Linked List Only Showing Last Value Entered - c++

I am trying to become a little more adept with linked lists. I've watched several videos and read multiple forum posts, but I am still having issues. I am trying to start with a simple linked list. However, with my current code only the last value prints. I would really appreciate if someone could explain to me what I have done wrong. Also, I usually define all my functions in the main.cpp file. However, it would not let me do this for my linked list. Also, is defining all functions in the main.cpp file a good practice or a habit I should break?
Thanks in advance :).
Below is my linkedlist file:
#pragma once
#include <iostream>
using namespace std;
class LinkedList {
struct node {
int data;
node *next;
};
public:
LinkedList() {
head = NULL;
}
node *newNode;
node *temp;
node *head;
void insertData(int value) {
newNode = new node;
newNode->data = value;
temp = newNode;
head = newNode;
temp->next = newNode;
temp = temp->next;
newNode->next = NULL;
}
void printList() {
node *print;
print = head;
while (print != NULL) {
cout << print->data;
print = print->next;
}
}
};
Here is my main.cpp file
#include <iostream>
#include "LinkedList.h"
using namespace std;
int main() {
LinkedList list;
list.insertData(1);
list.insertData(2);
list.insertData(3);
list.printList();
system("pause");
return 0;
}

void insertData(int value) {
newNode = new node;
newNode->data = value;
temp = newNode;
head = newNode;
temp->next = newNode;
temp = temp->next;
newNode->next = NULL;
}
In your code, head = newNode takes whatever head was pointing to, and throws it out in favor of newNode. Essentially, every time you try to insert a new value into the list, you discard the whole list.
Rather, your insert should do something like
void insertData(int value) {
newNode = new node; //Create ourselves a new node
newNode->data = value; //Put the proper value in
newNode->next = head; //Make it so the whole list is after our node
head = newNode; //Make our node the first in the list
}
There are some other things in your code I would advise changing, like the fact that you have newNode and temp as member variables when they could simply be local to the functions, and your lack of a destructor. But your print function should work with the modified insert (see it run here).

Related

C++ linked list nodes

I have very basic concepts of linked list with C++. Here I have nodes linked, but the idea is to delete the last node, how do I achieve this?
This is the portion of code aimed to delete the last node:
//deleting node here
age* temp = head;
temp->next->next;//!=NULL
temp = temp->next;
//temp->next = NULL;
delete temp;
#include<iostream>
using namespace std;
struct age{
int a;
age *next;
};
age *head,*current,*node1,*node2,*ona;
int main(){
//creating a first node
age *node1=new age();
head=node1;
node1->a=10;
//creating a second node
age *node2=new age();
node2->a=20;
//link nodes
node1->next=node2;
node2->next=NULL;
//insertion of node ona between node 1 and node 2
ona=new age;
ona->a=15;
ona->next=node1->next;
node1->next=ona;
//deleting node here
age* temp = head;
temp->next->next;//!=NULL
temp = temp->next;
//temp->next = NULL;
delete temp;
//displaying the otput
current=head;
while(current!=NULL){
cout<<current->a<<endl;
current=current->next;
}
}
I suggest to have a look here:
For plain C development: https://www.learn-c.org/en/Linked_lists
On this site all standard methods for handling linked lists are explained and you can find code snippets for every operation.
For CPP development: https://www.codesdope.com/blog/article/c-deletion-of-a-given-node-from-a-linked-list-in-c/
On this site you can find an example coded in CPP OOP style.
I change the C example a little bit to fit your code example:
void remove_last(age * head) {
/* if there is only one item in the list, remove it */
if (head->next == NULL) {
delete head;
head = NULL;
return;
}
/* get to the second to last node in the list */
node_t * current = head;
while (current->next->next != NULL) {
current = current->next;
}
/* now current points to the second to last item of the list, so let's remove current->next */
delete(current->next);
current->next = NULL;
}

LInked list not reversing

Bellow, I have some code that is supposed to display a linked list, reverse it, and then display the now reversed linked list, but it seems that it never displays. My only guess is that somehow the linked list is becoming null. What am I doing wrong? Both the reverse function and the function that should display the reversed array run, but there is no visual output after.
#include <iostream>
using namespace std;
class Node{
public:// creation of a simple Node class
int data;
Node* next;
};
class LinkedList{
public:
Node* head;
LinkedList() { head = NULL; }
void append( int x){
Node* temp = new Node;// allocate new node
Node* end = head;//used later
temp->data = x;//giving the node data
temp->next = NULL;//since this node will be last make the next of it NULL
if(head == NULL){// if list is empty then set new Node as the head
head = temp;
return;
}
while(end->next != NULL){// go until the last node
end = end->next;
}
end->next = temp;// change the next of the last node to the new node.
}
void reverse(){
Node* current = head;
Node* next = NULL;
Node* prev = NULL;
while(current != NULL){
next = current->next;// Store next
current->next = prev;// Reverse current node's pointer
prev = current;// Move pointers one position ahead.
current = next;
}
head = prev;
}
void display(){
while(head != NULL){// print data while not out of bounds
cout<<" "<<head->data;
head = head->next;
}
}
};
int main() {
LinkedList list;
list.append(1);
list.append(10);
list.append(32);
list.append(64);
list.append(102);
list.append(93);
list.display();
cout<<endl;
list.reverse();
cout<<"list reversed"<<endl;
list.display();
cout<<"reverse display ran"<<endl;
Turns out it was an oversight on my part, I should have set up a temporary variable that represented the head, in my current program I'm changing what head references in order to loop through the linked list, and thus setting head equal to null once it reaches the end of the list a correct way to write the display function would be:
void display(){
Node* temp = head;
while(temp != NULL){// print data while not out of bounds
cout<<" "<<temp->data;
temp = temp->next;
}
}
thanks to user Retired Ninja for reminding me that debuggers exist.

Why it is printing only 1st value of doubly linked list and than my program is crashing

I am trying to create a doubly linked list and then printing its value but the output is showing only first value and then the whole program is crashing.
I can't understand where is the problem in the code .
Input
3
1 2 3
Expected output
1 2 3
current output
1
#include<iostream>
#include<stdlib.h>
using namespace std;
class node //declation of node
{
public:
int data;
node *next;
node *prev;
};
node *makenode(node *head,int val) //function to create node
{
node *newnode=new node;
node *temp;
newnode->data=val;
newnode->next=0;
newnode->prev=0;
if(head==0) temp=head=newnode;
else
{
temp->next=newnode;
newnode->prev=temp;
temp=newnode;
}
return head;
}
void display(node *head) //display function
{
system("cls"); //clearing output screen
while(head!=0)
{
cout<<head->data<<" ";
head=head->next;
}
}
int main()
{
node *head;
head=0;
int val;
int s; //size of list
cout<<"ENTER THE SIZE OF LIST";
cin>>s;
system("cls");
for(int i=0;i<s;i++)
{
cout<<"ENTER THE "<<i+1<<" VALUE\n";
cin>>val;
head=makenode(head,val); //calling makenode and putting value
}
display(head); //printing value
return 0;
}
node *makenode(node *head,int val) //function to create node
{
node *newnode=new node;
node *temp; // #1
newnode->data=val;
newnode->next=0;
newnode->prev=0;
if(head==0) temp=head=newnode;
else
{
temp->next=newnode; // #2
Between the lines marked #1 and #2 above, what exactly is setting the variable temp to point to an actual node rather than pointing to some arbitrary memory address?
"Nothing", I hear you say? Well, that would be a problem :-)
In more detail, the line:
node *temp;
will set temp to point to some "random" location and, unless your list is currently empty, nothing will change that before you attempt to execute:
temp->next = newnode;
In other words, it will use a very-likely invalid pointer value and crash if you're lucky. If you're unlucky, it won't crash but will instead exhibit some strange behaviour at some point after that.
If you're not worried about the order in the list, this could be fixed by just always inserting at the head, with something like:
node *makenode(node *head, int val) {
node *newnode = new node;
newnode->data = val;
if (head == 0) { // probably should use nullptr rather than 0.
newnode->next = 0;
newnode->prev = 0;
} else {
newnode->next = head->next;
newnode->prev = 0;
}
head = newnode;
return head;
}
If you are concerned about order, you have to find out where the new node should go, based on the value, such as with:
node *makenode(node *head, int val) {
node *newnode = new node;
newnode->data = val;
// Special case for empty list, just make new list.
if (head == 0) { // probably should use nullptr rather than 0.
newnode->next = 0;
newnode->prev = 0;
head = newnode;
return head;
}
// Special case for insertion before head.
if (head->data > val) {
newnode->next = head->next;
newnode->prev = 0;
head = newnode;
return head;
}
// Otherwise find node you can insert after, and act on it.
// Checknode will end up as first node where next is greater than
// or equal to insertion value, or the last node if it's greater
// than all current items.
node *checknode = head;
while (checknode->next != 0 && (checknode->next->data < val) {
checknode = checknode->next;
}
// Then it's just a matter of adjusting three or four pointers
// to insert (three if inserting after current last element).
newnode->next = checknode->next;
newnode->prev = checknode;
if (checknode->next != 0) {
checknode->next->prev = newnode;
}
checknode->next = newnode;
return head;
}
You aren't actually linking anything together. This line: if(head==0) temp=head=newnode; is the only reason your linked list contains a value at all. The very first value sets head equal to it and when you print head you get that value. In order to properly do a linked list you need a head and tail pointer. The head points to the first element in the list and the tail points to the last. When you add an element to the end of the list you use tail to find the last element and link to it. It is easiest to make Linked List a class where you can encapsulate head and tail:
struct Node {
public:
int data;
node *next;
node *prev;
Node(int data) : data(data), next(nullptr), prev(nullptr) {} // constructor
};
class LinkedList {
private:
Node* head;
Node* tail;
public:
LinkedList() { head = tail = nullptr; }
// This function adds a node to the end of the linked list
void add(int data) {
Node* newNode = new Node(data);
if (head == nullptr) { // the list is empty
head = newNode;
tail = newNode;
}
else { // the list is not empty
tail->next = newNode; // point the last element to the new node
newNode->prev = tail; // point the new element to the prev
tail = tail->next; // point the tail to the new node
}
}
};
int main() {
LinkedList lList;
lList.add(1);
lList.add(2);
// etc...
return 0;
}

error : exception : Unhandled exception thrown: read access violation. temp was 0xDDDDDDDD

I am a beginner and am working on Linked list. I am trying to make a program which adds elements to the list, updates the list, dislays it and deletes it.I am getting an exception : read access violation. temp was 0xDDDDDDDD.
I think there is some problem with display() function. The debugger also does shows the same.
#include "stdafx.h"
#include "Node.h"
#include<iostream>
using namespace std;
Node::Node() //constructor
{
head = NULL;
}
Node::~Node() //destructor
{
}
void Node::addFirstNode(int n) //adding the first element in the list
{
node *temp = new node;
temp->data = n;
temp->next = NULL;
head = temp;
}
void Node :: addLast(int n) //Adding elements at the end of the list
{
node *last = new node;
last->data = n;
last->next = NULL;
node *temp = new node;
temp = head;
while (temp->next != NULL) {
temp = temp->next;
}
temp->next = last;
}
void Node::display() //Displaying the list
{
node *temp = head;
while (temp != NULL)
{
cout<<temp->data;
temp = temp->next;
}
}
//the main function:
#include "stdafx.h"
#include "Node.h"
#include<iostream>
using namespace std;
int main()
{
Node a;
a.addFirstNode(101); //Calling function : addFirstNode
a.addLast(102); //Calling function : addLast
a.addLast(103); //Calling function : addLast
a.addLast(104); //Calling function : addLast
a.display(); //Calling function : display
return 0;
}
The Node.h file is as below:
struct node
{
int data;
node *next;
};
class Node
{
private :
node *head;
public:
Node();
~Node();
void addFirstNode(int n);
void addLast(int n);
void display();
};
You should rename Node to better describe what it is, e.g. List.
In Node::addFirst(), replace temp->next = NULL; with temp->next = head; You don't want to throw away your list every time you add a Node to the beginning of it.
In Node::addLast(), replace node *temp = new node; with node *temp = head; You don't want to leak memory every time you add a Node to the end of it.

How do I make my Linked List Print backwards in C++

How do I make my program print the Linked List backwards? I got the printForward function working fine but the printBackwards function just doesn't seem to do anything. I think I'm on the right track but I'm a little stuck right now. I think the while loop isn't running because temp is NULL for some reason.
Any help would be great.
Thanks
List.h
#include <iostream>
using namespace std;
class LinkedList
{
private:
struct Node
{
int data;
Node * next;
Node * prev;
};
Node * head, *tail;
public:
LinkedList();
bool addAtBeginning(int val);
bool remove(int val);
void printForward() const;
void printBackward() const;
};
#endif
List.cpp
#include "List.h"
LinkedList::LinkedList()
{
head = NULL;
tail = NULL;
}
bool LinkedList::addAtBeginning(int val)
{
Node* temp;
temp = new Node;
temp->data = val;
temp->next = head;
head = temp;
return false;
}
bool LinkedList::remove(int val)
{
return false;
}
void LinkedList::printForward() const
{
Node* temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
cout << endl;
}
void LinkedList::printBackward() const
{
Node* temp = tail;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->prev;
}
cout << endl;
}
app.cpp
#include "list.h"
int main()
{
LinkedList aList;
aList.addAtBeginning(3);
aList.addAtBeginning(10);
aList.addAtBeginning(1);
aList.addAtBeginning(7);
aList.addAtBeginning(9);
aList.addAtBeginning(12);
aList.printForward();
aList.printBackward();
system("pause");
return 0;
}
I find it a bit odd that you only have an addAtBeginning method, and no method to add at the end, the latter which I would consider to be normal use of a linked list. That being said, I think the immediate problem here is that you never assign the tail to anything. Try this version of addAtBeginning:
bool LinkedList::addAtBeginning(int val)
{
Node* temp;
temp = new Node;
temp->data = val;
temp->next = head;
if (head != NULL)
{
head->prev = temp;
}
if (head == NULL)
{
tail = temp;
}
head = temp;
return false;
`}
The logic here is that for the first addition to an empty list, we assign the head and tail to the initial node. Then, in subsequent additions, we add a new element to the head of the list, and then assign both the next and prev pointers, to link the new node in both directions. This should allow you to iterate the list backwards, starting with the tail.
Update addAtBeginning function with given:
bool LinkedList::addAtBeginning(int val)
{
Node* temp;
temp = new Node;
temp->data = val;
temp->prev = temp->next = NULL;
// If adding first node, then head is NULL.
// Then, set Head and Tail to this new added node
if(head == NULL){
// If this linked list is circular
temp->next = temp->prev = temp;
head = tail = temp;
}else{ // If we already have at least one node in the list
// If this linked list is circular
temp->prev = head->prev;
temp->next = head;
head->prev = temp;
head = temp;
}
return false;
}
But remember, if you copy this function with the parts that it makes this list circular, you will get an infinite loop. So, either change print function or dont copy that parts.