Thread: How to link next node in linked list

I am looking help to change function in my code. currently function create node, assign data to node but it doesn't link node to next node in list

Code:

 #include <stdio.h>
#include <stdlib.h>


struct node 
{
    int data;
    struct node *nextNode;
};


void addNode(int value )
{
    struct node *newNode = NULL;
    
    newNode = malloc (sizeof(*newNode)); 
    
    if (newNode != NULL)
    {
        newNode -> data = value;
        
        newNode -> nextNode = NULL;
    }
}


int main() 
{


    addNode( 2 );
    
    return 0;
}

I know This line doesn't point to address of next node

Code:

 newNode -> nextNode = NULL;

can anyone explain theoritically what I need to link next node in linked list ?

Originally Posted by Salem

Well you need the idea of a current list you want to add a node to.

That's what I am trying to understand

Originally Posted by Salem

Normally I would write something like
list = addNode(list, 2);

I think When you call a function, you pass pointer 'list' and value as a argument.

Originally Posted by Salem

Yes, that's right.

What happens by doing this ?

Where does new node add in list ? at beginning of list or end of list ?

Originally Posted by Salem

> Where does new node add in list ? at beginning of list or end of list ?
You can do either.
Or you can splice the node into the middle of the list.

I want to add node at end of list. I think I have created one node of list

Code:

#include <stdio.h>
#include <stdlib.h>

struct node 
{
    int data;
    struct node *nextNode;
};




void addNode( struct node *list, int value )
{
    struct node *newNode = NULL;
    
    newNode = malloc (sizeof(*newNode)); 
    
    if (newNode != NULL)
    {
        newNode -> data = value;
        
        newNode -> nextNode = list;
    }
}




int main() 
{


    struct node *list = NULL;
    list = malloc (sizeof(*list)); 
    
    if (list != NULL)
    {
         addNode(list,  2 );
    }
    
   
    
    return 0;
}