0

i'm working on a program in c language in which I have to use linked-list and in this program I have to insert the new node at the start of the linked-list if the user pass the value of the place 0 and also insert the new node at the end of the linked-list if the user pass the value of the place 1 in the choice variable. But i'm not getting any output on console and my program ends by just writing Output: I can't figure out the problem in my code and here's my code.

/* 
  program for making nodes and adding them in memory as per 0
  and 1

  0 means that insert the number at front , in other words insert number after head 

  1 means insert number at the last place

  First you need to input a number and then enter the place you want to insert it by giving input as 0 and 1 
  
  *Recall what does 0 and 1 mean by looking at line 5-7 respectively. 
  
  Just like 

  5 0 6 1 7 0 8 1

*/

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

// declaring struct with typedef for ease of use
typedef struct node
{
  int data;
  struct node *next;
}node;

// declarations of functions use for this program respectively
void free_node(struct node *head);
void insert_at_beg(int num, struct node *head);
void insert_at_end(int num, struct node *head);
void print_node(struct node *head);

int main(void)
{
  struct node *head = NULL;
  int n;
  // taking input
  printf("Input number of nodes: ");
  scanf("%d",&n);

  int num, choice;

  printf("\nInput data for nodes->\n");
  // loop which takes value and choice 
  for (int i = 0; i < n; i++)
  {
    num = 0, choice = 0;
    printf("\nInput data for the %d node: ", i+1);
    scanf("%d",&num);

    do
    {
      printf("Input place for the %d node: ", i+1);
      scanf("%d",&choice);
    }
    while (choice != 1 && choice != 0);
    
    if (choice == 0)
    {
      // function to insert node at front of head
      insert_at_beg(choice, head);

    }
    else 
    {
      // function to insert node at last place
      insert_at_end(choice, head);

    }
  }
  
  // function to print nodes
  print_node(head);
  
  // function to free memory made by malloc()
  free_node(head);
}

// function to free the nodes 
void free_node(struct node *head)
{
  struct node *temp = head;
  while(temp != NULL)
  {
    free(temp);
    temp = temp->next;
  }
}

// function for inserting number at front
void insert_at_beg(int num, struct node *head)
{ 
  struct node *new_node = malloc(sizeof(node));

  if (new_node == NULL)
  {
    printf("Can't allocate memory.");
    exit (1);
  }

  new_node->data = num;  
  new_node->next = head;
  head = new_node;
}

// function for inserting node at end
void insert_at_end(int num, struct node *head)
{
  struct node *new_node, *last_node = NULL;
  new_node = malloc(sizeof(node));

  if (new_node == NULL)
  {
    printf("Can't allocate memory.");
    exit (1);
  }
  if (head == NULL)
  {
    new_node->data = num;
    new_node->next = NULL;
    head = new_node; 
  }
  
  last_node = head;

  new_node->data = num;
  new_node->next = NULL;

  while (last_node->next != NULL)
  {
    last_node = last_node->next;
  } 

  last_node->next = new_node;
}

//function for printing nodes
void print_node(struct node *head)
{
  printf("\nOutput: \n");
  struct node *temp = head;

  while(temp != NULL)
  {
    printf("%d ",temp->data);
    temp = temp->next;
  }
}
Vlad from Moscow
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Lovish Garg
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  • Sorry, I use a online IDE that's why I ca't run debugger in that. – Lovish Garg Oct 02 '20 at 16:30
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    We need a canonical dupe for question where people just didn't read enough to learn that C passes all arguments by copy (but pointers can be used to simulate passing by reference). Perhaps [What's the difference between passing by reference vs. passing by value?](https://stackoverflow.com/questions/373419/whats-the-difference-between-passing-by-reference-vs-passing-by-value) although that is probably too vague for C. – underscore_d Oct 02 '20 at 16:37

2 Answers2

1

You need to pass the pointer to the head node by reference to your functions insert_at_beg and insert_at_end and desirable to the function free_node.

Passing by reference in C means passing an object indirectly through pointer to it.

So for example the function insert_at_beg can look the following way. Pay attention to that such a function should not issue any message. It is the caller of the function that decides whether to output a message or not.

// function for inserting number at front
int insert_at_beg( struct node **head, int num )
{ 
    struct node *new_node = malloc( sizeof( struct node ) );
    int success = new_node != NULL;

    if ( success )
    {
        new_node->data = num;  
        new_node->next = *head;
        *head = new_node;
    }

    return success;
}

Correspondingly the function insert_at_end can look the following way

// function for inserting node at end
int insert_at_end( struct node **head, int num )
{
    struct node *new_node = malloc( sizeof( struct node ) );
    int success = new_node != NULL;

    if ( success )
    {
        new_node->data = num;
        new_node->next = NULL;
 
        while ( *head != NULL )
        {
            head = &( *head )->next;
        }

        *head = new_node;
    }

    return success;
}

The function free_node has undefined behavior because you are using the pointer temp to access the memory that was already freed.

free(temp);
temp = temp->next;

The function can be defined the following way

// function to free the nodes 
void free_node( struct node **head )
{
    while( *head != NULL )
    {
        struct node *temp = *head;
        head = &( *head )->next;
        free( temp );
    }
}

The functions can be called like

insert_at_end( &head, num );

or

if ( !insert_at_end( &head, num ) )
{
    printf( "There is no enough memory to insert the value %d\n", num );
}

The parameter of the function print_node should have the qualifier const because the list is not changed within the function

//function for printing nodes
void print_node( const struct node *head )
{
  printf("\nOutput: \n");
  const struct node *temp = head;
  //...
Vlad from Moscow
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0

C is a pass-by-value language -- values passed to functions are copied and changes to the parameter in the function do not affect the caller. So head never becomes non-null in main; the assignment to head in insert_at_end is local and does not update the head pointer in main.

Chris Dodd
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