Segmentation fault (core dumped) in Linked List using C

Question

I want an array of linked List and obviously each linked should have separate head node. Initially, as an example, I am starting with one array element. I am storing linkedlist into current[0] . But it is giving segmentation fault. If I use Node *current it will create a list and working fine. But, I want to store the list within array. What is wrong with the code ?

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

typedef struct Node {
    int data;
    struct Node *next;
} Node;

Node *current[20];

void insert_beg_of_list(Node *current[0], int data);

void print_list(Node *current[0]);

void insert_beg_of_list(Node *current[0], int data) {

    //keep track of first node
    Node *head = current[0];

    while(current[0]->next != head) {
        current[0] = current[0]->next;
    }
    current[0]->next = (Node*)malloc(sizeof(Node));
    current[0] = current[0]->next;
    current[0]->data = data;
    current[0]->next = head;
}

void print_list(Node *current[0]) {

    Node *head = current[0];
    current[0] = current[0]->next;
    while(current[0] != head){
        printf(" %d ", current[0]->data);
        current[0] = current[0]->next;
    }

}

int main() {

    Node *head = (Node *)malloc(sizeof(Node));
    head->next = head;  

    int data = 0 ;
    int usr_input = 0;
    int i;
    int m;
    int j;

        scanf("%d", &usr_input);

        for (i=0; i<usr_input; i++) {

            scanf("%d", &data);
            insert_beg_of_list(head, data);

        }

            printf("The list is ");
            print_list(head);
            printf("\n\n");

    return 0;
}

@BLUEPIXY Correct, but I bet OP is confusing global current array with the passed parameter... — LPs
– LPs, Commented May 4, 2017 at 12:23
with Node *current , it will not be stored within array element. In future, I want to use current [1], current [2] and so on. — cyberoy
– cyberoy, Commented May 4, 2017 at 12:35
currrent[0] = head;.... insert_beg_of_list(currrent[0], data); — BLUEPIXY
– BLUEPIXY, Commented May 4, 2017 at 12:40

Community · Accepted Answer · 2017-05-23 12:26:04Z

I think you have mixed the global array current's use. Change your code to this :

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

typedef struct Node {
    int data;
    struct Node *next;
} Node;

void insert_beg_of_list(Node *current, int data);

void print_list(Node *current);

void insert_beg_of_list(Node *current, int data) {

    //keep track of first node
    Node *head = current;

    while(current->next != head) {
        current = current[0]->next;
    }
    current->next = malloc(sizeof(Node));
    if (current->next == NULL)
        return;
    current = current->next;
    current->data = data;
    current->next = head;
}

void print_list(Node *current) {

    Node *head = current;
    current = current->next;
    while(current != head){
        printf(" %d ", current->data);
        current = current->next;
    }

}

int main() {

    Node *current[20];
    Node *head = malloc(sizeof(Node));
    if (head == NULL)
        return;

    head->next = head;  

    int data = 0 ;
    int usr_input = 0;
    int i;
    int m;
    int j;

    scanf("%d", &usr_input);

    for (i = 0; i < usr_input; i++) {
        scanf("%d", &data);
        insert_beg_of_list(head, data);
    }

    //assign the newly created pointer to a place in the array
    current[0] = head;

    printf("The list is ");
    print_list(head);
    printf("\n\n");

    return 0;
}

Keep in mind that the parameter current in your functions' prototypes and declarations is not the same as the array current created in your main function. I just left the name as it was.

NOTE : You should do something with head->next pointer, initialize it.

Also read this link on why not to cast the result of malloc and another one on why you should check its result.

Jabberwocky · Accepted Answer · 2017-05-04 13:00:26Z

0

You probably want this:

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

typedef struct Node {
  int data;
  struct Node *next;
} Node;

void insert_beg_of_list(Node *current, int data);
void print_list(Node *current);

void insert_beg_of_list(Node *current, int data) {

  //keep track of first node
  Node *head = current;

  while (current->next != head) {
    current = current->next;
  }
  current->next = (Node*)malloc(sizeof(Node));
  current = current->next;
  current->data = data;
  current->next = head;
}

void print_list(Node *current) {

  Node *head = current;
  current = current->next;
  while (current != head) {
    printf(" %d ", current->data);
    current = current->next;
  }
}

Node *NewList()
{
  Node *newnode = (Node *)malloc(sizeof(Node));
  newnode->next = newnode;
}

int main() {
  Node *arrayofheads[20];

  // We are using only arrayofheads[0] in this example

  arrayofheads[0] = NewList();

  int data = 0;
  int usr_input = 0;
  int i;

  scanf("%d", &usr_input);

  for (i = 0; i<usr_input; i++) {
    scanf("%d", &data);
    insert_beg_of_list(arrayofheads[0], data);
  }

  printf("The list is ");
  print_list(arrayofheads[0]);  printf("\n\n");

  return 0;
}

edited May 4, 2017 at 13:00

answered May 4, 2017 at 12:42

Jabberwocky

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3 Comments

cyberoy Over a year ago

I appreciate your answer. My interest is not just to create a list, but I want to create an array of pointers and store the list within array element. In future, I will use current [1], current [2] and so on

Jabberwocky Over a year ago

There is still room for improvement, especially you should totally encapsulate the list with functions, so you don't need to mess with head and next in the calling functions (main here). I just updated the answer accordingly.

cyberoy Over a year ago

Got it. Thanks a lot

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