Start pointer not updating while passing by reference [duplicate] - c++

This question already has answers here:
When I change a parameter inside a function, does it change for the caller, too?
(4 answers)
Closed 6 years ago.
This is push function.
void push(struct node **head)
{
struct node *temp;
temp = new node;
cout<<"enter the value";
cin>>temp->data;
temp->link=NULL;
if(*head==NULL)
{
*head=new node;
*head=temp;
}
else{
temp->link=*head;
*head=temp;
}
}
this is how i am calling push.
struct node *start=NULL;
push(&start);
this is node
struct node{
int data;
struct node *link;
};
now the problem: i don't think the list is updating. The start always remains the null. Don't know why.
edit:
void display(struct node **head)
{
struct node *temp;
temp=*head;
if(*head==NULL){
cout<<"\nthe head is NULL\n";
}
while(temp!=NULL)
{
cout<<temp->data;
temp=temp->link;
}
}
int main() {
struct node *start=NULL;
push(&start);
push(&start);
push(&start);
push(&start);
push(&start);
display(&start);
return 0;
}
input:
1
2
3
4
5
now display out should have been 5 4 3 2 1 but there is some mistake.

The answer is mentioned by paxdiablo in the comments: C++ has pass by reference. Example:
#include <iostream>
struct node
{
int data;
struct node *link;
};
void push(node*& head)
{
struct node *temp = new node;
std::cout << "enter the value";
std::cin >> temp->data;
temp->link = head;
head = temp;
}
int main()
{
node *start = NULL;
push(start);
return 0;
}

Alternative implementation:
void push(struct node** head_reference, int new_data)
{
struct node* a_node = new node;
a_node->data = new_data;
a_node->link = (*head_reference);
(*head_reference) = a_node;
}
int main() {
struct node* head = NULL;
push(&head, 10);
// rest of your code here
return 0;
}

Related

Why is this linked list program not giving any output?

So I made these two simple functions regarding linked lists. One adds a node at the front and the other just displays the linked list in a sequence front to end. I'm wondering why this code wouldn't give me any output.
#include <iostream>
using namespace std;
class Node
{
public:
int data;
Node *next;
};
Node *head;
void addFront(Node *head, int item)
{
Node *temp = new Node();
temp->data = item;
temp->next = head;
head = temp;
}
void traverse(Node *head)
{
Node *temp = head;
while(temp!=NULL)
{
cout << temp->data << " ";
temp = temp->next;
}
}
int main()
{
addFront(head, 1);
addFront(head, 2);
addFront(head, 3);
traverse(head);
}
You're operating on a copy of head pointer in addFront(). You have
to pass a pointer to pointer:
void addFront(Node **head, int item)
The entire code could look like this:
#include <iostream>
using namespace std;
class Node
{
public:
int data;
Node *next;
};
Node *head;
void addFront(Node **head, int item)
{
Node *temp = new Node();
temp->data = item;
temp->next = *head;
*head = temp;
}
void traverse(Node *head)
{
Node *temp = head;
while(temp!=NULL)
{
cout << temp->data << " ";
temp = temp->next;
}
}
int main()
{
addFront(&head, 1);
addFront(&head, 2);
addFront(&head, 3);
traverse(head);
}
Argument of addFront is input as well as output. (Node *head)
It needs to be passed as referenced

Merging List Alternatively

In the following code i am trying to merge 2 list alternatively and than printing them in reverse order. But my code is not giving correct output it is just merging the last element of the second list.
input:
1
3
1 3 5
3
2 4 6
Actual Output:
5 6 3 1
Expected Output:
5 6 3 4 1 2
Can someone please tell me whats the problem in my code....
#include<bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node *next;
};
void push(struct Node ** head_ref, int new_data)
{
struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node *head)
{
struct Node *temp = head;
while (temp != NULL)
{
cout<<temp->data<<' ';
temp = temp->next;
}
cout<<' ';
}
void mergeList(struct Node **head1, struct Node **head2);
int main()
{
int T;
cin>>T;
while(T--){
int n1, n2, tmp;
struct Node *a = NULL;
struct Node *b = NULL;
cin>>n1;
while(n1--){
cin>>tmp;
push(&a, tmp);
}
cin>>n2;
while(n2--){
cin>>tmp;
push(&b, tmp);
}
mergeList(&a, &b);
printList(a);
printList(b);
}
return 0;
}
void mergeList(struct Node **p, struct Node **q)
{
struct Node*temp1=*p,*temp2=*q,*t1,*t2;
while(temp1!=NULL)
{
if(temp2==NULL)
break;
t1=temp1->next;
t2=temp2;
temp1->next=t2;
t2->next=t1;
temp1=t1;
*q=temp2->next;
temp2=*q;
}
}
To be honest, I'm really not sure what exactly you are doing in the mergeList function. The code is pretty cancerous so I did not take the liberty of verifying the correctness. I have renamed a few variables and re-written the code, so you can take this as a reference point and see what's wrong with your code.
void mergeList(struct Node **p, struct Node **q)
{
struct Node *a = *p, *b = *q, *next_a, *next_b;
while(a != NULL)
{
if(b == NULL)
break;
next_a = a->next;
a->next = b;
next_b = b->next;
b->next = next_a;
a = next_a;
b = next_b;
}
}
Hope this helps. Cheers.

cannot convert 'list*' to 'Node*' in assignment

I'm trying to construct an adjacency list. The code I wrote is as follows.
struct Node
{
int dest;
struct Node* next;
};
struct list
{
struct list *head;
};
The class is defined as:
class Graph
{
private:
int vertix;
list *arr;
public:
Graph(int v)
{
vertix = v;
arr = new list [vertix];
for(int i=0;i<vertix;i++)
{
arr[i].head=NULL;
}
}
Node* getNewNode(int destination)
{
Node* newNode = new Node;
newNode->dest = destination;
newNode->next = NULL;
return newNode;
}
The errors are in these functions:
void addEdge(int src, int dest)
{
Node* newNode = getNewNode(dest);
newNode->next = arr[src].head;
arr[src].head = newNode;
newNode = getNewNode(src);
newNode->next = arr[dest].head;
arr[dest].head = newNode;
}
void print()
{
cout<<"Adjacency list of vertix: "<<endl;
for(int i = 0; i< vertix; i++)
{
Node *ptr = arr[i].head;
cout<< i << "-->";
while(ptr)
{
cout<< "-->"<<ptr->dest;
ptr=ptr->next;
}
cout<<endl;
}
}
};
The error messages I get are:
[Error] cannot convert 'list*' to 'Node*' in assignment
[Error] cannot convert 'list*' to 'Node*' in initialization
I'm not sure if it is just a typo, but instead of
struct list
{
struct list *head;
};
you should have
struct list
{
Node *head;
};
since the head of a list is a node, not another list. This causes the error in this line:
Node *ptr = arr[i].head;
because you are trying to assign the head of the list (which in your current code is a list*) to a Node*

Linked list Behaviour

Can Someone please explain the difference in Behaviour ?
#include <iostream>
using namespace std;
struct Node {
int data;
Node* next;
};
// only for the 1st Node
void initNode(Node *head,int n){
head->data = n;
head->next =NULL;
}
void insertFront(Node *head, int n) {
Node *newNode = new Node;
newNode->data = n;
newNode->next = head;
head = newNode;
}
void display(Node *head) {
Node *list = head;
while(list) {
cout << list->data << " ";
list = list->next;
}
cout << endl;
cout << endl;
}
int main()
{
Node *l1 = new Node;
initNode(l1,10);
display(l1);
insertFront(l1,5);
display(l1);
insertFront(l1,6);
display(l1);
insertFront(l1,7);
display(l1);
return 0;
}
Some how the nodes are not linking. The output is :
10
10
10
10
If The program is coded using pointer to a pointer as below then it works fine. what am is missing ?
#include <iostream>
using namespace std;
struct Node {
int data;
Node* next;
};
// only for the 1st Node
void initNode(Node *head,int n){
head->data = n;
head->next =NULL;
}
void insertFront(Node **head, int n) {
Node *newNode = new Node;
newNode->data = n;
newNode->next = *head;
*head = newNode;
}
void display(Node *head) {
Node *list = head;
while(list) {
cout << list->data << " ";
list = list->next;
}
cout << endl;
cout << endl;
}
int main()
{
Node *l1 = new Node;
initNode(l1,10);
display(l1);
insertFront(&l1,5);
display(l1);
insertFront(&l1,6);
display(l1);
insertFront(&l1,7);
display(l1);
return 0;
}
Output correct as expected :
10
5 10
6 5 10
7 6 5 10
In the first case, in function
void insertFront(Node *head, int n) {
Node *newNode = new Node;
newNode->data = n;
newNode->next = head;
head = newNode;
}
head is a copy of the pointer l1 used in the scope of main(). When head is modified, l1 is left unchanged. This is the reason why a pointer to l1 (&l1) is passed to the function in the second case void insertFront(Node **head, int n). Then *head is l1, not just a copy of l1.
First case is an example of passing argument by value, and the second case is passing an argument by reference What's the difference between passing by reference vs. passing by value?
For instance, the following function is basically useless :
void useless(int a){
a=42;
}
If int b=2;useless(b);cout<<b<<endl; is called it will print 2, not 42.
The following function is the right way to go :
void rightwaytogo(int*a){
*a=42;
}
Don't forget to write a function to delete the nodes of your linked list.

Simple linked list in C++

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