I need help to delete new node

My list contain with three nodes I want to delete new node. I do not understand how to delete new node

I want this output 123 13

Code:

---

#include<stdio.h>

#include<stdlib.h>


typedef struct node
{
    int n;
    struct node *next;
}node;

int main()
{
 node *current = NULL;
 node *new = NULL;
 node *last = NULL;
 
 node *head = NULL;
 
 current = malloc(sizeof(current));
 new = malloc(sizeof(new));
 last = malloc(sizeof(last));
 
  if (current == NULL)
      printf ("Memory not Allocated \n");
  else
      printf ("Memory Allocated \n");
      current->n = 1;
      current->next = new;
      new->n = 2;
      new->next= last;
      last->n = 3;
      last->next= NULL;
     
      free(new);
     
      for (head = current; head != NULL; head = head->next)
          {
              printf("%d\n", head->n);
          }

    return 0;
}

---

output
14090432
14104256
14093800
14095320

garbage value

Your problem is that you aren't setting head to point to the head of the linked list.

I suggest that you restrict yourself to having three node pointers: head, tail, and current. head points to the start node of the linked list, or is a null pointer if the linked list is empty. tail points to the end node of the linked list, or is a null pointer if the linked list is empty. current is used to iterate over the linked list, i.e., instead of changing head, you set current to head then change current to go to the next node. To build the linked list, you update head and tail in order to append the three nodes. (I'm assuming that you want to append, which is why I suggest having tail; if you only want to prepend, just having head will do.)

That said, if you're going to delete the newest node before printing the linked list, why create it in the first place?

Quote:

---

Originally Posted by laserlight

That said, if you're going to delete the newest node before printing the linked list, why create it in the first place?

---

We can add or delete anything in the list at any time. I made a list. Now I wanted to add something to it so I added a new node and after that I wanted to delete something so I was trying to delete the second node.

1. Create list
2. Add node
3. Delete node

Note : I do not want to create new functions to add or delete node , I want to do all three operations in main without creating other functions.

I know it's good to create a function, but for now I don't want to do that, that would be my next task

Quote:

---

Originally Posted by laserlight

Your problem is that you aren't setting head to point to the head of the linked list.

I suggest that you restrict yourself to having three node pointers: head, tail, and current.

---

I have changed pointer name as you suggested

Code:

---

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


typedef struct node
{
    int n;
    struct node *next;
}node;
int main()
{
 
 node *Current = NULL;
  node *Head = NULL;
 node *New = NULL;
 node *Tail = NULL;
 
 
 Head = malloc(sizeof(Head));
 New = malloc(sizeof(New));
 Tail = malloc(sizeof(Tail));
 
  if (Head == NULL)
      printf ("Memory not Allocated \n");
  else
      printf ("Memory Allocated \n");
      Head->n = 1;
      Head->next = New;
      New->n = 2;
      New->next= Tail;
      Tail->n = 3;
      Tail->next= NULL;
     
      for (Current = Head; Current!= NULL; Current = Current->next)
          {
              printf("%d\n", Current->n);
          }


    return 0;
}

---

Memory Allocated
1
2
3

Are you all doing the same course? If there are only 3 nodes you'd be better off using an array. Or, if you want to simulate a linked list just do:

Code:

---

#include <stdio.h>

typedef struct node {
    int n;
    struct node *next;
} node;

int main(void)
{
    node list[] = {
        { 1, &list[1] },    /* head */
        { 2, &list[2] },    /* silly */
        { 3, NULL }        /* tail */
    };
       
    for (node *curr = list; curr; curr = curr->next) {
        printf("%d\n", curr->n);
    }
   
    return 0;
}

---

Not much point in doing that though except to, perhaps, see how linked lists work

I guess what I'm saying is that unless you're using functions to construct the list there is essentially no point in having a linked list. By the way, in your version you're not freeing the memory you allocated using malloc(), so that's a problem

Or, if you want to be really cunning

Code:

---

#include <stdio.h>

typedef struct node {
    int n;
    struct node *next;
} node;

int main(void)
{
    node list[] = { { 1, &list[2] }, { 3, NULL }, { 2, &list[1] } };
       
    for (node *curr = list; curr; curr = curr->next)
        printf("%d\n", curr->n);
   
    return 0;
}

---

Quote:

---

Originally Posted by Hodor

Not much point in doing that though except to, perhaps, see how linked lists work

Edit: I guess what I'm saying is that unless you're using functions to construct the list there is essentially no point in having a linked list. By the way, in your version you're not freeing the memory you allocated using malloc(), so that's a problem

---

I know this is better way

Code:

---

#include <stdio.h>
typedef struct node {
    int n;
    struct node *next;
} node;


/* function to create linked list */
/* function to adding node in linked list*/
/* function to delete  node in linked list*/
/* function to print nodes in list
int main(void)
{
  /* create linked list with 4 nodes */
  /* print list
  /* add one node to list
  /* delete third node in list


    return 0;
}

---

output looks like something

list
1
2
3
4


updated list
1
2
3
4
5
delete third node in list
update list
1
2
4
5

I am trying following way

Code:

---

#include <stdio.h>
 
typedef struct node {
    int n;
    struct node *next;
} node;




int main(void)
{
  /* create linked list with 4 nodes
  /* print nodes in list
  /* add one node in list that is 5
  /* delete  node 3 in linked list
  /* print nodes in list


    return 0;
}

---

list
1
2
3
4


updated list
1
2
3
4
5
delete third node in list
update list
1
2
4
5

This is wrong:

Code:

---

Head = malloc(sizeof(Head));

---

It should have been:

Code:

---

Head = malloc(sizeof(*Head));

---

Quote:

---

Originally Posted by laserlight

This is wrong:

---

okay but compiler is not showing warning /error

I want to write code for algorithm

Code:

---

#include <stdio.h>
typedef struct node {
    int n;
    struct node *next;
} node;


/* function to create linked list */
/* function to adding node in linked list*/
/* function to delete  node in linked list*/
/* function to print nodes in list
int main(void)
{
  /* create linked list with 4 nodes */
  /* print list
  /* add one node to list
  /* delete third node in list


    return 0;
}

---

list
1
2
3
4


updated list
1
2
3
4
5
delete third node in list
update list
1
2
4
5

I do not understand how this can be done ?

I can create list, add node to list but I am having problem to delete the node I do not understand how this can be done

Have you gone through the process of deleting a node on paper; i.e. drawing a picture of a linked list and writing down the steps you take to solve the problem manually?

Quote:

---

Originally Posted by Player777

okay but compiler is not showing warning /error

---

It's a logic error. In theory, static analysis can uncover that particular kind of logic error because that particular argument to malloc happens to be determined at compile time, but compilers don't usually do enough static analysis to be able to detect that, so what could happen in practice is that you allocate enough space for a pointer, and then you get a segmentation fault or something when you try to write an entire object to the space allocated and end up writing beyond the memory allocated.

Quote:

---

Originally Posted by Hodor

Have you gone through the process of deleting a node on paper; i.e. drawing a picture of a linked list and writing down the steps you take to solve the problem manually?

---

Exactly that was my plan That's why I'm asking all operation in the main

I know that the pointer has to be free to delete the node but I am not able to delete the node even on paper

Quote:

---

Originally Posted by Player777

Exactly that was my plan That's why I'm asking all operation in the main

I know that the pointer has to be free to delete the node but I am not able to delete the node even on paper

---

You have to 1) find the node you want to delete (call it target) and at the same time keep track of the node previous to the target (call it prev); 2) change the previous node's next pointer to point to the the target node's next (e.g. prev->next = target->next); 3) at this point the target node is out of the list (nothing is pointing to target anymore) and you can free its memory. You'll have to handle the special case of target being the first (or only) node in the list

Personally I think that it's actually easier using functions than doing it all in main

What book are you using? Surely it has pictures of this