CODEWITHSUNDEEP
clinked-listsegmentation-faultconcatenation
user3330405
user3330405

Reputation: 3

Segmentation fault in a C function to concatenate two linked lists

I am trying to write a function that concatenates two lists to form a new list containing copies of all the nodes in list1 and list2, and returns a pointer to this new list.

Node *concat(Node *list1, Node *list2) {
  Node *head, *ptr;
  if (list1==NULL) {
    if (list2!=NULL)
      return copy(list2); 
/*copy refers to a function that copies a list & returns pointer to new list*/
    else
      return NULL;
  }
  else if (list2==NULL) {
    if (list1!=NULL)
      return copy(list1);
    else
      return NULL;
  }
  else {
  while (list1 != NULL) {
    if (head==NULL) {
      head=newNode(list1->airport);
      ptr=head;
    }
    else {
      Node *n=newNode(list1->airport);
      ptr->next=n;
      ptr=ptr->next;
    }
    list1=list1->next; }
  while (list2 != NULL) {
    Node *n1=newNode(list2->airport);
    ptr->next=n1;
    ptr=ptr->next;
    list2=list2->next;
  }
  }
  return head;
}

When I test this function using a program that prints

I get a segmentation fault.

I can't find the source of the segmentation of fault in the function, although I believe it is in the first portion, before while (list1 != NULL) , because it seems to work if neither of the lists are NULL.

Upvotes: 0

Views: 683

Answers (2)

user3407746
user3407746

Reputation: 26

Can't add much to the comments and answer you've received already, but have put comments in your code to aid your understanding of why it crashes.

Node *concat(Node *list1, Node *list2) {

/* These are uninitialised variables, which basically means they will 
 * take the whatever random values that are on the stack. It is unlikely
 * that the values will be NULL.
 */
Node *head, *ptr;  

if (list1==NULL) {
  if (list2!=NULL)
    return copy(list2); 
    /*copy refers to a function that copies a list & returns pointer to new list*/
  else
    return NULL;
}
else if (list2==NULL) {
  if (list1!=NULL)
    return copy(list1);
  else
    return NULL;
}
else {
while (list1 != NULL) {

/* When you get here, head isn't NULL so you go to the else block */

if (head==NULL) {
  head=newNode(list1->airport);
  ptr=head;
}
else {
  Node *n=newNode(list1->airport);

  /* Bang! Highly likely to crash here, as you are dereferencing a 
   * random value (in ptr) as though it is a valid pointer. 
   */
  ptr->next=n;
  ptr=ptr->next;
}
list1=list1->next; 
}
while (list2 != NULL) {
  Node *n1=newNode(list2->airport);
  ptr->next=n1;
  ptr=ptr->next;
  list2=list2->next;
}
}
return head;
}

Upvotes: 0

M Thotiger
M Thotiger

Reputation: 344

Since you already have a copy function which creates new list and returns head points so program can be modified as below (if NULL input is passed to a copy function it should return NULL, this assumption is made)

Node *concat(Node *list1, Node *list2) {

   Node *head = NULL, *ptr = NULL;

   if(list1 != NULL)
   {

      head = copy(list1);
   }

   if(list2 != NULL)
   {
      if(NULL == head)
      {
        return copy(list2);
      }
   }

   ptr = head;

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

   ptr->next = copy(list2);
   return head;

}

Upvotes: 0

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