Linked List Problem(add node)

[Kyel John David]

here's my code I can't seem to print the other values(5 and 2 ) my code only
prints out the number 8, what seems to be the problem with my code?

include<stdio.h>

include<stdlib.h>

struct node
{
int x;
struct node next;
};
void add(struct node root,int y)
{

struct node *conductor;
root = malloc(sizeof(struct node));
conductor=root;
conductor->next=0;
conductor->x=y;

}

void display()
{
struct node *conductor;
while(conductor->next!=NULL);
{
printf("%d\n",conductor->x);
conductor=conductor->next;
}
}
int main()
{
struct node root;
add(&root,5);
add(&root,2);
add(&root,8);
display(root);

system("\npause");
}

[Daniel]

Hi, I have also faced such type of problems so many time during prgramming so
I hope I will also get so many solutions for such type of problem. So thanks
for starting this thread. Thanks!

---

environmental sustainability software

[Jim Pattee]

You need to keep a pointer to the first node and start the display with that
node.
You are starting the display with the updated root variable which contains the
last node.

[Aris]

void add(struct node *root,int y)
{
   //why do you create "conductor" pointer root isn't ok??
   struct node *conductor;
   root = malloc(sizeof(struct node));
   conductor=root;
   conductor->next=0;
   conductor->x=y;
}

void display() // no arguments!!! 
{
   // you create "conductor" a local pointer. Where is this pointer
   // points??
   struct node *conductor;
   //while(conductor->next!=NULL) stops
   while(conductor->next!=NULL);//here because of ;
   {
      printf("%d\n",conductor->x);
      conductor=conductor->next;                           
   }
   // conductor->next is NULL because you don't change its value
   // anywhere....
}

int main() // its better int main(void)
{
   //you create a new node
   struct node root;

   // you pass &root as argument &root to add function
   // but inside add you dont use this node, you create a new
   // node with root = malloc(sizeof(struct node)); command
   // this root you create inside add function, is not linked with
   // the root that you pass as argument here.
   add(&root,5);
   add(&root,2);
   add(&root,8);
   // you pass root by value to display function but,you defined
   // display function
   //as void display() no arguments!!!

   display(root);

   system("\npause");

   return 0;
   // this command say's to Operating System that programms execution
   // end normaly.
}

// You don't link nodes each other anywhere
// You create a node in main, then you pass it by value in add()
// inside add() you lose any conection with main's node because you
// create a new with malloc and make next to point to NULL. So your
// are not linked.

// function() mean that function doesn't make check of the
// passing arguments.

/code]

[b] This programm could be like this
don't just copy it....
 try to understant it
[/b]

[code]
#include<stdio.h>
#include<stdlib.h>

struct node
{
   int x;
   struct node *next;
};

void add(struct node *node_ptr,  int y)
{
   while(node_ptr->next != NULL)
   {
      node_ptr = node_ptr->next;
   } //goes to the last node

   struct node *new_node = malloc(sizeof(struct node));
   new_node->next = NULL;
   new_node->x = y;

   node_ptr->next = new_node;

   return;   
}

void display(struct node * node_ptr)
{
   do
   {
      printf("%d\n",node_ptr->x);
      node_ptr=node_ptr->next;
   }
   while(node_ptr->next != NULL);

   printf("%d\n",node_ptr->x); // print the last one

   return;
}

int main()
{
   struct node *root = malloc(sizeof(struct node)); //first node
   root->next = NULL;
   root->x = 5;

   add(root,2);
   add(root,8);
   display(root);

   system("\n\npause");
   return 0;
}
[/code]

[Aris]

i wrote int main()
is better
int main(void)