node *head)
{
head = new Node;
head->next = NULL;
}
void insertNode(Node *head, int x)
{
Node *temp;
temp = new Node;
temp->data = x;
temp->next = head;
head = temp;
}
void traverse(Node *head)
{
Node *temp;
temp = head;
if(head == NULL)
{
cout << "End of list. " << endl;
}
else
{
while(temp != NULL)
{
cout << temp->data << " ";
temp = temp->next;
}
}
}
int main()
{
Node *head;
head = NULL;
insertNode(head, 5);
insertNode(head, 5);
traverse(head);
return 0;
}
Your head isn't being returned to main from insertNode. Note that even though head is a pointer, the pointer itself is a value and any changes to the pointer value are not reflected in main. The simplest solution is to pass back the updated value of head:
Node *insertNode(Node *head, int x)
{
...
return head;
}
And update it in main:
head = insertNode(head, 5);
The other common way of doing this is to pass a pointer to a pointer and update it directly:
void insertNode(Node **head, int x)
{
Node *temp;
temp = new Node;
temp->data = x;
temp->next = *head;
*head = temp;
}
And call it like this:
insertNode(&head, 5);
The way you have you initNode function written will result in memory leaks. You've passed in a pointer, but you need to pass in a reference to a pointer. (Same issue that James and casablanca mentioned for insertNode.)
Related
I am learning DSA, and was trying to implement linked list but the insertion function that i wrote is not
working in a for or while loop, its not the same when i call that function outside the loop, it works that way. I am not able to figure it out, please someone help me.
#include <iostream>
class Node {
public:
int data;
Node *next;
Node(int &num) {
this->data = num;
next = NULL;
}
};
class LinkedList {
Node *head = NULL;
public:
void insert(int num) {
Node *tmp;
if (head == NULL) {
head = new Node(num);
tmp = head;
} else {
tmp->next = new Node(num);
tmp = tmp->next;
}
}
void printList() {
Node *tmp = head;
while (tmp) {
std::cout << tmp->data << " ";
tmp = tmp->next;
}
std::cout << std::endl;
}
void reverseList() {
Node *curr = head, *prev = NULL, *nextNode;
while (curr) {
nextNode = curr->next;
curr->next = prev;
prev = curr;
curr = nextNode;
}
head = prev;
}
};
int main() {
LinkedList list1;
// This is not working
int num;
while (num != -1) {
std::cin >> num;
list1.insert(num);
}
// This is working
// list1.insert(1);
// list1.insert(2);
// list1.insert(3);
// list1.insert(4);
// list1.insert(5);
list1.printList();
list1.reverseList();
list1.printList();
return 0;
}
I expect this after insertion
Edit:
although #Roberto Montalti solved this for me, but before that I tried passing incrementing value using a for loop which worked but as soon as I pull that cin out it crashes. can someone tell me what's happening under the hood?
for (int i = 1; i <= 10; i++)
{
list1.insert(i);
}
When inserting the nth item (1st excluded) tmp is a null pointer, i don't understand what you are doing there, you are assigning to next of some memory then you make that pointer point to another location, losing the pointer next you assigned before, you must keep track of the last item if you want optimal insertion. This way you are only assigning to some *tmp then going out of scope loses all your data... The best way is to just keep a pointer to the last inserted item, no need to use *tmp.
class LinkedList
{
Node *head = NULL;
Node *tail = NULL;
public:
void insert(int num)
{
if (head == NULL)
{
head = new Node(num);
tail = head;
}
else
{
tail->next = new Node(num);
tail = tail->next;
}
}
...
}
You need to loop until you reach the end of the list and then add the new node after that. Like this.
void insert(int num) {
Node *tmp = head;
if (head == NULL) {
head = new Node(num);
}
else {
while (tmp->next != NULL) {
tmp = tmp->next;
}
tmp->next = new Node(num);
}
}
first of all you need to define a node for each of the tail and head of the list as follows
Node *h;
Node *t;
you may also separate the Node from the LinkedList class so you can modify easily
class Node{
public:
int data;
Node *next;
Node(int data, Node* next);
~Node();
};
Node::Node(int data, Node* next)
{
this->data= data;
this->next= next;
}
Node::~Node(){}
}
after that you can try to add these functions to your LinkedList class
so it can deal with other special cases such empty list or full, etc..
void addToHead(int data){
Node *x = new Node(data,h);
h=x;
if(t==NULL){
t=x;
}
void addToTail(int data){
Node *x = new Node(data,NULL);
if(isEmpty()){
h=t=x;
}
else
{
t->next=x;
t=x;
}
}
now for the insert function try this after you implemented the Node class and the other functions,
void insert(int v){
if(h==nullptr){addToHead(v); return;}
if(h->data>=v) {addToHead(v);return;}
if(t->data<=v) {addToTail(v); return;}
// In this case there is at least two nodes
Node *k=h->next;
Node *p=h;
while(k != nullptr){
if(k->data >v){
Node *z =new Node(v,k);
p->next=z;
return;
}
p=k;
k=k->next;
}
}
the idea of making all of this is not lose the pointer when it goes through elements in the Linked List so you don't end up with a run time error.
I hope this can be useful to you.
There was an issue with your insert function.
Read about segmentation fault here https://www.geeksforgeeks.org/core-dump-segmentation-fault-c-cpp/#:~:text=Core%20Dump%2FSegmentation%20fault%20is,is%20known%20as%20core%20dump.
for a quick workaround you can use this
using namespace std;
#include <iostream>
class Node
{
public:
int data;
Node *next;
Node(int num)
{
this->data = num;
next = NULL;
}
};
class LinkedList
{
Node *head = NULL;
public:
void insert(int num)
{
Node *tmp= new Node(num);
tmp->next=head;
head=tmp;
}
void printList()
{
Node *tmp = head;
while (tmp)
{
std::cout << tmp->data << " ";
tmp = tmp->next;
}
std::cout << std::endl;
}
void reverseList()
{
Node *curr = head, *prev = NULL, *nextNode;
while (curr)
{
nextNode = curr->next;
curr->next = prev;
prev = curr;
curr = nextNode;
}
head = prev;
}
};
int main()
{
LinkedList list1;
// This is not working
int num,i=0,n;
cout<<"Type the value of n";
cin>>n;
while (i<n)
{
cin >> num;
cout<<num<<" "<<&num<<endl;
list1.insert(num);
i++;
}
list1.printList();
list1.reverseList();
list1.printList();
return 0;
}
I need help in understanding why my Linked List approach doesn't work as expected.
#include <iostream>
using namespace std;
class Node {
public:
int Data;
Node* Next;
Node(int data) {
Data = data;
Next = NULL;
}
};
void insertNodeAtEnd(Node* HEAD, int data) {
Node* it = HEAD;
if (HEAD == NULL) { HEAD = new Node(data); }
else {
while (it->Next != NULL) { it = it -> Next; }
it -> Next = new Node(data);
}
}
void printLinkedList(Node* HEAD) {
Node* it = HEAD;
while (it != NULL) {
cout << it->Data << endl;
it = it -> Next;
}
}
int main() {
Node* HEAD = NULL;
// Node* HEAD = new Node(0);
insertNodeAtEnd(HEAD, 5);
insertNodeAtEnd(HEAD, 2);
insertNodeAtEnd(HEAD, 10);
printLinkedList(HEAD);
return 0;
}
The above main() function does NOT work (ie: no output, and the HEAD keeps resetting to NULL as soon as the control leaves insertNodeAtEnd()), and I've found similar questions here on SO which explain that this is because the pointer is being passed by value, and that makes partial sense to me.
Why does it work as expected when I replace Node* HEAD = NULL; with Node* HEAD = new Node(0); in the main() function, if the pointer is being passed as value?
How are nodes getting added if I initialise HEAD like Node* HEAD = new Node(0);, but not in the case where HEAD = NULL initially? I was able to get it to work properly by using pointer to pointer but I can't understand why this approach doesn't work. I am sorry if I haven't explained my question properly, please let me know if any clarification is required.
The underlying issue can be reduced to this code:
void insertNodeAtEnd(Node* HEAD, int data) {
//...
if (HEAD == NULL) { HEAD = new Node(data); }
//...
}
int main() {
Node* HEAD = NULL;
insertNodeAtEnd(HEAD, 5);
//...
You seem to assume that assigning to HEAD inside insertNodeAtEnd would change the HEAD variable inside of main. This is not true. Your pointer is passed by value, so the address is copied for the function. Changing this copied variable will not change the value of HEAD inside of main.
To fix this you could pass a pointer to a pointer instead, like this:
void insertNodeAtEnd(Node** HEAD, int data) {
//...
if (*HEAD == NULL) { *HEAD = new Node(data); }
//...
}
int main() {
Node* HEAD = NULL;
insertNodeAtEnd(&HEAD, 5);
//...
This pointer to a pointer is still passed by value, however the pointer that it points to will be the same as the on from main.
The problem come from your first insertion. You change the value of head which is reset when you quit the function. You can only change the value behind the pointer, not the pointer itself.
A solution for this would be to pass a pointer of pointer. Something like: (not tested)
void insertNodeAtEnd(Node** HEAD, int data) {
if (*HEAD == NULL) { *HEAD = new Node(data); }
else {
Node* it = *HEAD;
while (it->Next != NULL) { it = it -> Next; }
it -> Next = new Node(data);
}
}
int main() {
Node* HEAD = NULL;
// Node* HEAD = new Node(0);
insertNodeAtEnd(&HEAD, 5);
return 0;
}
As you don't change the pointer of pointer but only the value behind it (the actuual pointer to head) the change will be keep once you exit the function.
The answer has already been given by #Brotcrunsher. I am posting to help you implement a better solution, that separates the concept of a list and an element of a list, that incapsulates the methods used and that frees the resources it uses, when it goes out of scope:
#include <iostream>
using namespace std;
class Node {
public:
int Data;
Node* Next;
Node(int data = 0) {
Data = data;
Next = nullptr;
}
};
class List {
public:
Node* Head = nullptr;
void Insert(int data) {
if (Head == nullptr)
Head = new Node(data);
else {
Node* ptr;
for (ptr = Head; ptr->Next != nullptr; ptr = ptr->Next)
;
ptr->Next = new Node(data);
}
}
void Print() {
for (Node* ptr = Head; ptr != nullptr; ptr = ptr->Next)
cout << ptr->Data << endl;
}
~List() {
Node* ptr = Head;
while (ptr != nullptr) {
Node* tmp = ptr;
ptr = ptr->Next;
delete tmp;
}
}
};
int main() {
List list;
list.Insert(5);
list.Insert(2);
list.Insert(10);
list.Print();
return 0;
}
This question already has answers here:
Reason to Pass a Pointer by Reference in C++?
(7 answers)
Closed 1 year ago.
I wrote this piece of code in C++ for the function Insert At Tail in a linked list but when the list is empty it is not inserting the data.
This is the picture of it:- https://i.stack.imgur.com/wKkXk.png
I don't know why the lines from 35 to 39 are not executed.
This is my code:-
#include <iostream>
using namespace std;
class node
{
public:
int data;
node *next;
// Constructor
node(int d)
{
data = d;
next = NULL;
}
};
void display(node *head)
{
if (head == NULL)
{
cout << "The list is empty !!" << endl;
}
while (head != NULL)
{
cout << head->data << "->";
head = head->next;
}
cout << endl;
}
void Insert_At_Tail(node *head, int data)
{
if (head == NULL)
{
head = new node(data);
return;
}
node *tail = head;
while (tail->next != NULL)
{
tail = tail->next;
}
tail->next = new node(data);
return;
}
int main()
{
node *head = NULL;
int data;
cout << "Enter the data: ";
cin >> data;
Insert_At_Tail(head, data);
display(head);
return 0;
}
This is the snapshot of my output: https://i.stack.imgur.com/FFGj6.png
void Insert_At_Tail(node *head, int data)
In C++, by default function parametes are passed by value. This head parameter is a copy of the parameter that the caller passes in.
head = new node(data);
This sets the new head pointer. This is fine, but because this head is a copy of the original parameter, this does absolutely nothing, whatsoever, to the caller's passed-in head pointer. All this does is set the function's head parameter/variable. This has no effect on the head that was passed into this function.
You can do one of two things (your choice):
Pass the parameter by reference
return the new value of the head pointer from this function (which could be unchanged from what was passed in, if there were no changes to the head pointer), and have the caller make save the new head pointer.
The problem is that you don't change head at the caller. Either take a reference to the head
void insert_at_tail(node*& head, int data)
or better, return the new head:
void insert_at_tail(node *head, int data) {
if (!head) return new node(data);
node *tail = head;
while (tail->next != NULL) tail = tail->next;
tail->next = new node(data);
return head;
}
called like such:
head = insert_at_tail(head, data);
Even better is to wrap the whole thing into a class so you can write linked_list.insert_at_tail(data) and only have to mutate its members.
Currently, I am studying the linked list structure.
As I searched, linked list deletion function utilizing 'double pointer'.
In below code, the node is successfully deleted by double pointer in delete function.
#include <iostream>
using namespace std;
struct node
{
int data;
node* next;
};
class LinkedList
{
private:
node* head;
node* tail;
public:
LinkedList()
{
head = nullptr;
tail = nullptr;
}
void add_node(int n)
{
node* temp = new node;
temp->data = n;
temp->next = nullptr;
if(head == nullptr)
{
head = temp;
tail = temp;
}
else
{
tail->next = temp;
tail = tail->next;
}
}
node* gethead()
{
return head;
}
void display(node * head)
{
if(head == nullptr)
{
cout << "nullptr : No data" << endl;
return;
}
else
{
node* temp;
temp = head;
while(temp != nullptr)
{
cout << temp->data << endl;
temp = temp->next;
}
}
}
void del(node* head, int value)
{
if(!head)
{
return;
}
else
{
node** nd = &head;
while(*nd && (*nd)->data != value)
nd = &(*nd)->next;
if(*nd)
{
node* temp = *nd;
*nd = (*nd)->next;
delete temp;
}
else
{
cout << "No matching data in the node" <<endl;
}
}
}
};
int main()
{
LinkedList la;
la.add_node(10);
la.add_node(20);
la.add_node(30);
la.add_node(40);
la.add_node(50);
la.display(la.gethead()); //10 20 30 40 50
la.del(la.gethead(), 40);
la.display(la.gethead()); //10 20 30 50
return 0;
}
I just want to know why the below code is not working.
I am trying to simplify the code by not using a double-pointer.
#include <iostream>
using namespace std;
struct node
{
int data;
node* next;
};
class LinkedList
{
private:
node* head;
node* tail;
public:
LinkedList()
{
head = nullptr;
tail = nullptr;
}
void add_node(int n)
{
node* temp = new node;
temp->data = n;
temp->next = nullptr;
if(head == nullptr)
{
head = temp;
tail = temp;
}
else
{
tail->next = temp;
tail = tail->next;
}
}
node* gethead()
{
return head;
}
void display(node * head)
{
if(head == nullptr)
{
cout << "nullptr : No data" << endl;
return;
}
else
{
node* temp;
temp = head;
while(temp != nullptr)
{
cout << temp->data << endl;
temp = temp->next;
}
}
}
void del(node* head, int value)
{
if(!head)
{
return;
}
else
{
node* pp = head;
while((pp)->data != value)
pp = pp->next;
if(pp)
{
node* temp = pp;
pp = pp->next;
delete temp;
}
else
{
cout << "No matching data in the node" <<endl;
}
}
}
};
int main()
{
LinkedList la;
la.add_node(10);
la.add_node(20);
la.add_node(30);
la.add_node(40);
la.add_node(50);
la.display(la.gethead()); //10 20 30 40 50
la.del(la.gethead(), 40);
la.display(la.gethead()); //10 20 30 50
return 0;
}
Could you please give me advice about why the above code does not delete the node like the first code?
I think that the second code has to be worked because it deletes the node by utilizing pointer(which saves the address of the node).
Thanks in advance.
First of all, the term is "Pointer to pointer" and not "double pointer". Double pointer is a pointer to a variable of type double.
You should really realize that pointers are simply addresses in your memory.
When you are copying pointer (i.e.: node* pp = head;), you put the address that head points to in pp.
What it means? that both pp and head are pointing to the same address.
But, it is also important to remember that pp and head are variables, and thus are written in the memory as well. Saying that - pp and head also have addresses, but since pp and head are distinct variables, they have distinct addresses.
So the problem comes when you are trying to modify the content of head.
On the first occurence:
node** nd = &head;
while(*nd && (*nd)->data != value)
nd = &(*nd)->next;
You don't modify the content of head (or *nd) at all. And thus, you could used the version with pp.
You problem is with this piece of code:
pp = pp->next;
Here, pp is a copy of the address of the original linked list item, and since it is only a copy, then pp = pp->next does practicly nothing. In this case, you really should have written *nd = *nd->next, because this way change the original value and not a copy you created.
Note: even if you will change in your second example the variable pp to be a pointer-to-pointer, you code will still not be able to delete the first element in the list, since it is passed as a regular pointer to head, and thus only your first version is valid.
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