Printing Linked List in C++ - c++

My following code print just only first element. In print_list() function, it stops after printing first element. It says after first element, head->next is 0. Shouldn't point towards second element?
I want to simply print whole list.
#include<iostream>
#include<cstdlib>
using namespace std;
struct node {
int x;
node *next;
};
node* add_element(node*);
bool is_empty(node*);
void print_list(node*);
node* search(node*);
int main()
{
node *head;
head=NULL;
node* current=head;
for(int i=0;i<5;i=i+1)
{
if (current==NULL)
{
current=add_element(current);
head=current;
}
else{
current=add_element(current);
}
}
cout<<head->next<<endl;
// DOUBT: head->next gives NULL value. It should give me pointer to 2nd node
print_list(head);
}
node* add_element(node* current)
{
node* temp;
temp=new node;
temp->next=NULL;
cout<<"enter element"<<endl;
cin>>temp->x;
current=temp;
return current;
}
bool is_empty(node* temp)
{
return temp==NULL;
}
void print_list(node* temp)
{
if (is_empty(temp)==false)
{
cout<<"here temp(head)"<<temp->next<<endl;
while(temp!=NULL)
{
cout<<temp->x<<endl;
temp = temp->next;
}
}
}

Print function print first element because you have just one node in the Linked List! Actually the mistake is present in add_element(node*) function, you overwrite address of head node with new node (so having memory leak) as I marked below:
node* add_element(node* current)
{
node* temp;
temp = new node; <---" You allocated memory"
temp->next = NULL; <---" Set next NULL"
cout<< "enter element" << endl;
cin>> temp->x; <---" Assign a value in new node"
// Replace below two line with suggested
current = temp; <---"MISTAKE: Overwrite first node"
"temp next is NULL so losing address of other nodes"
return current; <--- "return first node"
}
Next of new node (so first node) is NULL hence the print function prints only first node's value.
Suggestion:
You should Correct as follows to add new node as a first node in linked list:
temp -> next = current; // new nodes next if present first node
return temp; // new code becomes first node
Be careful current should be NULL initially.
With my suggestion in add_element() function also change the for loop code in main() as follows:
for(int i=0; i < 5; i = i + 1){
current = add_element(current);
}
head = current;
And check the working code at Codepade (instead of user input I added value using y = 100 variable).
Edit To append new node:
You need to check whether new node is first node of not (read comments).
// returns first node address in linked list = head
node* add_element(node* head){
node *temp, *new_nd;
// Create new node
new_nd = new node;
new_nd->next = NULL;
cout<<"enter element"<<endl;
cin>>new_nd->x;
// Is new node is the first node?
if(!head)
return new_nd;
// move to last
temp = head;
while(temp->next) temp = temp->next;
// add new node at last
temp->next = new_nd;
// return old head
return head;
}
Also simply main() as below:
int main(){
node *head = NULL;
for(int i = 0; i < 5; i = i + 1){
head = add_element(head);
}
print_list(head);
}
check this working code.

Your problem is here:
node* add_element(node* current)
{
node* temp; //You created a new node
temp=new node; //You allocated it here
temp->next=NULL; //You set its next property to null
cout<<"enter element"<<endl; //
cin>>temp->x;
current=temp; //This should be current->next = temp. You are overwriting it!
return current; //And now you are returning essentially the temp object that
//You created and you set its next property to NULL
}
You are assigning the node you created in temp = new node to the current node that was passed in. What you want to do is assign the node you just created to the current node's next property. It should be current->next = temp

head->next is NULL because you set it so in add_element(). To have a linked list, you should set current->next = temp.
As you're using C++, you might consider using std::list instead of implementing your own linked list.

if (current==NULL)
{ current=add_element(current);
head=current;
}
else
{ current->next=add_element(current);
current=current->next;
}
The correct code.
You have to make a small correction in the loop.
You have to add a new node and then make it point to the next of the current node.
so the simplified code is current->next=add_element(current)
and then make current point to the new current.

Related

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.

trying to make a simpler insertion at end of linked list program.Help needed with minor issues

I have managed to create a simple insertion at beginnning of linked list program but now i am struggling with insertion at end of linked list.
The program seems to be able to take values from user but the output list is not coming correct.Could you help me out?
If possible keep along the lines of my program as i am beginner and won't be able to understand a completely different method.
Logic i used-
If list is empty then insert value at beginning else if list is not empty then travel along the list till the next value being pointed at is NULL and then enter the new value in place of NULL.
#include<iostream>
using namespace std;
struct node
{
int data;
node *next;
};
node *start=NULL;
void insertend(int x)
{
node* temp=new node;
if(start==NULL)
temp->data=x;
else
{
while(temp!=NULL)
{
temp=temp->next;
}
temp->next=x;
}
}
void display()
{
node* disp=new node;
while(disp!=NULL)
{
cout<<disp->data<<endl;
disp=disp->next;
}
}
int main()
{
int x;
char ch;
do
{
cout<<"Enter data";cin>>x;
cout<<endl;
insertend(x);
cout<<"Do you want to continue?(y/n)";cin>>ch;
cout<<endl;
}while(ch=='y');
cout<<"Your list:"<<endl;
display();
}
The entry point to your list is the variable start. But you never set it. Take for example the first item a user inputs. You will call insertend(), it will check that start == NULL, but then it never sets start. You must set start = temp or something similar. You have the same problem in the else section -- you loop through the nodes starting with temp, but you should be starting with start. And again in the function display(), you create a pointer to a node and start looping from it, but it will have no data -- you should use start as the starting point of your loop.
struct node{
int data;
node* next;
};
node *first = NULL, *last = NULL;
void insert(int x){
if(first == NULL){
first = new node;
first->data = x;
first->next = NULL;
}else if(last == NULL){
last = new node;
last->data = x;
first->next = last;
last->next = NULL;
}else{
node *n = new node;
n->data = x;
n->next = NULL;
last->next = n;
last = n;
}
}
As you can see I am keeping track of first and last node in the list. Insert function checks if there is anything in the list with if(first == NULL)part. If there isn't it creates the first node. Similar thing happens with the else if. Finally in the else block we create a new node with data x. Then point the node stored in variable last to our new node and set last to be that node.
Here is the display function:
void display()
{
node *disp =first;
while(disp->next != NULL){
cout << disp->data << " ";
disp = disp->next;
}
cout << disp->data;
}
I also recommend that you do a cleanup after your program is finished running since you are creating new nodes.
void cleanup(node* n)
{
if(n->next == NULL)return delete n;
cleanup(n->next);
delete n;
}
and then at the end of main call cleanup(first)
Hope this makes sense :) Have a nice day!

Why does temp->next!=NULL work while temp!=NULL doesn't in Linked List in C++

Why does the following code doesn't work? According to my knowledge, when the temp reaches NULL, (new)ing it should create a new node with temp pointing towards it. Oddly, changing the while condition to temp->next!=NULL works perfectly. Any explanation to this?
#include<iostream>
using namespace std;
class node{
int data;
node *next;
public:
node(){}
node(int value):data(value),next(NULL){}
void addNode(node *head,int value){
node *temp=head;
while(temp!=NULL){
temp=temp->next;
}
temp=new node(value);
temp->next=NULL;
}
void display(node *head){
node *temp=head;
while(temp!=NULL){
cout << temp->data << endl;
temp=temp->next;
}
}
};
int main(int argc,char** argv){
node *head=new node(15);
head->addNode(head,50);
head->addNode(head,22);
head->display(head);
cin.ignore();
return 0;
}
you have many mistakes in the pasted code.
if you want to add something to a linked list end like you want here, you first need to find the list end (like you try to do at your code), but you have corner case at start where head is NULL. lets say head is not NULL.
first we find the last node in the list
node * tmp = head;
while(tmp->next != NULL){
tmp = tmp->next
}
now we have the last node in the list. we allocate new one and fill it
tmp->next = new node;
tmp = tmp->next;
tmp->x = data;
tmp->next = NULL;
why your loop doesn't stop. might be that head is not intialized and you have a loop in the list (just a place in memory you travel)
adding with head can be null.
if(tmp != null){
while(tmp->next != NULL){
tmp = tmp->next
}
tmp->next = new node;
tmp = tmp->next;
} else {
head = tmp = new node;
}
Stackoverflow is not allowing me to post images..as I'm a new user.XD
First See this Image
So, when you completed your loop with:
while(temp!=NULL) you end up as shown in diagram.
Now when you enter the code:
temp=new node(value);
temp->next=NULL;
It allocates some memory to temp on heap and is in any way not related to your linked list. So, whenever you use linked list it is advised to allocate to only pointers that consist of your class variables(In this case it is next).
After all temp is just an temporary variable just pointing to nodes of a linked list and allocating memory to it will not change linked list in anyway.
So, always allocate like using temp->next. Not only for allocating but also shifting connections.
This is the reason your code doesn't work in expected way.
You're implementing a Queue as a linked list using pointers. When you're inserting a new node at the end of your Queue/linked list you need to locate the last node of the list so that your temp pointer (which you're using to traverse the list) points to the last node of the list and temp->next points to NULL. Then you can insert the new node by using temp->next = new node(value);You don't need the line temp->next = NULL; after that because that's already done in the constructor when the new node is created.
If you use the condition temp!=NULL then temp will traverse the entire list and end up pointing to the next of the last node instead of pointing to the last node/element of the list (which is what you need). temp won't have any connection with your Queue/linked list. That's why you need the condition
while(temp->next!=NULL){
temp = temp->next;
}
in your addNode function. I've modified your function below:
void addNode(node *head,int value){
node *temp=head;
while(temp->next!=NULL){
temp=temp->next;
}
temp->next=new node(value);
}

Linked List, insert at the end C++

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.

linked list C++ , question selflearning

#include <iostream>
using namespace std;
struct Node
{
int item; // storage for the node's item
Node* next; // pointer to the next node
};
Node* addNode(Node*& head, int data , int& count)
{
Node * q; // new node
q = new Node; // allocate memory for the new mode
q->item = data; // inserting data for the new node
q->next = head; // point to previous node ?? how would i do that? ( am i doing it correctly?)
count++; // keep track of number of node
head = q;
return q;
}
int main()
{
int a, count=0;
int data;
bool repeat;
Node *head= NULL;
//^^ assuming thats creating the first node ^^
do
{
cout << "please enter the data for the next node" <<endl;
cin >> data;
addNode(head, data, count);
cout << "do you wish to enter another node? (enter true or false)" << endl;
cin >>repeat;
}
while (repeat == true);
// assuming this is the print function
while(head != NULL)
{
cout << "output" << temp->item << endl;
cout << temp->next << endl;
}
system("pause");
return 0;
}
okey i tried adding a new element in the list how would i move the head around like a LIFO memory (stack) so the last element is on the very top..
any help would be appreciated ! The pointers and the nodes are messing with my brain lately ....
temp is an uninitialized pointer. So -
temp-> item = a; // temp is not initialized or pointing to a memory location
// that has Node object to use operator ->
First, temp needs to be allocated memory location using new.
temp = new Node;
temp -> item = a;
And now assign it head. Similarly allocate memory for the child nodes too in the while loop. And return all the resources acquired from child to head using delete before program termination.
You seem to have some misunderstandings here:
Your "head" is the start of the list. It's always the start.
You add append elements to a linked list by assigning them to the last node's next pointer.
Third, you're not allocating anything.
Node *head= new Node();
Node *temp = new Node();
cout<<"enter something into data"<<endl;
cin >> a ;
temp->item = a;
head->next = temp;
Now ... to add the next thing, you either need to keep track of the last node (tail), or traverse the list to find the last node.
Node *nextNode = new Node();
nextNode->item = 0.0;
Node *i;
for (i = head; i->next != null; i = i->next);
i->next = nextNode;
This is O(n) execution time. By keeping track of the tail you make it O(1):
Node *head= new Node();
Node *tail = head;
Node *temp = new Node();
cout<<"enter something into data"<<endl;
cin >> a ;
temp->item = a;
tail->next = temp;
tail = temp;
Node *nextNode = new Node();
nextNode->item = 0.0;
tail->next = nextNode;
tail = nextNode;
EDIT: As pointed out, if you want to prepend to the list, you would:
temp->next = head;
head = temp;
Since I'm not sure every answer completely answers it, here's a linked list implementation (written without testig:
// your (correct) structure
struct Node
{
float item; // storage for the node's item
Node* next; // pointer to the next node
};
Now we need two pointers somewhere to look after the list:
/* some pointers */
struct List
{
Node* head;
Node* tail;
};
Now we need to create some elements. As others have said, you can do that with new:
/* create some elements we want to link in */
Node* elem1 = new Node();
Node* elem2 = new Node();
Node* elem3 = new Node();
/* maybe even set their properties! */
elem1->item = 3.14;
elem2->item = 3.14;
elem3->item = 3.14;
Notice how I didn't try to add these elements to a list yet? That's because I've got a function in mind which looks like this:
void addtolist(List &list, Node* node)
{
/* if no head, initialise the list */
if ( list->head == NULL )
{
list->head = node;
list->tail = node;
}
else if ( list->head != NULL && list->tail != NULL )
{
/* access the tail element and set its
next to this ptr.
Move tail to this node */
list->tail->next = node;
list->tail = node;
}
else
{
/* probably raise an exception! */
}
}
You can call this by doing this:
List l;
addtolist(l, elem1); /* etc */
Deleting elements is somewhat more tricky, since you have to go to that element, remember its previous element, grab it's next element, join them up and delete the Node* you're on.
Now for traversing lists... your terminology HEAD|TAIL reminds me of Erlang and tail recursion, where the current element is referred to as the head and the remainder the tail. If I write:
Node* cur = l.head;
while ( cur != NULL )
{
// do something with cur.item ?
cur = cur->next;
}
You can see this happening. Replacing cur with head here would be harmless thanks to the List struct.
Finally, I've used a very C-like approach here, but there's scope for templates:
template<typename T>
struct node
{
T item; // storage for the node's item
Node<T>* next; // pointer to the next node
};
and encapsulating the List struct as a class:
template<typename T>
class List
{
protected:
Node<T>* head;
Node<T>* tail;
public:
void addtolist(Node<T>* node);
Node<T>* gethead();
Node<T>* gettail();
}
Which brings you a little bit closer to std::list.
Additionally note that you are doing an implicit cast from int to float on
temp-> item = a;
as a is an int, while temp->item is a double.
To solve your problem: You want to allocate a new structure before accessing temp, thus
temp = new Node();